Row Space, Column Space, Null Space

Definition

Let

• $A$ : $m\times n$ matrix
• $U$ : Subspace of $R^n$
• $V$ : Subspace of $R^m$

If $U=span[R(A)]$ then we call $R(A)$ row space of $A$

If $V=\mathrm{span}[C(A)]$ then we call $C(A)$ column space of $A$

If $N(A)$ : Solution space of $A\mathbf{x}=0$ (which is subspace of $R^n$) then we call $N(A)$ null space of $A$

Interpretation

By definition of span, any linear combination of columns of $A$ is in the column space of $A$. For example, if $\{{\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3\}$ column vectors of $A$, then ${\mathbf{a}}_1+3{\mathbf{a}}_2-4{\mathbf{a}}_3\in C(A)$.

ANY linear combination of columns of $A$, is in the column space of $A$.

And ANY linear combination of rows of $A$, is in the row space of $A$.

Finding basis of column/row space

Let $A$ : $m\times n$ matrix

To find basis of $C(A)$:

1. Obtain RREF of $A$
2. Find pivot column of $A$

To find basis of $R(A)$:

1. Obtain RREF of $A^T$
2. Find pivot column of $A^T$

These pivot columns obtained from $A$ or $A^T$ form the basis for $C(A)$ and $R(A)$ respectively.

Example: Checking if a vector is in column space

Let

$$A=\left[\begin{array}{ccccc}2& 1& 0& 5& 2\\ 0& -2& 0& -2& 0\\ 1& 2& 2& 4& 0\\ 0& -4& -2& -4& 1\end{array}\right]$$

Then,

$$C(A)=\mathrm{span}\left\{\left[\begin{array}{c}2\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}1\\ -2\\ 2\\ -4\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 2\\ -2\end{array}\right],\left[\begin{array}{c}5\\ -2\\ 4\\ -4\end{array}\right],\left[\begin{array}{c}2\\ 0\\ 0\\ 1\end{array}\right]\right\}$$

Suppose we have $\mathbf{v}={\left[\begin{array}{cccc}2& -1& 3& -4\end{array}\right]}^T$.

To check if $\mathbf{v}\in C(A)$, we have to check if there exists a solution $\mathbf{x}$ such that the linear combination $A\mathbf{x}=\mathbf{v}$ holds true.

Note

Basically, we’re trying to see if $\mathbf{v}$ is in the span of column vectors of $A$. Check the definitions.

So,

$$\begin{array}{rl}A\mathbf{x}& =\mathbf{v}\\ \left[\begin{array}{ccccc}2& 1& 0& 5& 2\\ 0& -2& 0& -2& 0\\ 1& 2& 2& 4& 0\\ 0& -4& -2& -4& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\right]& =\left[\begin{array}{c}2\\ -1\\ 3\\ -4\end{array}\right]\end{array}$$

Using Gaussian Elimination, we end up with

$$\left[\begin{array}{ccccc}1& \frac{1}{2}& 0& \frac{5}{2}& 1\\ 0& 1& 0& 1& 0\\ 0& 0& 1& 0& -\frac{1}{2}\\ 0& 0& 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\right]=\left[\begin{array}{c}1\\ \frac{1}{2}\\ \frac{5}{8}\\ -\frac{6}{8}\end{array}\right]$$

On the fourth row of $A$, we found that $0=-\frac{6}{8}$, which is a contradiction.

Thus, $A\mathbf{x}\ne \mathbf{v}$. Consequently, $\mathbf{v}$ is not in the column space of $A$.

Example: Finding basis of column/row space

Let

$$A=\left[\begin{array}{ccc}1& 2& 3\\ 2& 4& 5\\ 1& 2& 2\end{array}\right]$$

The RREF of $A$ and $A^T$ is:

$$\begin{array}{rl}\mathrm{RREF}(A)& =\left[\begin{array}{ccc}1& 2& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]\\ \mathrm{RREF}(A^T)& =\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]\end{array}$$

According to row space,

• $C(A)=\mathrm{span}\{[1,2,1{]}^T,[3,5,2{]}^T\}$
• $R(A)=\mathrm{span}\{[1,2,3],[2,4,5]\}$

Example: Finding null space

i.e., finding the solution space of $A^T$:

$$A^T=\left[\begin{array}{ccc}1& 2& 1\\ 2& 4& 2\\ 3& 5& 2\end{array}\right]$$

Row reduce $A^T$ to RREF:

$$\left[\begin{array}{ccc}1& 2& 1\\ 2& 4& 2\\ 3& 5& 2\end{array}\right]\stackrel{R_2-2R_1}{\to }\left[\begin{array}{ccc}1& 2& 1\\ 0& 0& 0\\ 3& 5& 2\end{array}\right]\stackrel{R_3-3R_1}{\to }\left[\begin{array}{ccc}1& 2& 1\\ 0& 0& 0\\ 0& -1& -1\end{array}\right]$$ $$\stackrel{R_2\leftrightarrow R_3}{\to }\left[\begin{array}{ccc}1& 2& 1\\ 0& -1& -1\\ 0& 0& 0\end{array}\right]\stackrel{-R_2}{\to }\left[\begin{array}{ccc}1& 2& 1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]\stackrel{R_1-2R_2}{\to }\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]$$

Columns 1 and 2 have pivots (basic variables x₁, x₂). Column 3 has no pivot (free variable x₃).

Set x₃ = 1:

• x₁ - 1 = 0 → x₁ = 1
• x₂ + 1 = 0 → x₂ = -1

Solution space:

$$\text{null}(A^T)=\text{span}\left\{\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]\right\}$$