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Column Space

Module by: Doug Daniels, Steven J. Cox. E-mail the authors

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Summary: This module defines precisely what a column space is, gives an example of one, and then a method for finding one given an arbitrary matrix.

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The Column Space

We begin with the simple geometric interpretation of matrix-vector multiplication. Namely, the multiplication of the n-by-1 vector xx by the m-by-n matrix A A produces a linear combination of the columns of A A. More precisely, if a j a j denotes the j jth column of A A, then

Ax=a1a2…anx1x2…xn=x1a1+x2a2+…+xnan A x a1 a2 … an x1 x2 … xn x1 a1 x2 a2 … xn an (1)
The picture that I wish to place in your mind's eye is that Ax A x lies in the subspace spanned by the columns of A A. This subspace occurs so frequently that we find it useful to distinguish it with a definition.

Definition 1: Column Space
The column space of the m-by-n matrix S is simply the span of the its columns, i.e. RaS≡ {Sx|x∈ R n } Ra S S x x R n . This is a subspace of ℜ m ℜ m . The notation Ra Ra stands for range in this context.

Example

Let us examine the matrix:

A=0100-10100001 A 0100 -1010 0001 (2)
The column space of this matrix is:
RaA={ x 1 0-10+ x 2 100+ x 3 010+ x 4 001|x∈ ℝ 4 } Ra A x ℝ 4 x 1 0 -1 0 x 2 1 0 0 x 3 0 1 0 x 4 0 0 1 (3)

As the third column is simply a multiple of the first, we may write:

RaA={ x 1 010+ x 2 100+ x 3 001|x∈ ℝ 3 } Ra A x ℝ 3 x 1 0 1 0 x 2 1 0 0 x 3 0 0 1 (4)

As the three remaining columns are linearly independent we may go no further. In this case, RaA Ra A comprises all of ℝ 3 ℝ 3 .

Method for Finding a Basis

To determine the basis for RaA Ra A (where A A is an arbitrary matrix) we must find a way to discard its dependent columns. In the example above, it was easy to see that columns 1 and 3 were colinear. We seek, of course, a more systematic means of uncovering these, and perhaps other less obvious, dependencies. Such dependencies are more easily discerned from the row reduced form. In the reduction of the above problem, we come very easily to the matrix

A red =-101001000001 A red -1010 0100 0001 (5)
Once we have done this, we can recognize that the pivot column are the linearly independent columns of A red A red . One now asks how this might help us distinguish the independent columns of A A. For, although the rows of A red A red are linear combinations of the rows of A A, no such thing is true with respect to the columns. The answer is: pay attention to the indices of the pivot columns. In our example, columns {1, 2, 4} are the pivot columns of A red A red and hence the first, second, and fourth columns of A A, i.e.,
0-10100001 0 -1 0 1 0 0 0 0 1 (6)
comprise a basis for RaA Ra A . In general:

Definition 2: A Basis for the Column Space
Suppose A A is m-by-n. If columns { c j |j= 1 , ... , r } j 1 , ... , r c j are the pivot columns of A red A red then columns { c j |j= 1 , ... , r } j 1 , ... , r c j of A A constitute a basis for RaA Ra A .

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