Students' answers to review problems for the second exam in Math 421:02, spring 2004

Students' answers to review problems for the second exam
in Math 421, spring 2004

Very few of these problems should require extensive computation. The actual exam will ask students for several of the definitions in problem 14.
The idea for the list of New Jersey Native Trees and its use here was copied from Professor B. Latka, whose linear algebra exams at Lafayette College featured the same device.

means no answer is available yet. The answers are all here! Very good job, people!
Various corrections have been made to the answers (6:48 PM, 4/8/2004). I currently believe the answers are right.

#1 C. Graziadio #2 J. Sirak #3 B. Cohen #4 C. Hunt #5 M. Woodrow #6 X. Sosa #7 S. Brynildsen
#8 L. Kohut #9 J. Malek
#10 J. Sirak
M. Woodrow
#11 J. Sirak #12 M. McNair #13 M. Mudrak #14 M. Woodrow
#15 A. Lu #16 T. Obbayi #17 M. Woodrow #18 N. Wilson #19 D. Ivanov #20 R. Wadhera #21 M. Woodrow

1. Suppose A=(a_ij) is a 200 by 200 matrix. All of the entries of A are 0 except
Diagonals Each a_ii=2 for 1<= i<=200.
Weirdos a_ij=10 when i=100 and j=150, and a_ij=x when i=150 and j=30.
What is the determinant of A?
Comment A matrix with relatively few non-zero entries is called a sparse matrix.
Answer
Multiply the 30th row by (-x/2) and add it to the 150th row to get rid of the x. This does not change any of the values on the diagonal. The matrix is now in upper triangular form so multiply along the diagonal to get the determinant. Since every value along the diagonal is still 2, the determinant is 2^200.
Sent by C. Graziadio at 2 Apr 2004 20:20:43

2. Suppose this matrix is in RREF:
(* 7 * * * * * *) (* * 0 1 * * * *) (* * * * 0 0 0 0)
Change each * to one of the following: 0 if the entry must be 0;
1 if it must be 1;
A if it can be anything (0 or not 0: no restrictions).
Explain your answers. What's the rank of this matrix?
Answer
[ 1 7 A 0 A A A A ] [ 0 0 0 1 A A A A ] [ 0 0 0 0 0 0 0 0 ]
Position (1,1) has to be a one because all leading entries in rows in RREF are one.
It also follows that positions (2,1) and (3,1) must be zero because in RREF all entries above and below a leading 1 entry must be zero.
Matrices in RREF also have an upper triangular look to it so, all entries in row 2, before position (2,4) must be zero.
So in row 3, all entries must be zero.
These are the properties of a matrix in RREF, therefore the other entries in rows one and two could be any value A.
The rank of this matrix = 2
Question from the Management
Why precisely do positions (2,2), (2,3), (3,2), and (3,3) have to be 0? Can this be explained in more detail than with the phrase, "an upper triangular look"?
Sent by J. Sirak at Sun, 4 Apr 2004 15:00:20

3. Suppose (1,2,3,0,0) and (0,0,1,1,1) are both solutions of one specific system of 5 linear equations in 5 unknowns.
a) If this system is written as AX=B where A is a matrix of coefficients, X is a column vector of unknowns, and B is a column vector of specific real numbers (you can't deduce much about the entries of A and B from what's given), then what are the dimensions of A and X and B?
b) Does the system AX=0 (here 0 is a column vector of 0's) have any solutions? Be as specific about the solutions as you can.
c) How many solutions does the original system AX=B have? Be as specific about the solutions as you can.
d) What is the determinant of A? What are the possible values of the rank of A?
Answer
The two vectors given will be called X₁=(1,2,3,0,0) and X₂=(0,0,1,1,1).
a) A is 5 by 5 from the fact that it is 5 eqns. in 5 unknowns X is 5 by 1 and B is also 5 by 1.
b) AX=0 has solutions for example A(X₁-X₂)=AX₁-AX₂=B-B=0 thus the vector X*=(1,2,2,-1,-1) is a solution to AX=0.
c) AX=B has infinitely many solutions of the form eX*+X_p where is any real number and Xp is any vector that solves the equation Ax=B. for example since AX*=0 and AX₂=B we have that A(eX*+X₂)=B where again e can be any real number.
d) since A is 5 by 5 and the system AX=B has more then one solution then inverse(A)=A^-1 does not exist because X not equal to A^-1*B. Since A does not have an inverse then detA=0 and therefore the rank of A is less then 5 and greater then zero
Sent by Ben Cohen at Mon, 29 Mar 2004 17:44:46

4. Suppose that A and B are both 2 by 2 matrices which have inverses, and that A^-1=
( 3 -1) (-8 1)
and B^-1=
(-2 1) (-1 7)
If C=A²B, what is the inverse of C?
Comment A small amount of thought can save much irritation.
Answer
C^-1 can be found easily by (B^-1)(A^-1)². Of course, (A^-1)²=
[ 17 -4] [-32 9]
so since (B^-1)=
[-2 1] [-1 7]
we see that (C^-1)=
[ -66 17] [-241 67]
Sent by C. Hunt Thu, 1 Apr 2004 19:28:13
Comment from the Management
I gave a problem much like this on a Math 250 exam. Math 250 is our introductory linear algebra course. I gave the question to see if students would realize that the inverses had to be "reversed", since matrix multiplication is not generally commutative. I did not think the problem was malicious, but believed (hoped?) it would be quick and easy. Instead, here is how about two-thirds of the students did the problem:
An alternate solution

Through row reduction it is found that A=
[-1/5 -1/5] [-8/5 -3/5]
and B=
[-7/13 1/13] [-1/13 2/13]
From these C can be found; C=
[ -67/325 17/325] [-241/325 66/325]
Again with row reduction, C^-1 =
[ -66 17] [-241 67]

Realize (I am sure you do!) that the row reductions take time and attention (it was on an exam!). And there are three of them necessary in this approach. So there are several ways to do this problem, some more efficient than others. Oh well.

5. Suppose that A is a 50 by 70 matrix whose rank is 40. What is the dimension of the subspace S of R^N consisting of the solutions of the homogeneous system AX=0? What is the dimension of the subspace S of R^M consisting of all possible Y's for which the equation AX=Y can be solved? What are N and M?
Answer
A is a 50 by 70 matrix with rank(A) = 40. This tells us that the Reduced Row Echelon Form of A is:
40 30 [ | ] 4 [ I40 | what ] 0 [ | ever ] [__________________|______] 1 [ all 0's ] 0 [ all 0's ]
So, for a homogeneous system Ax = 0, the dimension of the subspace is equal to the rank of A in Rⁿ where n = 70, because A is 50 by 70 so x is 70 by 1. Now, the dimension of the subspace S in R^m for all Y's that satisfy Ax = Y is 30, because it is the number of free variables in A, and m = 50 simply because A is 50 by 70, x is 70 by 1, so Y is 50 by 1 so Y's subspace will sit in R⁵⁰.
Sent by M. Woodrow at Tue Apr 6 15:22:56 2004

6. Find all values of x so that the matrix
(x 0 1 0) (2 0 1 1) (x 0 0 2) (0 1 1 1)
has an inverse.
Answer
A be the given matrix. A is invertible, that is has an inverse, if and only if det A is not 0. Calculating the determinant of A by taking cofactor expressions down the second column for simplicity yields
0 + 0 + 0 + (1)*(-1)^(6)*det (x 1 0 = 2x + -(4 - x) = 3x - 4. 2 1 1 x 0 2)
Therefore, x must equal 4/3, in order for the original matrix to NOT have an inverse.
Sent by X. Sosa at Tue, 06 Apr 2004 16:56:08

7.The matrix
(2 u v w) (0 -1 0 s) (0 0 3 t) (0 0 0 4)
is invertible for all u, v, w, s, and t. What is its inverse?
Comment This matrix is in upper-triangular form with an upper-triangular inverse. First use row operations to make the diagonal elements =1, then pivot and eliminate.
Answer

(2 u v w | 1 0 0 0) (0 -1 0 s | 0 1 0 0) (0 0 3 t | 0 0 1 0) (0 0 0 4 | 0 0 0 1)
Divide First Row by 2 Multiply Second Row by -1
(1 u/2 v/2 w/2 | 1/2 0 0 0) (0 1 0 -s | 0 -1 0 0) (0 0 3 t | 0 0 1 0) (0 0 0 4 | 0 0 0 1)
Divide Third Row by 3 Subtract u/2 * Second Row from First Row
(1 0 v/2 w/2+su/2 | 1/2 u/2 0 0) (0 1 0 -s | 0 -1 0 0) (0 0 1 t/3 | 0 0 1/3 0) (0 0 0 4 | 0 0 0 1)
Divide Fourth Row by 4 Subtract v/2 * Third Row from First Row
(1 0 0 w/2+su/2-vt/6 | 1/2 u/2 -v/6 0 ) (0 1 0 -s | 0 -1 0 0 ) (0 0 1 t/3 | 0 0 1/3 0 ) (0 0 0 1 | 0 0 0 1/4)
Subtract w/2+su/2-vt/6 * Fourth Row from First Row Add s * Fourth Row to Second Row Subtract t/3 * Fourth Row from Third Row
(1 0 0 0 | 1/2 u/2 -v/6 -w/8-su/8+vt/24) (0 1 0 0 | 0 -1 0 s/4 ) (0 0 1 0 | 0 0 1/3 -t/12 ) (0 0 0 1 | 0 0 0 1/4 )
Solution:
(1/2 u/2 -v/6 -w/8-su/8+vt/24) ( 0 -1 0 s/4 ) ( 0 0 1/3 -t/12 ) ( 0 0 0 1/4 )
may check answer by multiplying the original matrix by the solution and you will get the identity matrix of a 4 x 4 matrix
Sent by S. Brynildsen at Wed Mar 31 21:07:20 2004

8. TREE Are (-2,-4,8,-7), (1,2,1,1), and (1,2,-1,2) linearly independent?
Answer
For the vectors (-2,-4,8,-7), (1,2,1,1), and (1,2,-1,2) to be linearly independent, we can look at the equation C₁(-2,-4,8,-7) + C₂(1,2,1,1) + C₃(1,2,-1,2) = 0.
If the vectors are linearly independent, each component of the vector (-2C₁+C₂+C₃, -4C₁+2C₂+2C₃, 8C₁+C₂-C₃, -7C₁+C₂+2C₃) must be zero:
-2C₁ + C₂ + C₃ = 0 -4C₁ +2C₂ +2C₃ = 0 8C₁ + C₂ - C₃ = 0 -7C₁ + C₂ +2C₃ = 0
To solve this, we can write it was a matrix, and look at its reduced form (see SYCAMORE):
(-2 1 1) (1 0 -1/5) (-4 2 2)~(0 1 3/5) ( 8 1 -1) (0 0 0) (-7 1 2) (0 0 0)
This gives us the new equations
C₁-(1/5)C₃ = 0 C₂+(3/5)C₃ = 0
Since there are other equations with all zeros, lets plug in 1 for C₁:1 = (1/5)C₃ and C₂ = -(3/5)(5) We find that C₁ = 1, C₂ = -3 and C₃ = 5, and if we plug these solutions into the original equations, they work.
So since there are non-trivial solutions to this equation, the vectors (-2,-4,8,-7), (1,2,1,1), and (1,2,-1,2) are not linearly independent.
We can also prove this by trying to find one of these vectors as a linear combination of the others: To do this we should look at the reduced form (see HICKORY)
(-2 -4 8 -7) (1 2 0 3/2) ( 1 2 1 1)~(0 0 1 -1/2) ( 1 2 -1 2) (0 0 0 0)
The first 2 vectors are reduced to (1,2,0,3/2) and (0,0,1,-1/2). If we subtract the second one from the first, we get the vector (1,2,-1,2) which is the 3rd. Therefore, the vector (1,2,-1,2) is a linear combination of the first 2 vectors, and these vectors are not linearly independent.
Sent by L. Kohut at Fri, 2 Apr 2004 15:02:52

9. Is the sum of invertible matrices always invertible? Is the sum of singular matrices always singular?
Answer
a) Is the sum of invertible matrices always invertible? No, the sum of invertible matrices is not always invertible. For example when using one by one matrices, A=( 3) and B=(-3)... The sum of these matrices is zero, and zero is not invertible.
b) A square matrix is said to be nonsingular if it has an inverse. If it has no inverse, the matrix is called singular. The answer to this question is no. This can be proved by summing two singular matrices and obtaining an invertible matrix. For example,
(0 0)+(1 0)=(1 0) (0 1) (0 0) (0 1)
Comment from the Management
Of course the invertibility or non-invertibility of such small matrices is easy to check by using the determinant.
Sent by J. Malek at Tue, 6 Apr 2004 19:02:14

10. TREE What is the rank of A=
(0 2 5 2 0) (2 -1 1 0 1) (1 0 2 1 1) (1 -1 -1 -1 0)
? Find a basis for the subspace of all solutions of AX=0 in R^P. What is P?
Answer
By OAK,
[0 2 5 2 0 ] [1 0 0 -3 -3] [2 -1 1 0 1 ]~[0 1 0 -4 -5] [1 0 2 1 1 ] [0 0 1 2 2] [1 -1 -1 -1 1 ] [0 0 0 0 0]
The rank of this matrix = 3.
Sent by J. Sirak at Sun, 4 Apr 2004 16:13:29
Using OAK, the basis for all solutions to Ax = 0 is:
[x₁] [ 3x₄+3x₅] [ 3] [ 3] [x₂] [ 4x₄+5x₅] [ 4] [ 5] x =[x₃]=[-2x₄-2x₅]=x₄[-2]+x₅[-2] [x₄] [ x₄ ] [ 1] [ 0] [x₅] [ x₅ ] [ 0] [ 1]
So the basis for Ax = 0 is (3,4,-2,1,0) and (3,5,-2,0,1) in R^P, where P = 5.
Sent by M. Woodrow at Tue Apr 6 15:22:56 2004

11. Briefly explain why the vectors (3,4,-1), (2,2,1), (1,2,0), and (0,1,-4) in R³ must be linearly dependent. Do this in two ways:

Use a very short argument based on dimension.
Use the fact that a specific homogeneous system must have a non-trivial solution.

Answer
(i) Set up a matrix of the 4 vectors such that you get the following :
[3 2 1 0] A = [4 2 2 1] [4 1 0 -4]
We see that we get a 3 x 4 matrix. In order for the vectors to be linearly independent, rank = # columns = dim(A). In this case, we can easily see that the maximum possible rank of the matrix will be 3. However, the number of columns exceeds this rank. A free variable is present, indicating that that the vectors are in fact linearly dependent ( in this case, in the x4 variable ). In general, when the # columns > # rows , the vectors will be dependent.
(ii) By placing the matrix A into RREF we get the following :
[1 0 0 (-7/6)] RREF of A = [0 1 0 (2/3) ] [0 0 1 (13/6)]
Then solving the homongeneous system. We get the following 3 equations :
x₁ = (7/6) x₄ x₂ = (-2/3) x₄ x₃ = (-13/6) x₄ x₄ = x₄
So a general solution can be the following :
[ 7/6] [ -2/3] [-13/6] x₄ [ 1 ]
where x₄ can be any value and hence there are non-trivial solutions.
Sent by J. Sirak at Wed, 31 Mar 2004 13:56:10
Comment from the Management
These solutions are very good. Perhaps they can be written a bit more efficiently. For example:
i) R³ has dimension 3, therefore the maximum number of linearly independent vectors in R³ is 3. Here we have 4 vectors, and (4>3) therefore the vectors must be linearly rdependent.
ii) If we express the linear independence condition as a homogeneous system of linear equations, we will have 3 equations with 4 variables. If the number of variables is greater than the number of equations, then we have seen that there is a non-trivial solution.
Further note
Ms. Sirak's solutions are actually much the same as mine, and in some ways, her solutions are morally superior (!) because she finds explicit linear combinations and she also finds explicit non-trivial solutions of homogeneous systems, whereas I just assert that, using results in class, we can do this. Certainly I would prefer my approach, a qualitative one, if I had 150 vectors in R¹³⁰, but many people would like hers better.

12. TREE Verify that (1,2,0,2,1) is in the linear span of (1,1,0,0,2) and (2,1,2,-2,1) and (4,2,5,-4,0).
Answer
To see that (1,2,0,2,1) is in the linear span of (1,1,0,0,2) (2,1,2,-2,1) and (4,2,5, -4,0), look at the tree corresponding to BIRCH. In RREF the matrix has a row of 0's on the bottom where (1,2,0,2,1) was positioned, which means that the vector is linearly dependent on the other 3 vectors. To verify this you can find the 3 constants multiplied against the 3 spanning vectors of R^5 to form the vector under question.
1 = A + 2B + 4C 2 = A + B + 2C 0 = 2B + 5C 2 = -2B - 4C 1 = 2A + B
Then
(1-B)/2 = A (-2/5*B) = C
Substitute A and C into the first equation to get a value for B
1=(1-B)/2 + 2B +4(-2/5*B) B=-5 A=3 C=2
Which proves (1,2,0,2,1) is a linear combination and in the span of (1,1,0,0,2) (2,1,2, -2,1) and (4,2,5, -4,0).
Sent by M. McNair at Tue, 30 Mar 2004 17:41:38
Comment from the Management
The problem could also be done using PINE. This RREF shows that the rank of the system x₁(1,1,0,0,2)+x₂(2,1,2, -2,1)+x₃(4,2,5, -4,0)+x₄(1,2,0,2,1)=0 is 3. It also shows that this system is row equivalent to
x₁+3x₄=0 x₂-5x₄=0 x₃+2x₄=0
so that if x₄=1, we will get the vector as a linear combination of the other vectors, with coefficients -3 and 5 and -2. These are x₄, so writing the vector as a linear combination means adjusting all the signs.
I think Mr. McNair's solution is fine. I just wanted to show an alternate method.

13. Suppose A is a 3 by 3 matrix, and det(A)=2. Compute det(A⁵) and det(5A) and det(A^-1) and det(AA^t).
Answer
Since det(A^5) is equivalent to det(A*A*A*A*A) which equals det(A)*det (A)*det(A)*det(A)*det(A), det(A^5)=2^5=32.
Multiplying A by 5 really means multiplying by
(5 0 0) (0 5 0) (0 0 5)
and the determinant of this matrix is 125, so det(5A)=125*2=250
det(A^-1)=1/2 because A*A^-1=I and taking det of both sides gives detA*det(A^-1)=detI, where detI=1.
Lastly, since detA=det(A^t), det(AA^t)=2*2=4
Sent by Michele Mudrak at Mon, 29 Mar 2004 00:13:48

14. Be prepared to discuss the meaning of
homogeneous ... inhomogeneous ... basis ... linear independence ... linear combination ... subspace ... dimension ... spanning ... rank ... eigenvalue ... eigenvector ... transpose ... matrix addition and multiplication ... symmetric ... diagonalization ... life.
Answers

homogeneous
A system of linear equations AX=B, where A is an n by m matrix of coefficients and X is an m by 1 matrix of unknowns and B is an n by matrix of constants is called homogeneous if B is equal to the zero vector.
inhomogeneous
A system of linear equations AX=B, where A is an n by m matrix of coefficients and X is an m by 1 matrix of unknowns and B is an n by matrix of constants is called inhomogeneous if B is not equal to the zero vector.
basis
A collection of vectors v₁, v₂, ..., v_t in a subspace S of Rⁿ is called a basis if v₁, v₂, ..., v_t are linearly independent and also span S.
linear independence
A collection of vectors v₁, v₂, ..., v_t in Rⁿ is called linearly independent if none of the vectors is a linear combination of the others.
linear combination
A linear combination of a collection of vectors v₁, v₂, ..., v_t is n₁v₁ + n₂v₂ + ...+ n_tv_t where each n_j is a scalar (real or complex number).
subspace
A collection of vectors S in Rⁿ is called a subspace if 1) The zero vector is in S; 2) The sum of any vectors in S is in S; 3) The product of any vector in S with any real number is also in S.
dimension
If S is a subspace of Rⁿ, then the dimension of S is equal to the number of vectors in any basis for the subspace.
spanning
A collection of vectors v₁, v₂, ..., v_t in a subspace S of Rⁿ is called a spanning set if every vector in S is a linear combination of v₁, v₂, ..., v_t.
rank
If A is an n by m matrix, then the rank of A is the number of nonzero rows in A_R where A_R is the row echelon form of A.
eigenvalue
If A is an n by n matrix, then a number is an eigenvalue of A if there is a nonzero n by 1 matrix (vector) X such that AX=X.
eigenvector
If A is an n by n matrix, then a vector v in Rⁿ is an eigenvector of A if X is not the zero vector and AX=X for some number . X is called an eigenvector associated to the eigenvalue .
transpose
If A is an n by m matrix, then a matrix B is the transpose of A, written A^t, if B = [a_ji].
matrix addition and multiplication
If both A and B are n by m matrices, then A+B, the result of matrix addition is n n by m matrix equal to [a_ij + b_ij].
If A is an n by m matrix, and B is an m by p matrix, then AB, the result of matrix multiplication is an n by p matrix equal to [SUM a_ikb_kj ; from k=1 to m].

symmetric
A matrix A is symmetric if A = A^t.
diagonalization
Suppose A is an n by n matrix. Then a diagonalization of A is P^-1AP if there even exists an n by n nonsingular matrix P making this product a diagonal matrix.
Sent by M.Woodrow at Wed, 31 Mar 2004 21:52:10
Comment from the Management
What a lot of work!

15. TREE Consider the system
3x₁+5x₂-1x₃=a 2x₁+6x₂+1x₃=b 4x₁+4x₂-3x₃=c
a) Find a specific vector (a,b,c) in R³ so that the system has no solution. Explain.
b) Find a specific vector (a,b,c) in R³ so that the system has at least one solution. Find all solutions for that vector. Explain.
Answer
I) rewrite the equations into an augmented matrix
(3 5 -1 : a) (2 6 1 : b) (4 4 -3 : c)
II)RREF of the augmented matrix (see WILLOW)=
(1 0 -11/8 :3a/4-5b/8) (0 1 5/8 :-a/4+3b/8 ) (0 0 0 :-2a+b+c)
III)
X(1)-11/8X(3)= 3a/4-5b/8 X(2)+5/8X(3)= -a/4+3b/8 0= -2a+b+c
a) a specific vector (a,b,c) in R^3 so that the system has no solution is when -2a+b+c does not equal to zero (the system is inconsistent): (a,b,c)=(2,1,1).
b) a specific vector (a,b,c) in R^3 for the system to have at least one solution occurs when 0= -2a+b+c, the system is consistent: (a,b,c)=(1,1,1).
Sent by A. Lu at Tue, 30 Mar 2004 19:29:07

16. Suppose A is a 4 by 4 matrix whose inverse is
(2 3 1 -1) (1 1 1 0) (2 0 2 3) (4 1 2 2)
a) Find all solutions of AX=B if B=(2,1,1,0).
b) Find all solutions of the homogeneous system AX=0.
Answer
X = A^-1*B
(2 3 1 -1) (2) ( 8) X = (1 1 1 0) (1) = ( 4) (2 0 2 3) (1) ( 6) (4 1 2 2) (0) (11)
For number 16B, A^-1*0 = 0
Sent by T. Obbayi at Tue, 06 Apr 2004 17:20:37
Comment from the Management
Of course there is another way to do this problem. That is, you could find A if you wished. I have been informed that A must be
(-1/7 -1/7 -3/7 4/7) ( 6/7 -8/7 4/7 -3/7) (-5/7 16/7 -1/7 -1/7) ( 4/7 -10/7 5/7 -2/7)
and you could then write the system explicitly as AX=B. Of course, to get a solution you'd then turn around and compute A^-1! So why bother finding A when you have what you need already!

17. Give an example of a specific matrix which has exactly 20 non-zero entries and whose characteristic polynomial is (L-5)³(L+2)⁵(L²+4). Answer
The characteristic polynomial tells us:
1) our matrix is 10 by 10 because the polynomial is degree 10
2) the diagonal entries; three terms are 5, five terms are -2, and the other two are +/-2i
so the matrix, with 20 non-zero entries, will look like:
[5 1 1 1 1 1 1 1 1 1] [0 5 0 0 0 0 0 0 0 0] [0 0 5 0 0 0 1 0 0 0] [0 0 0 -2 0 0 0 0 0 0] [0 0 0 0 -2 0 0 0 0 0] [0 0 0 0 0 -2 0 0 0 0] [0 0 0 0 0 0 -2 0 0 0] [0 0 0 0 0 0 0 -2 0 0] [0 0 0 0 0 0 0 0 2i 0] [0 0 0 0 0 0 0 0 0 -2i]

Sent by M. Woodrow at Tue Apr 6 15:22:56 2004
Comment from the Management
Of course there are many other answers to this question. I could have made the question a bit more difficult by requesting a matrix with only real entries. How could we get that? In class we considered the 2 by 2 example
( 0 1) (-1 0)
which had +/-i eigenvalues. Using this, we could modify the previous answer to the following:
[5 1 1 1 1 1 1 1 1 1] [0 5 0 0 0 0 0 0 0 0] [0 0 5 0 0 0 1 0 0 0] [0 0 0 -2 0 0 0 0 0 0] [0 0 0 0 -2 0 0 0 0 0] [0 0 0 0 0 -2 0 0 0 0] [0 0 0 0 0 0 -2 0 0 0] [0 0 0 0 0 0 0 -2 0 0] [0 0 0 0 0 0 0 0 0 2] [0 0 0 0 0 0 0 0 -2 0]

>
18. TREE a) Explain why the vectors (2,-3,5), (-7,1,-2), and (3,-5,7) are a basis of R³.
b) Write a linear combination of the vectors in a) whose sum is equal to the vector (2,1,3). Answer
a) Looking at ASH, we see that the rref of the matrix consisting of the three vectors as its columns is I₃. Therefore, the vectors are linearly independent. Clearly, the vectors also span R³ since any vector in R³ can be written as a combination of i,j,k (unit vectors). Since the vectors span R³ and are linearly independent, they form a basis of R³.
b) In order for (2,1,3) to be a linear combination of the three vectors from part a, we must find the coefficients such that x₁(2,- 3,5)+x₂(-7,1,-2)+x₃(3,-5,7)=(2,1,3). We can solve this as the augmented matrix ASH with (a,b,c) replaced by (2,1,3). Then we get x₁=133/25 (note that the coefficient of the b term should be positive), x₂=-6/25, x₃=-86/25.
Sent by N. Wilson at Mon, 5 Apr 2004 15:06:41

19. a) Diagonalize A=
(2 2) (1 1)
That is, find the eigenvalues and eigenvectors of A, and find a matrix T so that TAT^-1 is diagonal.
b) 3⁵=243. Use this fact and the diagonalization of A found in a) to compute A⁵.
Answer
a) First find eigenvalues by setting det(A-*I₂)=0, where I₂ is a 2x2 identity matrix and is the eigenvalue. This is equivalent to setting the determinant of C=
( (2-) 2 ) ( 1 (1-) )
to zero. This characteristic polynomial therefore is (2-)(1-)-2=0, or eigenvalues ² -3=0. The eigenvalues are =3 and =0.
Now find eigenvectors that correspond to each eigenvalue by plugging values into C and solving C*X=0 for non-zero X, where X is a vector in R². For =0:
0=( 2 2 ) (x₁) ( 1 1 ) (x₂)
Standard procedure is to reduce matrix C, so reduced row echelon form of C is
(1 1) (0 0)
from which X can be determined:
(x₁)=x₂(-1) (x₂) ( 1)
Therefore (-1, 1) is one of the eigenvectors of A that corresponds to the eigenvalue 0. Similarly, (2,1) is one of the eigenvectors of A that corresponds to eigenvalue 3. Defining D=
(0 0) (0 3)
to be the diagonal matrix, one must define T =
(-1 2) ( 1 1)
Since the first column of D contains the eigenvalue 0, the eigenvector corresponding to eigenvalue 0 must be the first column of T and because the eigenvalue 3 is in the second column of the diagonal matrix D, the eigenvector corrsponding to eigenvalue 3 must be in the second column of T.
To find the inverse of T, reduce (T | I₂):
(-1 2 | 1 0)~(1 0 | -1/3 2/3 ) (1 1 | 0 1) (0 1 | 1/3 1/3 )
The inverse of T is T^-1=
(-1/3 2/3) ( 1/3 1/3)
One can verify that indeed, D=T^-1*A*T
b) Since D=T^-1*A*T, A=T*D*T^-1 so that Aⁿ = (T*D*T^-1)ⁿ = T*(Dⁿ)*T^-1. Therefore A⁵ = T*(D⁵)*T^-1, and D⁵ =
(0 0) (0 3⁵)
then (D⁵)*T^-1=B=
( 0 0 ) ( 81 81 )
and T*B=
([2/3]*3⁵ [2/3]*3⁵)=(162 162) ([1/3]*3⁵ [1/3]*3⁵) ( 81 81)
Which is the same result maple gives when told to evaluate A⁵.
Sent by D. Ivanov at Thu, 1 Apr 2004 00:32:20

20. If A=
(2 1 1 0) (1 1 0 0) (1 0 1 0) (0 0 0 -2)
then A has eigenvalues -2, 0, 1, and 3 with associated eigenvectors (0,0,0,1), (-1,1,1,0), (0,-1,1,0), and (2,1,1,0), respectively. Find an orthogonal matrix W so that WAW^-1 is diagonal.
Comment from the Management
The Management made an error. I have generally been writing W^-1AC=the diagonal matrix. So what I asked for here is the inverse of the matrix constructed before. I am sorry. This was not intentional.
Answer
All of the eigenvectors are already orthogonal considering that the dot product of any two of them is 0. After much frustration from looking at the book, and the class notes, I noticed that the book, and your procedure gave the matrix B so that inv(B)*A*B=the diagonalized matrix. However this problem wants W*A*inv(W), and recalling last lecture's QotD, AB does not necessarily equal BA. So, normally we would make the matrix: B =
[0 -1 0 2] [0 1 -1 1] [0 1 1 1] [1 0 0 0]
which arose from the eigenvectors, then normalize it by dividing each column by it's length (ex. B(i,1)/sqrt(B(1,1)^2 + B(2,1,)^2 + B(3,1)^2 + B(4,1)^2)) ) this in turn would give us a diagonalized answer for inv(B)*A*B...BUT we want W*A*inv(W).
Yeah, at this point, I just went to your office hours, and the part that asks for W*A*inv(W) was a typo; it should be inv(W)*A*W. So the answer is W =
[0 -1/sqrt(3) 0 2/sqrt(6)] [0 1/sqrt(3) -1/sqrt(2) 1/sqrt(6)] [0 1/sqrt(3) 1/sqrt(2) 1/sqrt(6)] [1 0 0 0 ]

Sent by R. Wadhera (The h is silent...but deadly) at Sun, 4 Apr 2004 17:02:28

21. Explain why A=
( 0 0) (231 0)
can't be diagonalized: that is, there is no invertible matrix T so that TAT^-1 is diagonal.
Answer
First solve:
[- 0] det(A-I)=det[231-]=²
So the only eigenvalue of A is =0 Now find the associated eigenvectors by solving Ax=x which yields:
[ 0 0][x₁] [0] [231 0][x₂]=[0]
from this we find x₁ = 0 and x₂ is free so: (x₁,x₂)=(0,x₂)=x₂(0,1).
Our associated eigenvector when =0 is (0,1), BUT in order for A, a 2 by 2 matrix, to be diagonalizable it must have two linearly independent eigenvectors so A cannot be diagonalized.
Sent by M. Woodrow at Tue Apr 6 15:22:56 2004

Students' answers to review problems for the second exam in Math 421, spring 2004

Students' answers to review problems for the second exam
in Math 421, spring 2004