4 Linear algebra

4.17 Column space basis

We can now calculate a basis for the nullspace of a m×n matrix A by putting it into RREF and reading off the fundamental solutions to A𝐱=𝟎. In this section we consider the problem of finding a basis of the column space C(A), which we defined in Definition 3.6.2 to be the span of the columns of A.

By Proposition 3.2.1, the set of vectors A𝐯 is exactly the set of linear combinations of the columns of A. Therefore the column space C(A) is equal to the image of TA:𝔽n𝔽m given by TA(𝐱)=A𝐱.

It would be nice to be able to solve this problem using RREF, because it is very easy to find a basis for the column space of a RREF matrix like

R=(012030001400000).

The columns containing a leading entry, in this example columns 2 and 4, are easily seen to be a basis for the column space of R. Unfortunately doing row operations can change the column space of a matrix, so knowing the column space of R does not immediately give you the column space of A.

One solution for this would be to introduce column operations and column reduced echelon form, and re-prove all the things about row operations and row reduced echelon form. Instead we are going to stick with the row operations we already know and use the transpose to convert columns into rows.

We defined the column space of a matrix as the span of its columns. The row space is defined similarly.

Definition 4.17.1.

Let A be a m×n matrix. The row space of A is defined to be the span of the rows of A.

Theorem 4.17.1.

Let A be m×n, let E be a m×m invertible matrix, and let F be a n×n invertible matrix. Then the row space of EA equals the row space of A and the column space of AF equals the column space of A.

Proof.

We will do the second part only as the first one can be proved similarly. By Corollary 3.2.4, the columns of AF are linear combinations of the columns of A, that is, elements of the subspace C(A). The span C(AF) of the columns of AF is therefore also contained in C(A).

Applying the same argument again with AF in place of A and F1 in place of F, the column space C(AFF1), that is, the column space of C(A), is contained in C(AF). ∎

Since doing a row operation to a matrix is the same as left multiplication by an elementary matrix (Theorem 3.8.1), this shows that doing row operations to a matrix doesn’t change its row space.

Theorem 4.17.2.

Let R be a m×n RREF matrix. Then the nonzero rows of R are a basis for the row space of R.

Proof.

Certainly the nonzero rows span the row space, so we only need show they are linearly independent. Let the nonzero rows be 𝐫1,,𝐫l, and let the leading entry in row i occur in column ci. Suppose i=1lai𝐫i=𝟎. Pick any 1il and consider the entry in column ci of this sum. On the right we have 0. On the left ai𝐫i has a ai in column ci, and all the other 𝐫js have zeros in column ci because R is in RREF. Thus we have ai=0 for 1il, so the rows are linearly independent. ∎

The columns of A are the transposes of the rows of AT, so we can get a basis for the column space of A by forming the matrix AT, doing row operations until we reach a RREF matrix, then taking the transposes of the nonzero rows of this RREF matrix.

Example 4.17.1.

Let A=(123456789). To find a basis of C(A) we take the transpose of A to get

AT=(147258369)

Doing row operations, we reach the RREF matrix

R=(101012000).

The nonzero rows (101) and (012) are a basis for the row space of R, which equals the row space of AT, so their transposes

(101),(012)

are a basis for the column space of A.