# Find a Basis For the Null Space of a Given $2\times 3$ Matrix

## Problem 132

Let
$A=\begin{bmatrix} 1 & 1 & 0 \\ 1 &1 &0 \end{bmatrix}$ be a matrix.

Find a basis of the null space of the matrix $A$.

(Remark: a null space is also called a kernel.)

Contents

## Solution.

The null space $\calN(A)$ of the matrix $A$ is by definition
$\calN(A)=\{ \mathbf{x} \in \R^3 \mid A\mathbf{x}=\mathbf{0} \}.$ In other words, the null space consists of all solutions $\mathbf{x}$ of the matrix equation $A\mathbf{x}=\mathbf{0}$.

So we first determine the solutions of $A\mathbf{x}=\mathbf{0}$ by Gauss-Jordan elimination. The augmented matrix is
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0
\end{array} \right].
\end{align*}
Subtracting $R_1$ from $R_2$, we reduce the augmented matrix to the reduced row echelon form matrix as follows.
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0
\end{array} \right] \xrightarrow{R_2-R_1}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array} \right].
\end{align*}
Thus the solution $\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ of $A\mathbf{x}=\mathbf{0}$ satisfies $x_1+x_2=0$, or equivalently $x_1=-x_2$.
Substituting the last equality, we see that solutions are of the form
$\mathbf{x}=\begin{bmatrix} -x_2 \\ x_2 \\ x_3 \end{bmatrix}=x_2\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}+x_3\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}.$ Therefore the null space is
\begin{align*}
-1 \\
1 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} \text{ for any } x_2, x_3 \in \R \right \}\6pt] &= \mathrm{Sp} \left\{ \, \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \, \right\}. \end{align*} Thus, the set \left\{ \, \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \, \right\} is a spanning set for the null space \calN(A). Now, we check that the vectors \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} are linearly independent. Consider a linear combination \[a_1\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}+a_2\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} =\mathbf{0}. This is equivalent to
$\begin{bmatrix} -a_1 \\ a_1 \\ a_2 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$ Hence we must have $a_1=a_2=0$.
Since the linear combination equation has only the zero solution, the vectors $\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ are linearly independent.

Therefore the set $\left\{ \, \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\, \right\}$ is a linearly independent spanning set, thus it is a basis for the null space $\calN(A)$.

Let $V$ be the following subspace of the $4$-dimensional vector space $\R^4$. \[V:=\left\{ \quad\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\...