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2.F Row space and column space

In this section we learn a quick way of finding a basis of a subspace in $F^{n}$ , which can also be used as a time-efficient test for linear independence. The trick is to use matrices instead of systems of linear equations (which should remind you of MATH105).

The following definitions consider the rows and columns of a matrix as vectors in $F^{n}$ , by interpreting the entries as coordinates for the standard basis.

Definition 2.46:

Let $A\in\operatorname{M}_{{n\times m}}({F})$ be a matrix.

•

The row space of $A$ is the subspace in $F^{m}$ spanned by its rows,
•

The column space of $A$ is the subspace in $F^{n}$ spanned by its columns.

Example 2.47.

Consider the matrix $A:=\begin{bmatrix}1&2&3\\ 4&5&6\end{bmatrix}\in\operatorname{M}_{{2\times 3}}({{\mathbb{R}}})$ . Then the row space of $A$ is

\operatorname{span}_{{\mathbb{R}}}\{(1,2,3),(4,5,6)\}\subset{\mathbb{R}}^{3}.

The column space of $A$ is

\operatorname{span}_{{\mathbb{R}}}\{(1,4),(2,5),(3,6)\}\subset{\mathbb{R}}^{2}.

Theorem 2.48.

Let $A\in\operatorname{M}_{{n\times m}}({F})$ . Then

i.

Row operations don’t change the row space of $A$ ,
ii.

Column operations don’t change the column space of $A$ .

Proof.

One proves this by taking each type of e.r.o. separately, and assuming one has a general matrix, and a general e.r.o. of that type. Then one needs to prove that each new row (after the e.r.o.) is a linear combination of the old rows (before the e.r.o.). Argue similarly for e.c.o.’s. ∎

Theorem 2.49.

Let $A\in\operatorname{M}_{{n\times m}}({F})$ , and assume $A_{r}$ is an echelon form of $A$ (see Definition 1.9(iii)). Then the non-zero rows of $A_{r}$ form a basis for the row space of $A$ .

Proof.

We obtain $A_{r}$ from $A$ through a sequence of e.r.o.’s, so by Theorem 2.48, $A$ and $A_{r}$ must have equal row spaces. Therefore the non-zero rows of $A_{r}$ span the row space of $A$ . To prove they form a basis, we need to prove linearly independence.

Let $\vec{v_{1}},\cdots,\vec{v_{k}}$ be the non-zero rows of $A_{r}$ , and assume $\sum\alpha_{i}\vec{v_{i}}=\vec{0}$ . Since $A_{r}$ is in echelon form, the left-most non-zero coordinate in $\vec{v_{1}}$ is zero for all the other $\vec{v_{i}}$ . Therefore $\alpha_{1}=0$ . Similarly, since $A_{r}$ is in echelon form, the left-most coordinate in $\vec{v_{2}}$ is zero for all the other $\vec{v_{i}}$ , $i\geq 3$ ; hence $\alpha_{2}=0$ . Continuing in this way (i.e. by induction), we see $\alpha_{i}=0$ for all $i$ . Therefore the sequence $\vec{v_{1}},\cdots,\vec{v_{k}}$ is linearly independent. ∎

Notice that the matrix $A_{r}$ in the above Theorem does not need to be reduced row echelon form; so there are multiple correct bases.

Exercise 2.50:

Let $A=\begin{bmatrix}-3&1&3\\ 1&1&1\\ -2&0&1\end{bmatrix}\in\operatorname{M}_{{3}}({{\mathbb{R}}})$ . If $W\subset\mathbb{R}^{3}$ is the row space of $A$ , find a basis for $W$ . [Hint: Use Theorem 2.48 and Theorem 2.49.]

[End of Exercise]

Example 2.51.

Find a basis for the row space of $\begin{bmatrix}1&1&-2\\ 2&1&-3\\ -1&0&1\\ 0&1&-1\end{bmatrix}\in\operatorname{M}_{{4\times 3}}({\mathbb{R}})$ .

Solution 1: The reduced row echelon form is $\begin{bmatrix}1&0&-1\\ 0&1&-1\\ 0&0&0\\ 0&0&0\end{bmatrix}$ . By Theorem 2.49 a basis for this subspace is $(1,0,-1),(0,1,-1)$ ; in particular, it is two dimensional.

Solution 2: We could have instead used the algorithm from Theorem 2.36, but it takes a bit longer. That procedure results in $(1,1,-2),(2,1,-3)$ for a basis of the row space; these are the first two rows of the matrix.

Theorem 2.52.

Let $\vec{v_{1}},\cdots,\vec{v_{r}}\subset F^{n}$ be a sequence of vectors. Let $A\in\operatorname{M}_{{r\times n}}({F})$ be the matrix whose rows are the vectors in the sequence, and let $A_{r}$ be an echelon form of $A$ . The sequence is linearly independent if and only if $A_{r}$ has no zero rows.

Proof.

Follows directly from Theorem 2.49. ∎

Example 2.53.

Is the following sequence of vectors linearly independent in ${\mathbb{R}}^{4}$ ?

(3,1,0,-1),(2,1,1,1),(-1,1,-1,-8),(1,0,0,1).

Solution: Form the matrix of row vectors, and row reduce it.

\begin{bmatrix}3&1&0&-1\\ 2&1&1&1\\ -1&1&-1&-8\\ 1&0&0&1\end{bmatrix}\rightarrow\cdots\rightarrow\begin{bmatrix}1&0&0&1\\ 0&1&0&-4\\ 0&0&1&3\\ 0&0&0&0\end{bmatrix}.

There is a zero row, so by Theorem 2.52 the original sequence is linearly dependent. Another way to see this is to use Theorem 2.49, which shows the subspace spanned by these 4 vectors is only 3 dimensional, therefore they must be linearly dependent.