Lecture 4
Span of a Set of Vectors

We have already considered linear combinations of a fixed collection of vectors. It is useful to consider ALL such linear combinations, that is, all possible choices of coefficients for the combinations.

Def. For a nonempty set S = {u₁, . . . , u_k} of vectors in Rⁿ, the span of S is the set of all linear combinations of u₁, . . ., u_k in Rⁿ (in other words, ALL vectors of the form c₁ u₁ + · · · + c_k u_k for any choice of coefficients).
This span is denoted span S = span{u₁, . . ., u_k}.

Note that the span of 0 is {0}, while the span of any other single vector consists of infinitely many vectors (all multiples of the fixed one). Geometrically, in R², the span of any nonzero vector is the line through that vector. The span of two nonparallel vectors in R² is all of R².

When is a given vector in the span of a given set of vectors? We have already seen that a column vector of length n is a sum of multiples of the columns of an m x n matrix if and only if the corresponding linear system has a solution.

Theorem Let S = {u₁, . . ., u_m} be a set of vectors in Rⁿ, and let A be the matrix with these vectors as columns. Then v is in the span of S if and only if the equation Ax= v is consistent.

Examples

Example: check the statement about two nonparallel vectors in R² spanning all of R², as an application of this method.

Def. If V is the subset of Rⁿ which is the span of the set of vectors S in Rⁿ, then we say that V is the span of S (and write V = span(S)), and S spans V.

Example: find the span of a pair of vectors in R³.

How can we determine whether all of R^m is the span of a given set of vectors? Form the matrix with these vectors as its columns, and use what we already know,

Theorem The following statements about an m x n matrix A are equivalent.
(a) The span of the columns of A is R^m.
(b) The equation A x = b is consistent for every b in R^m.
(c) The rank of A is m.
d) The reduced row echelon form of A has no zero rows.

Notice that (c) and (d) are clearly equivalent (since A has m rows, and the rank is the number of nonzero rows in row echelon form), and these are the easiest conditions to check. Also, (a) and (b) are clearly equivalent, by the definition of "span" and the meaning of consistency. We already know that (b) and (d) are equivalent, since if there is no zero row then we know that the equations are consistent regardless of the right hand side, and if there is a zero row then we can choose a right hand side which has a nonzero entry in that row (and for which there is then no solution to the corresponding equation).

What is true about the span of a set of vectors S in Rⁿ, from an algebraic point of view? In many ways, even if this span is not all of Rⁿ, it has very similar properties. In particular, if you add two vectors in the span of S, or take a scalar multiple of a vector in the span of S, the result is still in the span of S. By combining these statements repeatedly, we see that the span of any collection of vectors in the span of S is still in the span of S.

Theorem For any finite subset S of Rⁿ, the following statements are true.
(a) S is contained in span(S).
(b) If S' is any finite set contained in span(S), then span(S') is also contained in span(S)
(c) For any vector z in Rⁿ, span(S) = span(S È {z}) if and only if z is in the span of S.

The first statement is clear, and the second statement is a summary of what we discussed above. For the third statement, if z is not in the span of S, then, since z is in span (S È, {z})) the latter clearly is different from span(S). On the other hand, if z is in span(S), then part (b) implies that span(S È{z}) is contained in span(S), and the fact that span(S) is contained in span(S È {z}) is obvious.

250 Lecture Index