Mr. Dwyer kindly had his work on the previous QotD put on the board. Please, students, you should be able to find eigenvalues and eigenvectors of very simply matrices!

Example 1
Last time we saw that the matrix A=

(5 2)
(3 4)

Philosophy or hope or ...
I have a basis of eigenvectors. Maybe I can rethink multiplication by A into three stages. Here I will start with a vector described in terms of the standard basis (1,0) and (0,1).

1. Change information to be in terms of the basis of eigenvectors.
2. Multiply by A.
3. Change back to the standard basis information.

I want to change (1,0) as a column vector to (1,1) and change (0,1) as a column vector to (2,-3). So I need to find a matrix M which does the following:

(a b)(1) is (1) and (a b)(0) is ( 2).
(c d)(0)    (0)     (c d)(1)    (-3)

(1  2)
(1 -3)

How can I undo switching information from the standard basis to the basis of eigenvectors? "Clearly" I should find M^-1:

(1  2 | 1 0)~(1  2 |  1 0)~(1 0 | 3/5  2/5 )
(1 -3 | 0 1) (0 -5 | -1 1) (0 1 | 1/5 -1/5 )

A multiplication
I computed M^-1AM. I noted that this could be computed in two ways, as (M^-1A)M or as M^-1(AM). Matrix multiplication is associative so you can regroup as you wish. It is not always commutative (recalled the first linear algebra QotD!) so you can't change the order of the matrices you are multiplying.

      M-1 A                                                  M-1AM 
(3/5  2/5)(5 2) is (21/5 14/5) and then (21/5 14/5)(1  2) is (7 0).
(1/5 -1/5)(3 4)    ( 2/5 -2/5)          ( 2/5 -2/5)(1 -3)    (0 2)

Why do this?
This is a very appropriate question (voiced by Mr. Novak, I believe). Well, computations with diagonal matrices are very easy. For example,

(7 0)(7 0) is just (49 0).
(0 2)(0 2)         ( 0 4)

(750 0 )
( 0 250)

Why do this?
Why? We can now efficiently compute, say, 17A⁹⁸-56A¹⁴+.002A⁷ because
17A⁹⁸-56A¹⁴+.002A⁷=M(17D⁹⁸-56D¹⁴+.002D⁷)M^-1 and 17D⁹⁸-56D¹⁴+.002D⁷ is the same as

(17·798-56·714+.002·77            0         )
(        0             17·298-56·214+.002·27)

Why do this?
Everybody (well, all engineers, at least!) wants to solve differential equations. So let us look at a linear system of ordinary differential equations (the more ways you know to solve the darn things, the better):

{dx/dt}=5x(t)+2y(t)
{dy/dt}=3x(t)+4y(t)

(SUMn=0infinity(7t)n/(n!)      0)=(e7t 0)
(0       SUMn=0infinity(2t)n/(n!))=(0  e2t)

(1  2)(e7t 0 )(3/5  2/5)
(1 -3)( 0 e2t)(1/5 -1/5)

([[(3/5*exp(7*t)+2/5*exp(2*t))*x(0)+(2/5*exp(7*t)-2/5*exp(2*t))*y(0)], 
[(3/5*exp(7*t)-3/5*exp(2*t))*x(0)+(2/5*exp(7*t)+3/5*exp(2*t))*y(0)]])

Let's go on to the dreadful second example. Here I wanted to analyze the matrix C given by

( 0 -1 1)
(-1  1 2)
( 1  2 1)

Analysis of C
This didn't go badly but I was running out of time and getting nervous. It probably showed. Sigh. Now the characteristic polynomial of C is

   (-  -1   1)
det(-1 1-   2)
   ( 1   2 1-)

>charpoly(C,x);
                          3      2
                         x  - 2 x  - 5 x + 6

Doing it correctly ...
We noticed that the eigenvectors were all perpendicular (in mathspeak: orthogonal). That was nice. The matrix for the first part of the philosophy (eigenvectors into columns) was P=

( 2 -1 0)
(-1 -1 1)
( 1  1 1)

( 2 -1 1)
(-1 -1 1)
( 0  1 1)

(6 0 0)
(0 3 0)
(0 0 2)

( 2/sqrt(6) -1/sqrt(3)    0     )
(-1/sqrt(6)  1/sqrt(3) 1/sqrt(2))
(-1/sqrt(6)  1/sqrt(3) 1/sqrt(2))

The major result
We can diagonalize the symmetric matrix C using a matrix of normalized eigenvalues. The eigenvalues are orthogonal, and the resulting matrix used for changing bases has the wonderful property that its inverse is its transpose. A matrix P such that P^-1=P^t is called orthogonal.

This is always true for symmetric matrices:

Every symmetric matrix has a basis of orthogonal eigenvalues, which, when normalized, form an orthogonal matrix. The matrix of eigenvalues and its inverse/transpose diagonalize the symmetric matrix.

This result is not easy to prove, and is usually the triumph (?) of our Math 250 course.

HOMEWORK
Please read sections 8.8, 8.10, and 8.12 of the text. The following problems almost surely can be done easily by Maple or Matlab or any grown-up calculator. I strongly urge students to do these problems by hand and get the correct answers.
8.8: 13, 19. 8.10: 15 8.12: 5, 13, 39.

1. If A and B are n-by-n matrices, then det(AB)=det(A)det(B).
   Comments This is best understood if one realizes that det also measures a volume distortion factor, so A in effect "maps" Rⁿ to Rⁿ by matrix multiplication, and det(A), it turns out, is the way n-dimensional volumes are stretched. So multiplying by A and then by B concatenates the effects. Notice, as we observed in class, that det(A+B) is not ... Mr. Shah's example: A=I₂ and BA=2I₂ so A+B=3I₂ and det(A)=1 and det(B)=4 and det(A+B)=9.
2. If A is an n-by-n matrix, with A^t its transpose, then det(A)=det(A^t).
   Comments Here the transpose is the result of flipping the matrix over its main diagonal: rows and columns get interchanged. Algebraically, the ij^th entry in A^t is the ji^th entry in A. The reason these determinants are the same is that when the flip is done, the signs on each rook arrangement is preserved. (-1)^# counts stuff in each upper-right quadrant. But each upper-right quadrant is counted exactly when the matrix element is in the lower-left quadrant of the other element. And transposing lets upper right change to lower left. Whew -- this is actually a correct explanation, but perhaps not totally clear!

>A:=matrix(2,2,[5,2,3,4]); [5 2] A := [ ] [3 4] >plot([5*cos(t)+2*sin(t),3*cos(t)+4*sin(t),t=0..2*Pi], thickness=3,color=black);
>B:=evalm(A^2); [31 18] B := [ ] [27 22] >plot([31*cos(t)+18*sin(t),27*cos(t)+22*sin(t),t=0..2*Pi], thickness=3,color=black);

>eigenvects(A);
                [2, 1, {[1, -3/2]}], [7, 1, {[1, 1]}]

(a b)
(c d)

Ms. Horn asked why I looked at the unit circle. Well, if you give me all information about what happens on the unit circle, then I will know what happens everywhere. For example, I could ask what happens to (15,-48). Call sqrt({15}²+{48}²), w. Then (15,-48)=w(15/w,-48/w). If everything is linear, and if I know what happens to the unit circle then the w factor on the outside will just multiple what happens to (15/w,-48/w), a vector on the unit circle. So it you know that, then you know "everything"

VOCABULARY TIME! If A is an n-by-n matrix, an n-dimensional vector X is called an eigenvector of A is X is not zero and if there is a scalar so that AX=X. The scalar is called an eigenvalue and X is an eigenvector associated to that eigenvalue.

Understanding eigenvalues and eigenvectors is important if you are going to any sort of extensive computation with matrices (by hand, by computer, by thinking). As I observed in class, the requirement that an eigenvector be not equal to 0 is very important. Without it, any value is an eigenvalue! Eigen{value|vector} is also called characteristic {value|vector} and proper {value|vector}, depending on what you read and what you work with.

Example
As requested by Mr. Huertas, I worked out the details, the indispensable details, of course, for the example which generated the pictures above. So A=

(5 2) 
(3 4)

5x1+2x2=x1
3x1+4x2=x2

(5-)x1+2x2=0
3x1+(4-)x2=0

(5-  2)
(3  4-)

VOCABULARY TIME! If A is an n-by-n matrix, and we let In be the n-by-n identity matrix (1's on the diagonal with 0's elsewhere) then the characteristic polynomial of A is det(A-In). This is a polynomial of degree n. The roots of the characteristic polynomial are the eigenvalues of A.

The roots of ²-9+14=0 are 7 and 2. Nice people like Ms. Julien factor this polynomial. People like me, who wear suspenders and a belt on their pajamas, use the quadratic formula. By the way, there is no general formula for the roots of a polynomial, and frequently numerical approximation schemes are needed. Knowing these values of doesn't end our task. I would like to find eigenvectors for each of these eigenvalues.

=7
Go back to the homogeneous system

(5-)x1+2x2=0
3x1+(4-)x2=0

-2x1+2x2=0
3x1-3x2=0

Facts about eigenvectors
Any non-zero multiple of an eigenvector is also an eigenvector. Any non-zero sum of eigenvectors associated to the same eigenvalue is also an eigenvector. In practice, I suggest you choose some useful way of specifying an eigenvector. Different computational packages may give different answers. Sometimes this can be annoying.

=2
Now plug in =2 in the homogeneous system. The result is

3x1+2x2=0
3x1+2x2=0

Why do this, and why do the pictures look the way they did?
Well we now know that (1,1) is an eigenvector associated to the eigenvalue 7 and (2,-3) is an eigenvector associated to the eigenvalue 2. I can create a vector, let's see: 5(1,1)+3(2,-3)=(11,-4). Weird, but o.k. Now I can ask what effect multiplication by A has on this vector (actually on the transpose of the vector, changing it from a row vector to a column vector, but few people bother being that linguistically pure). I could compute the product of the matrix A with the vector, or just notice that due to linearity, the effect is 5 multiplied by what A does to (1,1) plus 3 multiplied by what A does to (2,-3). Since the affected vectors are eignevectors, I can tell you the result easily: 5·7(1,1)+3·2(2,-3). There is no purpose to doing the arithmetic, and in fact if I did the arithmetic, things would be less comprehensible. Notice, please, that (1,1) and (2,-3) are linearly independent. Since they are two linearly independent vectors in R² they form a basis of R² and every vector in R² can be written as a linear combination of these two. If I am going to do extensive computations with A, it is much better to write things in terms of this basis than the standard basis, (1,0) and (0,1). In fact, let's see: suppose I want to "find" the result of multiplying 5(1,1)+3(2,-3) by the matrix A⁵. This will be 5·7⁵(1,1)+3·2⁵(2,-3). This is also 7⁵(5(1,1)+3·.002(2,-3) ). The .002 is the approximate value of (2/7)⁵. So this vector points almost totally in the direction of the eigenvector. Hey! The ellipses above will get narrower and narrower very fast!

Material science, physics, stress, strain, and many other things I don't know about
I mentioned that finding eignevalues and eigenvectors was important in understanding the structure of materials. This went nowhere since I got a number of questions and had to admit ignorance very soon. I went to Google and found over 25,000 pages listed in response to the query crystal structure eigenvalues. Oh well ...

I haven't exactly lied to you ...
Things don't have to be as nice as the behavior of the example I analyzed. You won't always get a basis of eigenvectors. Here are two simple examples.

#1
If A=

(2 17)
(0  2)

(0 17)
(0  0)

#2
If A=

(0 -1)
(1  0)

My next example became the QotD since I was informed that the poor chemical engineers had an example following the class. How they suffer!
Anyway, I asked for eigenvalues and eigenvectors of the matrix A=

(2 17)
(0  3)

The matrix A	The result of the Maple command eigenvects(A)
(5 2) (3 4)	[2, 1, {[1, -3/2]}], [7, 1, {[1, 1]}]
(2 17) (0 2)	[2, 2, {[1, 0]}]
(0 -1) (1 0)	[I, 1, {[1, -I]}], [-I, 1, {[1, I]}]
(2 17) (0 3)	[3, 1, {[17, 1]}], [2, 1, {[1, 0]}]

HOMEWORK
Read sections 8.8, 8.10, and 8.12 of the text. We will finish our course's formal study of linear algebra next time.

The Oxford English Dictionary lists the first appearance of paradigm in 1483 when it meant "an example or pattern", as it does today.

But you must know for the purposes of this course some standard computational methods of evaluating determinants. So I'll tell you about row operations and cofactor exponasions.

Row operations and their effects on determinants
The row operation	What it does to det
Multiply a row by a constant	Multiplies det by that constant
Interchange adjacent rows	Multiplies det by -1
Add a row to another row	Doesn't change det

Examples
Suppose A is this matrix:

( -3  4  0 18 ) 
(  2 -9  5  6 )
( 22 14 -3 -4 )
(  4  7 22  5 )

( -6  8  0 36 ) 
(  2 -9  5  6 )
( 22 14 -3 -4 )
(  4  7 22  5 )

( -3  4  0 18 ) 
( 22 14 -3 -4 )
(  2 -9  5  6 )
(  4  7 22  5 )

( -3  4  0 18 ) 
(  2 -9  5  6 )
( 24  5  2  2 )
(  4  7 22  5 )

Silly examples (?)
Look:

   ( 1 2 3 )    ( 2 3 4 )
det( 4 5 6 )=det( 3 3 3 ) (row2-row1)
   ( 7 8 9 )    ( 3 3 3 ) (row3-row2)

 
   (   1   4   9  16 )    (  1  4   9  16 )                (  1  4  9 16 )
det(  25  36  49  64 )=det( 24 32  40  48 ) (row2-row1)=det( 24 32 40 48 )
   (  81 100 121 144 )    ( 56 64  72  80 ) (row3-row2)    ( 32 32 32 32 ) (row3-row2)
   ( 169 196 225 256 )    ( 88 96 104 112 ) (row4-row3)    ( 32 32 32 32 ) (row3-row2)

There are all sorts of tricky things one can do with determinant evaluations, if you want. Please notice that the linear systems gotten from, say, the finite element method applied to important PDE's definitely give coefficient matrices which are not random: they have lots of structure. So the tricky things above aren't that ridiculous.

Use row operations to ...
One standard way of evaluating determinants is to use row operations to change a matrix to either upper or lower triangular form (or even diagonal form, if you are lucky). Then the determinant will be the product of the diagonal terms. Here I used row operations (actually I had Maple use row operations!) to change this random (well, the entries were produced sort of randomly by Maple) to an upper-triangular matrix.

[1 -1 3 -1] And now I use multiples of the first row to create 0's 
[4  4 3  4] below the (1,1) entry. The determinant won't change:
[3  2 0  1] I'm not multiplying any row in place, just adding
[3  1 3  3] multiples of row1 to other rows.
                              
[1 -1  3 -1] And now multiples of the second row to create 0's 
[0  8 -9  8] below the (2,2) entry. 
[0  5 -9  4] 
[0  4 -6  6]

[1 -1    3  -1] Of course, multiples of the third row to create 
[0  8    9   8] 0's below the (3,3) entry.
[0  0 -27/8 -1] 
[0  0  -3/2  2]

[1 -1    3   -1 ] Wow, an upper triangular matrix!
[0  8   -9    8 ]
[0  0 -27/8  -1 ]
[0  0    0  22/9]

Minors
If A is an n-by-n matrix, then the (i,j)^th minor of A is the (n-1)-by-(n-1) matrix obtained by throwing away the i^th row and j^th column of A. For example, if A is

[1 -1 3 -1]
[4  4 3  4]
[3  2 0  1]
[3  1 3  3]

>minor(A,2,3);
       [1 -1 -1]
       [3  2  1]
       [3  1  3]

Evaluating determinants by cofactor expansions
This field has a bunch of antique words. Here is another. It turns out that the determinant of a matrix can be evaluated by what are called cofactor expansions. This is rather weird. When I've gone through the proof that cofactor expansions work, I have not really felt enlightened. So I will not discuss proofs. Here is the idea. Suppose A is an n-by-n matrix. Each (i,j) position in this n-by-n matrix has an associated minor which I'll call M_ij. Then:

• For any i, det(A)=SUM_j=1ⁿ(-1)^i+ja_ijdet(M_ij). This is called expanding along the i^th row.
• For any j, det(A)=SUM_i=1ⁿ(-1)^i+ja_ijdet(M_ij). This is called expanding along the j^th column.

Here: let's try an example. Suppose A is

[1 -1 3 -1]
[4  4 3  4]
[3  2 0  1]
[3  1 3  3]

> det(minor(A,1,1));                                 
                                      -3
> det(minor(A,1,2));
                                       6
> det(minor(A,1,3));
                                      -16
> det(minor(A,1,4));
                                      -21

Recursion and strategy
You should try some examples, of course. This is about the only way I know to learn this stuff. If I had to give a short definition of determinant, and if I were allowed to use recursion, I think that I might write the following:
Input A, an n-by-n matrix.
     If n=1, then det(A)=a₁₁
     If n>1, then det(A)=SUM_j=1ⁿ(-1)^j+1a_1jdet(M_1j) where M_1j is the (n-1)-by-(n-1) matrix obtaining by deleting the first row and the j_th column.
This is computing det(A) by repeatedly expanding along the first row. I've tried to write such a program, and if you have the time and want some amusement, you should try this also. The recursive nature rather quickly fills up the stack (n! is big big big) so this isn't too practical. But there are certainly times when the special form of a matrix allows quick and efficient computation by cofactor expansions.

More formulas
You may remember that we had a
A decision problem Given an n-by-n matrix, A, how can we decide if A is invertible?
Here is how to decide:
       A is invertible exactly when det(A) is not 0.
Whether this is practical depends on the situation.
There was also a
Computational problem If we know A is invertible, what is the best way of solving AX=B? How can we create A^-1 efficiently?
Well, this has an answer, too. The answer is on page 383 of the text. The inverse of A is the constant (1/det(A)) multiplied by the adjoint of A. I have to look this up. The adjoint is the transpose (means: flip over the main diagonal, or, algebraically, interchange i and j) of the matrix whose entries are (-1)^i+jdet(M_ij).

I think this is hideous and the only example I have seen worked out in detail (enough to be convincing!) is n=2. So here goes:

A is (a b) and M11=d and M12=c and M21=b and M22=a
     (c d)
Then the adjoint (put in +/- signs, put in transpose) is ( d -b)
                                                         (-c  a).
Since the det is ad-bc, the inverse must be 
( d/(ad-cd) -b/(ad-bc) )  
(-c/(ad-cd)  a/(ad-bc) )

So there will be times you might need to decide between using an algorithmic approach and trying to get a formula. Let me show you a very simple example where you might want a formula. This example itself is perhaps not too realistic, but maybe you can see what real examples might look like.

Suppose we need to understand the linear system
2x+7y=6
Qx+3y=Q
Well, if the parameter Q is 0, then (second equation) y=0 and so (first equation) x=3. We could think of Q as some sort of control or something. I tried inadequately to convey a sort of physical problem that this might model, but the effort was perhaps not totally successful. What happens to x and y when we vary Q, for example, move Q up from 0 to a small positive number? I don't think this is clear. But we can in fact find a formula for x and y. This is sort of neat, actually. You may remember such a formula from high school, even.

   det( 6 7 )     det( 2 6 )              
      ( Q 3 )        ( Q Q ) 
x= ----------  y= ----------    
   det( 2 7 )     det( 2 7 )
      ( Q 3 )        ( Q 3 )

Cramer's Rule
I think this is Theorem 8.23 (page 392) of the text. It discusses a formula for solving AX=B where A is an n-by-n matrix with det(A) not equal to 0, and B is a known n-by-1 matrix, and X is an n-by-1 matrix of unknowns. Then x_j turns out to be det(A_j)/det(A), where A_j is the matrix obtained by replacing the j^th column of A by B.

Well, the QotD was computing part of an example when n=3 of this.
3x-5y+z=2
2x+5y-z=0
x-y+z=7
What is z? According to Cramer's Rule, z is

   (3 -5 2)
det{2  5 0)
   (1 -1 7)
------------
   (3 -5  1)
det{2  5 -1)
   (1 -1  1)

>det(matrix(3,3,[3,-5,2,2,5,0,1,-1,7]));
                                      161

> det(matrix(3,3,[3,-5,1,2,5,-1,1,-1,1]));
                                      20

> A:=matrix(3,4,[3,-5,1,2,2,5,-1,0,1,-1,1,7]);
                                [3    -5     1    2]
                                [                  ]
                           A := [2     5    -1    0]
                                [                  ]
                                [1    -1     1    7]
> rref(A);
                             [1    0    0    2/5]
                             [                  ]
                             [               29 ]
                             [0    1    0    -- ]
                             [               20 ]
                             [                  ]
                             [               161]
                             [0    0    1    ---]
                             [               20 ]

This class meeting itself seemed rather disorganized to me: lots of formulas. I apologize for not showing the formulas more effectively. I hope you will learn the more important ones, and remember that there are alternative ways of solving linear systems.

HOMEWORK
Please still read 8.4 and 8.5 and 8.7. Do these problems:
8.4: 25
8.5: 11-14, 21, 31
8.7: 1
Also, here are two problems for you to do. You will get a prize!

AX=B, where A is an n-by-n matrix
Rank considerations	rank(A)=n	rank(A)<n
Name(s)	Full rank; regular; non-singular; invertible	Singular
Homogeneous equation	AX=0 has only the trivial solution.	AX=0 has infinitely many solutions in addition to the trivial solution.
Inhomogeneous equations	AX=Y has a unique solution for all Y.       If you know A-1, then X=A-1Y	There are Y's for which AX=Y has no solution; for all other Y's, AX=Y has infinitely many solutions

There are really two problems here.

1. A decision problem Given an n-by=n matrix, A, how can we decide if A is invertible?
2. Computational problem If we know A is invertible, what is the best way of solving AX=B? How can we create A^-1 efficiently?

Algorithmically ...
Suppose A is an n-by-n matrix. We will augment the heck (??!) out of A with an n-by-n identity matrix, I_n. Here I_n is 1 on the diagonal, and has 0 entries off the diagonal. Then we have (A|I_n). Now run RREF on this. What I would like to see as the end result is (I_n|Some "stuff"). If this occurs, then A is invertible, and Some "stuff" is A^-1, the inverse of A. If we can't get I_n on the left, then A is not invertible.

How much work is computing A^-1 this way? Here I used the word "work" in a rather elementary sense. I tried to convince people that a really rough bound on the number of arithmetic operations (add, multiply, divide, etc.) to find A^-1 using this method is, maybe, 6n³. This actually isn't too bad. Finding A^-1 this way is a problem which can be computed in polynomial time. Such problems are actually supposed to be rather nice computationally. In fact, the degree in n can actually be lowered substantially, I believe. People worry about this in the real world.

Well, this works. But there are alternate ways of thinking about the problem. Particularly, we can solve the decision probloem by computing the determinant of A, det(A). This is a number. If it is 0, then A is singular. If it is not 0, then A is invertible. If det(A) is not 0, it is then possible to write a formula for A^-1, and we can also write a formula (called Cramer's rule) for finding the unique solution to AX=B. The problem is that the formulas needed are quite complicated.

The official definition of determinant

First imagine that we have an n-by=n chessboard. Recall that a rook (the thing that looks like a castle) on a chessboard can move freely on rows and columns. (Recently published and not yet read by me: Birth of the Chess Queen which "sees the rise of female power throughout the centuries reflected in the history of the chess queen".) Anyway, I asked students for the largest number of rooks which could be put on a chessboard so that they don't attackone another. We thought for a while, and then decided that we could put one rook in each column and each row: thus a rook arrangement could have n rooks. In the diagrams below, I will use * for a rook placed on the board and 0 to indicate an empty position on the board.

How many different rook arrangements are there? Well, there are n places to put a rook in the first column. Once a rook is placed there, a whole row is eliminated for further rook placement. So there are n-1 places (non-attacking places!) to put a rook in the seoond column. And there n-2 in the third column, etc. Some thought should convince you that there are n! (n factorial) different rook arrangements. n! grows very fast with n. Everyone should know, even just vaguely, the Stirling approximation to n!. This says that n! is approximately (n/e)ⁿsqrt(2Pi n). In fact, Maple tells me that 20! is exactly
2432 90200 81766 40000 while the Stirlling approximation is
2422 78684 67611 33393.1 (about)
But I am not particularly interested in specific values of factorials. The important fact is that the factorial function is superexponential, and grows much faster, for example, than any polynomial.

Each rook arrangement has a sign attached to it, either a + or -. This sign is gotten by computing (-1)^# where what matters is the parity (even or oddness) of the number. What's the number? For each rook, look in the rook's "first quadrant", up and to the right. Count the rooks there. Then take the total or all of the "first quadrant" rooks for each of the rooks. That's the number, #. Here are two examples.

(* 0 0 0 0) First quadrant has 0 rooks.      (0 0 0 0 *) First quadrant has 0 rooks.
(0 0 * 0 0) First quadrant has 0 rooks.      (0 0 0 * 0) First quadrant has 1 rook.
(0 * 0 0 0) First quadrant has 1 rook.       (* 0 0 0 0) First quadrant has 2 rooks.
(0 0 0 0 *) First quadrant has 0 rooks.      (0 * 0 0 0) First quadrant has 0 rooks.
(0 0 0 * 0) First quadrant has 1 rook.       (0 0 * 0 0) First quadrant has 1 rook.
1+1=2=#.                                     1+2+1=4=#.

Now for each rook arrangement, take the product of the terms in the matrix A which are in the rook places: the product of n entries of A. Then prefix this by the sign, (-1)^# as mentioned above. And, then, finally, take the sum of this strange signed product over all n! rook arrangements. This sum is det(A).

I then worked out what the definition would be for 2-by-2 and 3-by-3 matrices. This turns out to be the same as the traditional formulas which I think students all know. How about 4-by-4? IMPORTANT I don't know any simple way to remember the definition in this case. I know no reason to try to remember the formula. The n=2 and n=3 cases are too darn simple! The formulas and scemes remembered from "childhood" will not work for n>3.

Determinants when n=2
The matrix A is

(a b)
(c d)

(* 0)
(0 *)

(0 *)
(* 0)

Determinants when n=3
The matrix A is

(a b c)
(d e f)
(g h i)

(* 0 0)
(0 * 0)
(0 0 *)

(* 0 0)
(0 0 *)
(0 * 0)

(0 * 0)
(* 0 0)
(0 0 *)

(0 0 *)
(* 0 0)
(0 * 0)

(0 * 0)
(0 0 *)
(* 0 0)

(0 0 *)
(0 * 0)
(* 0 0)

Determinants when n=4 (enough already!)
Here is what Maple says:

> A:=matrix(4,4,[a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p]);
                                 [a    b    c    d]
                                 [                ]
                                 [e    f    g    h]
                            A := [                ]
                                 [i    j    k    l]
                                 [                ]
                                 [m    n    o    p]

> det(A);
a f k p - a f l o - a j g p + a j h o + a n g l - a n h k - e b k p +
e b l o + e j c p - e j d o - e n c l + e n d k + i b g p - i b h o - 
i f c p + i f d o + i n c h - i n d g - m b g l + m b h k + m f c l - 
m f d k - m j c h + m j d g

There are lots and lots and lots of formulas for determinants. They all depend on various det properties. Here are some.

If you multiply the elements in one row of a matrix by a constant, then the determinant of that matrix is multiplied by that constant. That's because the entries in the sum for the det (see the n=4 examploe above) each have one peice from the one row, so we can factor out the constant.

Here is a more subtle property. If you interchange adjoining rows of a matrix, the determinant is multiplied by -1. This happens because the total # changes by 1 (either adds one or subtracts one) in every case, and then this factors out to make the sign change for the det.

Example 403.67
What is the determinant of the following matrix?

( 3   5 17   2   0 )
(-2   4  8  -1   4 )
( 7   2 -4  -4  31 )
( 6   2  8   8  12 )
( 6 -12 -24  3 -12 )

( 3   5 17   2   0 )
(-2   4  8  -1   4 )
( 7   2 -4  -4  31 )
( 6   2  8   8  12 )
(-2   4  8  -1   4 )

( 3   5 17   2   0 )
(-2   4  8  -1   4 )
( 7   2 -4  -4  31 )
(-2   4  8  -1   4 )
( 6   2  8   8  12 )

( 3   5 17   2   0 )
(-2   4  8  -1   4 )
(-2   4  8  -1   4 )
( 7   2 -4  -4  31 )
( 6   2  8   8  12 )

Example 309.67
What is the determinant of the following matrix?

( 3 -2  7   8 99  6 )
( 0  3 3/4 -1 Pi .7 )
( 0  0  3  -9 77 .4 )
( 0  0  0   3 -3 13 )
( 0  0  0   0  3 -9 )
( 0  0  0   0  0  3 )

QotD
What is the determinant of

( 0 0 0 0 E )
( 0 0 0 D 0 )
( 0 0 C 0 0 )
( 0 B 0 0 0 )
( A 0 0 0 0 )

HOMEWORK
What I am doing is generally not in the text. Please read the textbook's relevant sections: 8.4 and 8.5 and 8.7.

Here is my version of the diagram in HTML:

For m linear equations in n unknowns AX=B
Two cases: B=0 and B not 0. Let rank(A)=r.

                             AX=0
                              |        
                              |
                              |
                             \ /
                              v
            -----------------------------------
            |                                 |         
            |                                 |
            |                                 |
           \ /                               \ /
            v                                 v
     Unique sol'n: X=0             Infinite number of sol'ns.
     rank(A)=n                     rank(A)<n, n-r 
                                   arbitrary parameters  
                                   in the sol'n

                                  AX=0, B not 0
                                         |        
                                         |
                                         |
                                        \ /
                                         v
                       -----------------------------------
                       |                                 |         
                       |                                 |
                       |                                 |
                      \ /                               \ /
                       v                                 v
                  Consistent                      Inconsistent 
                  rank(A)=rank(A|B)               rank(A)<rank(A|B)
                       |        
                       |        
                       |        
                      \ /
                       v
       ----------------------------------
       |                                |
       |                                |
       |                                |
      \ /                              \ /
       v                                v          
  Unique solution               Infinite number of sol'ns    
  rank(A)=n                     rank(A)<n-r 
                                arbitrary parameters  
                                in the sol'n

gloss
1. [Linguistics][Semantics]
   a. an explanatory word or phrase inserted between the lines or in
   the margin of a text.

AX=0: m equations in n unknowns; B=0
I	II
Unique sol'n: X=0 rank(A)=n	Infinite number of solutions. rank(A)<n, n-r arbitrary parameters in the solutions

When B is not zero then we've got:

AX=0: m equations in n unknowns; B not 0
Consistent rank(A)=rank(A|B)		Inconsistent rank(A)<rank(A|B)
III	IV	V
Unique solution rank(A)=n	Infinite number of solutions rank(A)<n-r arbitrary parameters in the solutions	No solutions

By the way, I do not recommend that you memorize this information. No one I know has done this, not even the most compulsive. But everyone I know who uses linear algebra has this installed in their brains. As I mentioned in class, I thought that a nice homework assignment would be for students to find examples of each of these (in fact, there have already been examples of each of these in the lectures!). The problem with any examples done "by hand" is that they may not reflect reality. To me, reality might begin with 20 equations in 30 unknowns, or maybe 2,000 equations ....

So I just gave 6 examples which we analyzed in class. These examples are (I hope!) simple (m=number of equations; n=# of variables; r=rank(A), where A is the coefficient matrix):

1. 2x+3y=0
   m=1; n=2; r=1. We are in case I. The solutions are x=[3/2]t and y=-[2/3]t, or all linear combinations of ([3/2],-[2/3]), a basis of the 1 dimensional solution space.
2. 2x+3y=0
   -5x+7y=0
   m=2; n=2; r=2. r=2 since
   

   ( 2  3)~( 1  3/2)~( 1 0 )
   (-5  7) ( 0 29/2) ( 0 1 )

   I think you could already have seen that the rows of this 2-by=2 matrix were linearly independent just by looking at it (don't try "just looking at" a random 500-by-300 matrix!), so r=2. This is case II. There is a unique solution, x=0 and y=0, the trivial solution.
3. 2x+3y=0
   -5x+7y=0
   4x+5y=0
   m=2; n=2; r=2. r=2 since r is at least 2 using the previous row reduction, and r can be at most 2 since the number of variables is 2. Again, this is case II. There is a unique solution, x=0 and y=0, the trivial solution.
4. 2x+3y=1
   m=1; n=2; r=1. We are in case IV. The solutions are (-1,1)+t([3/2],-[2/3]), an infinite 1-parameter family of solutions. Where did this come from? Here we first searched for a particular solution of 2x+3y=1, and we guessed x_p=-1 and y_p=1. (Don't try this with a big system: use row reduction or, better, use a computer!). Then I looked at 2x+3y=0 and used list of all solutions of the associated homogeneous system given in example A, above: x_h=[3/2]t and y_h=-[2/3]t. Now what? We use linearity:
   2x_p+3y_p=1
   2x_h+3y_h=0 (a list of all solutions)
   2(x_p+x_h)+3(x_p+y_h)=1.
5. 2x+3y=1
   -5x+7y=2
   m=1; n=2; r=2. Since here rank(A)=rank(A|B) and further row reduction shows that

    
   ( 2  3 | 1)~( 1  3/2 | 1/2)~( 1 0 | 1/2 - (3/2)·(9/29))
   (-5  7 | 2) ( 0 29/2 | 9/2) ( 0 1 |       9/29        )

   we are in case III, with exactly one solution which row reduction has produced: x=1/2-(3/2)·(9/29)) and y=9/29.
6. 2x+3y=1
   -5x+7y=2
   4x+5y=3
   Now m=1; n=2; r=2. But look:

   ( 2  3 | 1) ( 1  3/2 | 1/2) ( 1 0 | 1/2 - (3/2)·(9/29))
   (-5  7 | 2)~( 0 29/2 | 9/2)~( 0 1 |       9/29        )
   ( 4  5 | 3) ( 0  -1  |  1 ) ( 0 0 |     1-(9/29)      )

   I am lazy and I know that 1-(9/29) is not 0, so the row reduction showed that rank(A)=2<rank(A|B)=3: case V, with no solutions.

Heuristic, the word

adj. 
1. allowing or assisting to discover.
2. [Computing] proceeding to a solution by trial and error.
"heuristic method"[Education] a system of education under which pupils
are trained to find out things for themselves.

Heuristic linear algebra (?)
When I had Maple create random 3-by 5 matrices for the RREF samples I gave students last time, almost all of the matrices created had rank 3. In fact, lots of experimentation shows that matrices want to be the highest possible rank. There are reasons for most examples created in courses or examples which occur in nature which have lower than the highest possible rank. In fact, as I remarked in class, if you measure a physical process at 17 different points, and do this 300 times, and create a 300-by-17 matrix, and then do row reduction, there should be a reason why the resulting rank is not 17. (Of course, in reality, there's all sorts of sources of measurement and analytic error, but ... I don't do reality.) Now looking at the possible alternatives I through V for linear systems, the most interesting occur when A is square and has maximal rank. That's what I will concentrate on right now.

Another silly example
The system

3x+4y=1
5x+6y=0

(3 4 | 1)~(1  4/3 |  1/3)~(1 0 |  -3)
(5 6 | 0) (0 -2/3 | -5/3) (0 1 | 5/2)

And another
The system

3x+4y=0
5x+6y=1

(3 4 | 0)~(1  4/3 | 0)~(1 0 |   2 )
(5 6 | 1) (0 -2/3 | 1) (0 1 | -3/2)

More, even?
The system

3x+4y=578
5x+6y=394

Solved 'em all, at least for this coefficient matrix!
The system

3x+4y=u
5x+6y=v

( -3    2 )(u)
( 5/2 -3/2)(v)

( -3    2 )
( 5/2 -3/2)

( 3 4 )
( 5 6 )

Definition Suppse A is an n-by-n matrix. Then the inverse of A, usually written A^-1, is an n-by-n matrix whose product with A is the n-by-n identity matrix, I_n (a square matrix with diagonal entries equal to 1 and offdiagonal entries equal to 0).

Now if

A=( 3  4 ) and, say, C= ( -3    2 ) you can check that AC=(1 0)=I2
  ( 5  6 )              ( 5/2 -3/2)                       (0 1)

X=(x) and B=(u)
  (y)       (v)

An algorithm for finding inverses
If A is a square matrix then augment A by I_n, an identity matrix of the same size: (A|I_n). Use row reduction to get (I_n|C). Then C will be A^-1. If row reduction is not successful, A does not have an inverse (this can happen: rank(A) could be less than n).

Vocabulary
A matrix which has no inverse is called singular. A matrix which has an inverse is called invertible or non-singular or regular.

A Math 250 exam problem, which was the QotD
I gave a problem very much like this on a Math 250 exam. So:
Suppose A is a 3-by-3 matrix and you know that A^-1 is

(3 2 -1)
(0 2  5)
(7 3  7)

( 2)
(-1)
( 2).

( 1)
(13)
(32)

The exam comes back
I returned the first exam. In this course, I try to help you learn in several different ways (lectures, QotD's, this diary, office hours, e-mail). Another important aspect of what I do in the course is write exams and then report the grades. I try to give exams which are written and graded fairly. I would be very happy if everyone got A's, especially because I could feel this is a partial consequence of my own efforts. At the same time, I should report results corresponding to actual student achievement. I hope you will look at the answers supplied and the remarks on grading and let me know if you have any questions about the exam, the answers, or my grading.

A vector w is a linear combination of v₁, v₂, ..., v_n if there are scalars a₁, a₂, ..., a_n so that v=SUM_j=1ⁿa_jv_j.

We study At(t+1)+B(t+2)(t+3)+C(t+4)(t+5)=t². If this equation is valid, then
(A+B+C)t²+(A+5B+9C)t+(6B+20C)=t²
Which leads to the system

 
1A+1B+1C=1
1A+5B+9C=0
0A+6B+20C=0

( 1 1  1 |  1 ) ( 1 1  1 |  1 ) ( 1 0 -1 |  5/4 ) ( 1 0 0 |  46/32 )
( 1 5  9 |  0 )~( 0 4  8 | -1 )~( 0 1  2 | -1/4 )~( 0 1 0 | -20/32 )
( 0 6 20 |  0 ) ( 0 6 20 |  0 ) ( 0 0  8 |  6/4 ) ( 0 0 1 |   6/32 )

The answer to the original question is "Yes" and I am fairly confident that
(46/32)t(t+1)+(-20/32)(t+2)(t+3)+(6/32)(t+4)(t+5)=t².

Now I asked another question: can t be written as a linear combination of t(t+1) and (t+2)(t+3) and (t+4)(t+5)? I also asked if we could answer this question without any additional computation. In fact, the answer is "Yes". All we need to do is change the column augmenting the coefficient matrix so that it looks like

0
1
0

1
0
0

Now I asked another question: are the three polynomials t(t+1) and (t+2)(t+3) and (t+4)(t+5) linearly independent? Here we consider the question: if At(t+1)+B(t+2)(t+3)+C(t+4)(t+5)=0, must A and B and C all be 0? Again, row reduction tells me that A=0 and B=0 and C=0 since row operations on a column with all 0 entries leave the column all 0's, so that, yes, indeed, these polynomials are linearly independent.

Problem 2
Are the functions cos(t) and [cos(t)]² and [cos(t)]³ linearly independent?

The vectors v₁, v₂, ..., v_n are linearly independent if whenever SUM_j=1ⁿa_jv_j=0 then all of the a_j's must be 0.

So we need to study
A cos(t)+B[cos(t)]²+C[cos(t)]³=0
Now it isn't totally clear what to do. In the previous case, we actually relied upon 1 and t and t² as a "natural" way to look at polynomials, and used this way to obtain a system of linear equations. Now one way to get a system of linear equations is to "specialize" (?) the functional equation by evaluating it at various t's and hope there will be enough information to make some conclusion about the question.

If we let t=Pi/2 then the equation becomes A·0+B·0+C·0=0. But for any values of A and B and C, this is 0=0, so we get no information!

Now t=0 gives A+B+C=0. And t=Pi gives -A+B-C=0. And t=Pi/4 gives (1/sqrt(2))A+(1/2)B+(1/[2sqrt(2)])C=0. And t=2Pi gives the same equation as t=0 (because of 2Pi periodicity of cosine). Etc. and etc. How long should we do this? How many t's, and how to choose them? Well, so far we have this linear system:
A+B+C=0
-A+B-C=0
(1/sqrt(2))A+(1/2)B+(1/[2sqrt(2)])C=0
If we convert the coefficient matrix to RREF and get what we want (a diagonal matrix with 1's) then we know that the functions are linearly independent. I wouldn't think of working with fewer than three equations for this reason. I can also hope that working with these three equations will give me enough information. Of course, I will admit honestly, if I didn't know these functions too well, I could have chosen t=0 and t=2Pi and t=4Pi to get a linear system, and the resulting RREF would certainly not have enough information for me to decide. Oh well. So here we go (I don't need a column of 0's for augmentation since the row reduction won't change them at all):

(     1      1        1      ) ( 1        1                1       ) ( 1 0       1       ) ( 1 0 0 )
(    -1      1       -1      )~( 0        2                0       )~( 0 1       0       )~( 0 1 0 )
( 1/sqrt(2) 1/2 1/[2sqrt(2)] ) ( 0 (1/2)-(1/sqrt(2)) -1/[2sqrt(2)] ) ( 0 0 -1/[2sqrt(2)] ) ( 0 0 1 )

I remarked in class that I could have asked if the three functions [cos(t)]² and [sin(t)]² and cos(2t) are linearly independent. Of course, this example is a bit silly, but I would like to illustrate what could happen. If I then write the equation
A[cos(t)]²+B[sin(t)]²+Ccos(2t)=0
and from there, by specializing at t=blah and t=blahblah etc., deduce a system of linear equations and try to show that the only solution is the trivial solution (following the example above) I will not be successful. Any such system will have rank (number of non-zero rows in its RREF) at most 2. I need rank 3 to conclude that A=0 and B=0 and C=0. This all happens because, in fact, the functions [cos(t)]² and [sin(t)]² and cos(2t) are linearly dependent. One of the early trig formulas (double angle stuff) tells us that
1[cos(t)]²-1[sin(t)]²-cos(2t)=0
I want to be honest with you and remark that the logic and computations supporting the logic can get complicated.

Problem 3
Are the vectors (4,3,2) and (3,2,3) and (-4,4,3) and (5,2,1) in R³ linearly independent? Now I wrote the vector equation:
A(4,3,2)+B(3,2,3)+C(-4,4,3)+D(5,2,1)=0 (this is (0,0,0) for this instantiation of "linear independence") which gives me the system:
4A+3B-4C+5D=0 (from the first components of the vectors)
3A+2B+4C+2D=0 (from the second components of the vectors)
2A+3B+3C+1D=0 (from the third components of the vectors)
and therefore would need to row reduce

( 4 3 -4 5 )
( 3 2  4 2 )
( 2 3  3 1 )

What can these RREF's look like? Let me write all of the RREF's possible whose first column (as here) has some entry which is not zero.

( 1 0 0 * ) ( 1 0 * 0 ) ( 1 0 * * ) ( 1 * 0 0 ) ( 1 * 0 * ) ( 1 * * 0 ) ( 1 * * * )     
( 0 1 0 * ) ( 0 1 * 0 ) ( 0 1 * * ) ( 0 0 1 0 ) ( 0 0 1 * ) ( 0 0 0 1 ) ( 0 0 0 0 )
( 0 0 1 * ) ( 0 0 0 1 ) ( 0 0 0 0 ) ( 0 0 0 1 ) ( 0 0 0 0 ) ( 0 0 0 0 ) ( 0 0 0 0 )

A homogeneous system with more variables than equations always has an infinite number of non-trivial solutions.

Homily The background for these words includes the upcoming course exam scheduled for the next regular class time and the lack of response from the majority of students to my request that students do the review problems and send me the result to post on the web. I see this as a lack of interest or ability or both. The students in this class are mostly juniors and seniors, and know the flaws and features of a Rutgers education. No matter how well-intentioned or skilled your instructors have been, by now you should know that your best teachers will be yourselves, using your own skills and efforts. That will almost surely continue for the rest of your lives. Most of your instruction will come through your own practice. Students cannot rely upon instructors to do everything, nor can students anticipate exams (or jobs!) which only test abilities that have been acquired casually, abilities that are either dimly remembered or invented on the spot. Most students in the class have considerable time and experience, and must do much more themselves to learn this material than they already have.

The QotD
I gave students individualized 3 by 5 matrices whose entries were "random" integers created by Maple in the range from -3 to 3. I told students that the matrices should be converted to RREF, and that students doing this successfully would have 5 points added to their first exam score. The results showed that some students did not know what RREF was (that's after five hours of class time relying on RREF). Many other students seemed to have serious difficulty correctly completing a limited amount of rational number arithmetic. My homily above would have included more excoriation if I had known these results ahead of time.

/excoriate/
1. a. remove part of the skin of (a person etc.) by abrasion.
   b. strip or peel off (skin).
2. censure severely.

/homily/ 
1. a sermon.
2. a tedious moralizing discourse.

I began by writing the following on the side board, so that I would not mess up things again.

A linear combination of vectors is a sum of scalar multiples of the vectors. A collection of vectors is spanning if every vector can be written as the linear combination of vectors in the collection. A collection of vectors is linearly independent if, whenever a linear combination of the vectors is the zero vector, then every scalar coefficient of that linear combination must be zero.

I wanted to suffer no further obloquy in this class. My online dictionary gives me this as a definition:

/obloquy/ n. 
1. the state of being generally ill spoken of.
2. abuse, detraction.

The language and ideas of linear algebra are used everywhere in applied science and engineering. Basic calculus deals fairly nicely with functions defined by formulas involving standard functions. I asked how we could understand more realistic data points. I will simplify, because I am lazy. I will assume that we measure some quantity at one unit intervals. Maybe we get the following:

Tabular presentation of the data	Graphical presentation of the data
"Independent" variable (x) "Dependent" variable (y) 1 5 2 6 3 2 4 4 5 -2 6 4

Piecewise linear interpolation We could look at the tabular data and try to understand it, or we could plot the data because the human brain has lots of processing ability for visual information. But a bunch of dots is not good enough -- we want to connect the dots. O.k., in practice this means that we would like to fit our data to some mathematical model. For example, we could try (not here!) the best fitting straight line or exponential or ... lots of things. But we could try to do something simpler. We could try to understand this data as just points on a piecewise linear graph. So I will interpolate the data with line segments, and even "complete" the graph by pushing down the ends to 0. The result is something like what's drawn to the right. I will call the function whose graph is drawn F(x).

Well, this interpolation function is not differentiable at various points, but it is continuous and it certainly is rather simple. But in fact it can be written in even a simpler form. Let me show you.

Meet the tent centered at the integer j
First, here is complete information about the function T_j(x) which you are supposed to get from the graph of T_j(x): (j should be an integer)

This is a peculiar function. It is 0 for x<j-1 and for x>j+1. It has height 1 at j, and interpolates linearly through the points (j-1,0), (j,1), and (j+1,0). I don't much want an explicit formula for T_j(x): we could clearly get one, although it would be a piecewise thing (but as several students observed, T_j(x) could be written quite easily in Laplace transform language!).

We can write F(x) in terms of these T_j(x)'s fairly easily. Moving (as we were accustomed in Laplace transforms!) from left to right, we first "need" T₁(x). In fact, consider 5T₁(x) and compare it to F(x). I claim that these functions are exactly the same for x<=1. Well, they are both 0 for x<=0. Both of these functions linearly interpolate between the points (0,0) and (1,5), so in the interval from 0 to 1 the graphs must be the same (two points still do determine a unique line!). Now consider 5T₁(x)+6T₁(x) and F(x) compared for x's less than 2. Since T₂(x) is 0 for x<1, there is no chain in the interval (-infinity,1]. But between 1 and 2, both of the "pieces" T₁(x) and T₂(x) are non-zero. But the sum of 5T₁(x)+6T₁(x) and F(x) match up at (1,5) and (2,6) because we chose the coefficients so that they would. And both of the "tents" are degree 1 polynomials so that their sum is also, and certainly the graph of a degree 1 polynomial is a straight line, so (again: lines determined by two points!) the sum 5T₁(x)+6T₁(x) and F(x) totally agree in the interval from 1 to 2. Etc. What do I mean? Well, I mean that
F(x)= 5T₁(x)+6T₁(x)+2T₁(x)+4T₁(x)+ -2T₁(x)+4T₁(x).
These functions agree for all x.

Linear combinations of the tents span these piecewise linear functions
If we piecewise linearly interpolate data given at the integers, then the resulting function can be written as a linear combination of the T_j's. Such linear combinations can be useful in many ways (for example, the definite integral of F(x) is the sum of constants multiplied by the integrals of the T_j(x), each of which has total area equal to 1!). The T_j(x)'s are enough to span all of these piecewise linear functions.

But maybe we don't need all of the T_j's. What if someone came up to you and said, "Hey, you don't need T₃₃(x) because:"
T₃₃(x)=53T₁₂(x)-4T₅(x)+9T₁₄(x)
Is this possibly correct? If it were correct, then the function T₃₃(x) would be redundant (extra, superfluous) in our descriptions of the piecewise linear interpolations, and we wouldn't need it in our linear combinations. But if
T₃₃(x)=53T₁₂(x)-4T₅(x)+9T₁₄(x)
were correct, it should be true for all x's. This means we can pick any x we like to evaluate the functions, and the resulting equation of numbers should be true. Hey: let's try x=33. This is not an especially inspired choice, but it does make T₃₃'s value equal to 1, and the value of the other "tents" in the equation equal to 0. The equation then becomes 1=0 which is currently false.
Therefore we can't throw out T₃₃(x). In fact, we need every T_j(x) (for each integer j) to be able to write piecewise linear interpolations.
We have no extra T_j(x)'s: they are all needed.

Let me rephrase stuff using some linear algebra language. Our "vectors" will be piecewise linear interpolations of data given at integer points, like F(x). If we consider the family of "vectors" given by the T_j(x)'s, for each integer j, then:

• This family spans every thing. Every piecewise linear function is a sum of scalar multiples of the T_j(x)'s.
• This family is linearly independent. We need each member, each specific T_j(x), in order to be able to write any F(x) as a linear combination of the T_j(x)'s.

I emphasized that one reason I wanted to consider this example first is because we use linear algebra ideas constantly, and presenting them in a more routine setting may discourage noticing this. My second major example does, however, present things in a more routine setting, at least at first.

My "vectors" will be all polynomials of degree less than or equal to 2. So one example is 5x²-6x+(1/2). Another example is -(Pi)xx²+0x+223.67, etc. What can we say about this stuff?

I claim that every polynomial can be written as a sum of 1 and x and x² and the Kadakia polynomial, K(x)=3x²-9x+4 (well, it was something like this). Since the class showed signs of irritation (perhaps because the instructor is from the planet Zornix, where conversations are done with odors), I offered to verify that, say, the polynomial 17x²+44x-98 could indeed be written as a sum of 1 and x and x² and the (fabulous) Kadakia polynomial, K(x)=3x²-9x+4. Thus I need to find numbers filling the empty spaces in the equation below, and the numbers should make the equation correct.
17x²+44x-98=[ ]1+[ ]x+[ ]x²+[ ]K(x)
Indeed, through great computational difficulties I wrote
17x²+44x-98=-781+91x+2x²+5K(x)
(I think this is correct, but again I am relying upon carbon-based computation, not silicon-based computation!)

We discussed this and concluded that the span of 1 and x and x² and K(x) is all the polynomials of degree less than or equal to 2. But, really, do we need all of these? That is, do we need the four numbers (count them!) we wrote in the equation above, or can we have fewer? Well, 1 and x and x² and K(x) are not linearly independent. They are, in fact, linearly dependent. There is a linear combination of these four which is zero. Look:
1K(x)+-3x²+-44x+981=0.
so that K(x)=3x²+44x+-981
and we can "plug" this into the equation
17x²+44x-98=-781+91x+2x²+5K(x)
to get 17x²+44x-98=-781+91x+2x²+5(3x²+44x+-981 )

Groucho Marx declared:

A child of five would understand this.       Send someone to fetch a child of five.

If we distribute the sum out as any child would, we get the remarkable equation
17x²+44x-98=17x²+44x+-98
As remarked, one trouble with linear algebra is that it seems too easy and too obvious when we deal with familiar situations. Let me unfamiliarize (!?) this situation a bit.

Some weird polynomials to change our point of view
Look at these polynomials of degree 2:
   P(x)=x(x-1)
   Q(x)=x(x-2)
   R(x)=(x-1)(x-2)
Why these polynomials? Who would care about such silly polynomials?

Are these polynomials linearly independent?
This was the QotD. I remarked that I was asking students to show that if
A P(x)+B Q(x)+C R(x)=0
for all x, then the students would need to deduce that A=0 and B=0 and C=0. I also remarked that I was interested in the logic which was used more that any thing else.

• What I expected from students in their answers
  Well, take the equation A P(x)+B Q(x)+C R(x)=0 and plug in x=0. Then we get (since A(0)=0 and B(0)=0 and R(0)=2) that 2C=0 so that C must be 0. Similarly, x=1 gives B=0 and x=2 gives C=0. So the linear combination is the trivial linear combination, and the polynomials are linearly independent.
• What I was doing while the students were writing
  At the board I computed P(x)=x²-x and Q(x)=x²-2x and R(x)=x²-3x+2 (o.k., I had an arithmetic error, but my intention was there). Then I wrote the equation
  A P(x)+B Q(x)+C R(x)=0
  as the equation A (x²-x)+B (x²-2x)+C (x²-3x+2)=0
  and then I further distributed and got:
  (A+B+C)x²+(-A-2B-3C)x+(2C)=0
  which results in the linear system

   1A+1B+1C=0
  -1A-2B-3C=0
         2C=0

  and then I row reduced (by hand!) the coefficient matrix:

  ( 1  1  1 ) ( 1  1  1 ) ( 1  0  1 ) ( 1  0  0 )
  (-1 -2 -3 )~( 0 -1 -2 )~( 0  1  2 )~( 0  1  0 ) 
  ( 0  0  2 ) ( 0  0  2 ) ( 0  0  2 ) ( 0  0  1 )

  This shows that the original system was row equivalent to the system A=0 and B=0 and C=0 (remember that "row equivalent"<==>"same solution set"), therefore there are no solutions to the equation A P(x)+B Q(x)+C R(x)=0 except the trivial solution. And therefore P(x) and Q(x) and R(x) are linearly independent: none of them are "redundant".

But can I describe all of the deg<=2 polys this way?
I assert that every polynomial of degree less than or equal to 2 can be described as a linear combination of P(x) and Q(x) and R(x). How would I verify this claim? Please note that I am more interested in the logic than the computational details here!
I should be able to write x² as sum of P(x) and Q(x) and R(x). This means I should be able to solve the equation
A P(x)+B Q(x)+C R(x)=x².
Just as above, this leads to an augmented matrix which looks like:

( 1  1  1 | 1 )   ( 1  0  0 | FRED    )
(-1 -2 -3 | 0 )~~~( 0  1  0 | STANLEY ) 
( 0  0  2 | 0 )   ( 0  0  1 | MARISSA )

( 1  1  1 | 0 )
(-1 -2 -3 | 1 )
( 0  0  2 | 0 )

VOCABULARY TIME! A collection of vectors is a basis if the vectors are linearly independent and if every vector is a linear combination of these vectors (that is, the span of these vectors is everything>

So P(x) and Q(x) and R(x) are a basis of the polynomials of degree less than or equal to 2.

Why would we look at P(x) and Q(x) and R(x)?
Suppose again I have data points, let's say (0,13) and (1,-78) and (2,37). I could linearly interpolation as we did above. If I want to be a bit more sophisticated, and get maybe something smoother, I could try to get a polynomial Ax²+Bx+C which fits these data points. Here is the polynomial: [37/2]P(x)+[-78/-1]Q(x)+[13/2]R(x). How was this remarkable computation done? Well, I know certain function values

The function	Its values when x=0 and x=1 and x=2
P(x)	0	0	2
Q(x)	0	-1	0
R(x)	2	0	0

so when I "plug in" x=0 and x=1 and x=2 in the linear combination A P(x)+B Q(x)+C R(x) and I want to get 13 and -78 and 37, respectively, the structure of the table makes it very easy to find A and B and C. If we want to interpolate quadratically, I would get a function defined by a simple formula using this basis. In fact, these functions are very useful in quadratic interpolation, and in the use of splines, a valuable technique for numerical approximation of solutions of ordinary and partial differential equations.

HOMEWORK
Please read the textbook: chapter 7.6 and the first sections of chapter 8.

   3x₁+2x₂+x₃-x₄=A
   4x₁-1x₂+5x₃+x₄=B
   2x₁+5x₂-3x₃-3x₄=C

These questions will be important:

1. For which values of A and B and C does this system have any solutions?
2. If A and B and C have values for which there are solutions, describe the solutions in some effective way.

(3  2  1 -1 | A)
(4 -1  5  1 | B)
(2  5 -3 -3 | C)

• Add one equation to another.
• Multiply an equation by a non-zero number.
• Interchange equations.

There are all sorts of things one can do with elementary row operations, but one very useful goal is to change the coefficient matrix into reduced row echelon form, RREF. This means:

• Any initial non-zero entry in each row is 1.
• The other column entries for an initial non-zero entry of 1 will be 0.
• Initial 1's to the right always occur in lower rows. That is, if a_ij is an initial 1 and a_mn is another initial 1, then if i<m, j must be <n.

 3 by 4         2 by 7

(1 0 0 0)   (1 0 0 0 0 4 4)
(0 0 1 5)   (0 0 0 0 1 2 3)
(0 0 0 0)

(1 0 0)  (1 0 *)  (1 * *)  (1 0 0)  (0 1 0)  (0 0 1)  (0 0 0) 
(0 1 0)  (0 1 *)  (0 0 0)  (0 0 1)  (0 0 1)  (0 0 0)  (0 0 0) 
(0 0 1)  (0 0 0)  (0 0 0)  (0 0 0)  (0 0 0)  (0 0 0)  (0 0 0)

I then tried to use row operations on the augmented matrix of my system of linear equations so that the coefficient matrix was in RREF. I did this by hand. What follows is taken from a Maple session (I did remove my errors!). Maple does have a rref command, but it also allows row operations one at a time. I will comment in this manner about each command.

>with(linalg):
This loads the linear algebra package.
>M:=matrix(3,5,[3,2,1,-1,A,4,-1,5,1,B,2,5,-3,-3,C]);
                        [3     2     1    -1    A]
                        [                        ]
                   M := [4    -1     5     1    B]
                        [                        ]
                        [2     5    -3    -3    C]
This command just creates a matrix of specified size
with the listed entries.
>mulrow(M,1,1/3);
                   [1    2/3    1/3    -1/3    A/3]
                   [                              ]
                   [4    -1      5      1       B ]
                   [                              ]
                   [2     5     -3      -3      C ]
Multiply row 1 by a third. Creates an initial 1.
>addrow(%,1,2,-4);
              [1     2/3     1/3     -1/3       A/3   ]
              [                                       ]
              [                                4 A    ]
              [0    -11/3    11/3    7/3     - --- + B]
              [                                 3     ]
              [                                       ]
              [2      5       -3      -3         C    ]
Add -4 times row 1 to row 2. This "pivots" and makes the
(2,1) entry of the coefficient matrix 0.
>addrow(%,1,3,-2);
              [1     2/3      1/3     -1/3       A/3   ]
              [                                        ]
              [                                 4 A    ]
              [0    -11/3    11/3     7/3     - --- + B]
              [                                  3     ]
              [                                        ]
              [                                 2 A    ]
              [0    11/3     -11/3    -7/3    - --- + C]
              [                                  3     ]
Add -2 times row 1 to row 3. So the (3,1) entry becomes 0.
>mulrow(%,2,-3/11);
              [1    2/3      1/3     -1/3       A/3   ]
              [                                       ]
              [                       -7     4 A   3 B]
              [0     1       -1       --     --- - ---]
              [                       11     11    11 ]
              [                                       ]
              [                                2 A    ]
              [0    11/3    -11/3    -7/3    - --- + C]
                                                3
Makes another leading 1.
>addrow(%,2,1,-2/3);
              [                               A     2 B]
              [1     0        1      1/11    ---- + ---]
              [                               11    11 ]
              [                                        ]
              [                       -7     4 A   3 B ]
              [0     1       -1       --     --- - --- ]
              [                       11     11    11  ]
              [                                        ]
              [                                2 A     ]
              [0    11/3    -11/3    -7/3    - --- + C ]
              [                                 3      ]
Makes the (1,2) entry equal to 0.
>addrow(%,2,3,-11/3);
                [                          A     2 B ]
                [1    0     1    1/11     ---- + --- ]
                [                          11    11  ]
                [                                    ]
                [                 -7      4 A   3 B  ]
                [0    1    -1     --      --- - ---  ]
                [                 11      11    11   ]
                [                                    ]
                [0    0     0     0      -2 A + B + C]
And now the (3,2) entry is 0. The coefficient matrix 
is now in RREF.

Now back to the questions:

1. For which values of A and B and C does this system have any solutions?
2. If A and B and C have values for which there are solutions, describe the solutions in some effective way.

What if I know -2A+B+C=0? Let's choose some values of A and B and C which will make this true. How about A=7 and B=34 and C=-20. Then -2A+B+C=-2(7)+34-20=0 (I hope). The RREF system then becomes (inserting these values of A and B and C [I did this in my head so there may be ... errors]):

                [                          75  ]
                [1    0     1    1/11     ---- ]
                [                          11  ]
                [                              ]
                [                 -7       74  ]
                [0    1    -1     --    - ---- ]
                [                 11       11  ]
                [                              ]
                [0    0     0     0         0  ]

Be sure to look carefully at the signs, to check on what I've written. The equations have been written this way so that you can see that x₃ and x₄ are free. That is, I can give any values for these variables. Then the other variables (x₁ and x₂) will have their values specified by what is already given. So: we select A and B and C satisfying the compatibility condition. Then there will be a two-parameter family of solutions to the original system of linear equations. Notice that we get solutions exactly when the compatibility condition is satisfied: there are solutions if and only if (as math folks might say) the compatibility condition is correct.

The logic here is actually "easy". Since all the computational steps we performed are reversible, I know that the assertions I just made are correct. What is more wonderful is that the general situation will always be much like this.

What RREF does in general
Take your original augmented matrix, and put the coefficient matrix into RREF. Then you get something like

(BLOCK OF 1's & 0's WITH|  JJJJJJ  UU  UU  NN  N  K K | Linear stuff )
(THE 1's MOVING DOWN AND|    JJ    UU  UU  N N N  KK  | from original)
(TO THE RIGHT.          |  JJJJ    UUUUUU  N  NN  K K |  right sides )
(--------------------------------------------------------------------)
(                    MAYBE HERE SEVERAL               |   More such  )
(                       ROWS OF 0'S                   | linear stuff )

I then started to look at the vocabulary of linear algebra, and made a few dreadful mistakes. My idea was to give examples of some fundamental definitions, and show how considering candidates for examples satisfying the definitions turns out to be studying systems of linear equations. Well, what follows is what I should have done, maybe.

Linear combinations
First, I asked if w=(3,1,2,3) was equal to a sum of scalar multiples of the vectors v₁=(3,3,-1,2) and v₂=(5,2,-3,2) and v₃=(-2,4,5,6). So now:

VOCABULARY TIME! A vector w is a linear combination of v1, v2, ..., and vN if there are scalars a1, a2, ..., and aN so that w=SUMj=1Najvj.

In this case, the vector equation w=a₁v₁+a₂v₂+a₃v₃. For the vectors listed above, we can look at each of the components of the vectors in R⁴ and get a system of linear equations:
3a₁+3a₂-2a₃=3 (from the first components)
1a₁+3a₂+4a₃=1 (from the second components)
2a₁-1a₂+5a₃=2 (from the third components)
3a₁+2a₂+6a₃=3 (from the fourth components)
If this system has a solution, then w is equal to a linear combination of v₁ and v₂ and v₃. If this system does not have a solution, then w is not equal to such a linear combination.

VOCABULARY TIME! A system of linear equations is called consistent if it has at least one solution. Otherwise the system is called inconsistent.

VOCABULARY TIME! If a vector w is equal to a linear combination of vectors v1, v2, ..., and vN then w is in the span of v1, v2, ..., and vN.

Linear independence
When we're given a collection of vectors, we can ask if any of them are "extra", that is, not needed because they are already linear combinations of other vectors. The non-obvious but useful example I should have started with here is the collection of these three functions: sin(t), cos(t), and 5sin(t-4). Are any of these not necessary? Since I know that 5sin(t-4)=5[sin(t)cos(-4)+cos(t)sin(-4)]= {5cos(-4)}sin(t)+{5sin(-4)}cos(t) I can immediately see that, in the language of linear algebra, 5sin(t-4) is redundant.

/redundant/
1. superfluous; not needed.
2. that can be omitted without any loss of significance.

5sin(t-4) is a linear combination of sin(t) and cos(t), with scalars equal to 5cos(-4) and 5sin(-4). How can we systematize this idea of redundancy, or the idea of not being redundant?

VOCABULARY TIME! The vectors v1, v2, ..., and vN are linearly dependent if there is some non-trivial linear combination of them which is equal to 0. That is, for some scalars a1, a2, ..., and aN which are not all 0, we know that SUMj=1Najvj=0 (the zero vector).

The functions sin(t) and cos(t) and 5sin(t-4) are linearly dependent since
5cos(-4)sin(t)+5sin(-4)cos(t)+-15sin(t-4)=0.
The scalar coefficients are 5cos(-4) and 5sin(-4) and -1.

Here is a more routine example, in some sense. Consider the following vectors in R⁵: v₁=(2,3,4,5,7) and v₂=(8,2,3,3,4) and v₃=(-2,7,-3,5,5) and v₄=(0,9,9,1,1). Are these vectors linearly dependent? We are asking for solutions to
a₁v₁+a₂v₂+a₃v₃+a₄v₄=0
so that not all of the a_j's are 0.

We can translate that vector equation in R⁵ into a system of 5 linear equations:
2a₁+8a₂-2a₃+0a₄=0 (from the first components)
3a₁+2a₂+7a₃+9a₄=0 (from the second components)
4a₁+3a₂-3a₃+9a₄=0 (from the third components)
5a₁+3a₂+5a₃+1a₄=0 (from the fourth components)
7a₁+4a₂+5a₃+1a₄=0 (from the fifth components)

This is a homogeneous system of equations and we want to know if there is a non-trivial solution (where at least one of the variables is not 0).

VOCABULARY TIME! A system of linear equations is homogeneous if the right-hand sides are all 0. A solution to such a system is non-trivial if it is not the all 0's solution.

It is not difficult to see if the example I just wrote is linearly dependent. Look:

> with(linalg):
Warning, the protected names norm and trace have been redefined and
unprotected
> M:=matrix(5,4,[2,8,-2,0,3,2,7,9,4,3,-3,9,5,3,5,1,7,4,5,1]);
                                [2    8    -2    0]
                                [                 ]
                                [3    2     7    9]
                                [                 ]
                           M := [4    3    -3    9]
                                [                 ]
                                [5    3     5    1]
                                [                 ]
                                [7    4     5    1]

> rref(M);
                              [1    0    0    0]
                              [                ]
                              [0    1    0    0]
                              [                ]
                              [0    0    1    0]
                              [                ]
                              [0    0    0    1]
                              [                ]
                              [0    0    0    0]

The QotD
I asked if t^{some odd number} is a linear combination of various t^{even number}'s. I expected a solution using simple symmetry: switching t to -t for any non-zero t leads to a contradition.

Here is a more explicit solution. I asked if t⁶⁷ could be written as a linear combination of t², t¹⁶, t³⁴, t⁹⁸, and t¹¹². If we can write
t⁶⁷=At²+Bt¹⁶+Ct³⁴+Dt⁹⁸+Et¹¹²
for some constants A, B, C, D, and E, then consider:
t=1:1=A+B+C+D+E
t=-1:-1=A+B+C+D+E
which would imply 1=-1. Therefore the original assumption is incorrect.

HOMEWORK
Please read the first sections of chapter 8. Hand in the following problems on Tuesday:
8.1: 19, 23, 29 and 8.2: 11, 15. Also, are the functions e^t and e^2t and e^3t linearly independent? Explain your answer, please.
I will try to get the solutions back to you by Thursday.
I will go to a lecture on Monday from 3 to 4 but will be in my office for office hours after 4.

Goals Students should know the definition of the Laplace transform. They should be able to compute simple Laplace transforms "by hand" and should be able to use a table of Laplace transforms to solve initial value problems for constant coefficient ODE's. They should be able to write "rough" data in terms of Dirac and Heaviside functions. They should be able to recognize and use simple results regarding Laplace transforms.

I hope that students can do problems covering this material, especially since

We will have our first formal exam in two weeks, on Tuesday, October 11. More information about the exam (such as its style and material to be covered) will be given very soon.

Today begins the portion of the course related to linear algebra. I tried to give some idea of the importance of this material. One way was to list the math department courses devoted to linear algebra. Another way I tried was to give a comparative description of linear algebra in various settings appropriate to engineering.

Setting: Calc 3 + statics Here vectors are forces and directed line segments in R2 and R3. These objects may represent forces or fluid flow or heat flow or ... and we add them and multiply them by scalars (in this case, just real numbers). Adding them gives resultants and various components of the vectors have names like flux. We can draw nice simple pictures whose geometry is frequently appealing. Object: understand the geometry and physics of simple problems

Setting: Laplace transforms If we take the Laplace transform of systems of ODE's, as in section 4.6 of the text, we get collections of equations involving the Laplace transforms of unknown functions with rational functions (quotients of polynomials) as coefficients. For example, if you've done some problems in that section of the book, you would expect several equations similar to what follows: [(s-1)/(s2+4)]X(s)+ {s/(s+2)3]Y(s)=1/(s+7)+2s -7 and then we want to find expressions for one of the unknown functions (X(s) or Y(s)) in terms only of s, and then take the inverse Laplace transform. The initial conditions are all invisibly part of equations like the one I just wrote. The vectors here are X(s) and Y(s), and the scalars are rational functions such as [(s-1)/(s2+4)] and {s/(s+2)3] and 1/(s+7)+2s. Object: solve initial value problems for systems of ODE's

Setting: Fourier series We will look at this in the last part of the course. Fourier series look like 3sin(5t)+5cos(7t): sums of constants multiplied by trig functions. The applicable trig functions here will be sine and cosine (not cosecant, thank goodness!). The vectors are functions like sin(5t) and cos(7t), and the scalars in this setting are usually real constants. These series are an extremely useful way to analyze certain partial differential equations, PDE's, and their boundary value problems. There also turns out to be a complex version of these sums, involving things like (5+2i)e3i t. Here the scalars are complex numbers, while the vectors are functions like e3i t. Object: solve boundary value problems for classical PDE's such as the wave equation and the heat equation

Setting: Digital signal processing Here the sums are sort of square waves which are the vectors, and the scalars essentially turn out to be (this seems ludicrous when first encountered!) just 0 (off) and 1 (on). Extremely elaborate ideas and algorithms having to due with storage and efficient transmission and transformation of signals are expressed in this language. A chief algorithm is the Fast Fourier Transform (connected with such tools as CAT scans and magnetic resonance), and one way of looking at this algorithm is that it involves writing a matrix as a product in a particularly efficient way. Object: analysis, storage, and transmission of signals

Wassup?
I will first examine how to strip irrelevancies from systmes of linear equations. I will want a way to detect essential aspects of these systems (RREF: reduced row echelon form). Then I'll need to describe these "essential" aspects. I will also want to discuss how to invert matrices in several ways. Another topic which is wonderfully useful is diagonalization of matrices, which will provide great computational efficiencies when it can be used.

Vocabulary
There are a bunch of terms which students should learn. Included are: homogeneous, inhomogeneous, consistent, inconsistent, basis, linear independence, linear combination, spanning, subspace, dimension, rank, eigenvalue, eigenvector, matrix addition and matrix multiplication, symmetric, diagonalization. I will certainly ask for definitions of some of these on exams.

Silly linear algebra questions
We discussed these questions. I hope that after we are done, students will be able to find, understand, and support (say, in writing!) answers to these questions.

1. Suppose (1,2,3,4) and (-9,3,2,2) are solutions to seven linear equations in four unknowns. Are there other solutions?
   Linear equation The unknowns occur only as sum of scalar multiples of the first powers.
   System A collection of linear equations.
   Solution When the "values" are "plugged in", the equations are correct.
   Answer It turns out that the answer is "Yes, there are always other solutions." This does not depend on the number of equations.
2. Someone comes up to you on the street and says that there is a system of 218 linear equations in 234 variables which has exactly 568 solutions. Could this possibly be correct?
   Answer In fact, this situation could not possibly be correct! We will discover why.
3. Now we have a homogeneous system of linear equations. Here the right-hand side, not involving the variables, is 0. A homogeneous system always has what is called the trivial solution: all the variables=0. For example, 3x₁+5x₂+2x₃=0. Using the values x₁=0 and x₂=0 and x₃=0 we certainly get a solution (check:0=0). Are there other solutions? I think that x₁=1 and x₂=1 and x₃=-4 make the equation correct. This is a non-trivial solution, because there is at least one variable which is not 0. That's the setting for the following almost ludicrous question:
   Suppose we have a linear system with 235 equations and 517 unknowns. Does this linear system have a non-trivial solution?
   Answer In fact, there is actually, always (!) a solution to this homogeneous system which is not all 0's. I don't think this is obvious. .

We discussed, debated, and perhaps did not bet about answers to these questions. This was difficult, although a number of students carefully slept through it.

I discussed how to get a matrix, and how substituting solutions of a linear system, one into the other, leads to the definition of matrix multiplication. The definition of matrix multiplication did not occur without many examples. The abbreviations I use are now generally accepted, principally because they are needed in practice. We reviewed how to get the matrix product.

Matrix addition is both commutative and associative. Matrix multiplication, however, is associative and, since products in different orders might matter, we need to be careful that we don't use commutativity because

The QotD
Find two 2-by-2 matrices A and B so that AB is not the same as BA. (This turns out not to be too difficult since almost any A and B will work, but not, as we remarked, an A whose entries are all 0's.)

Homework
Look at the syllabus. Start to read chapter 8.

Maintained by greenfie@math.rutgers.edu and last modified 9/4/2004.