# Determine Null Spaces of Two Matrices

## Problem 242

Let
$A=\begin{bmatrix} 1 & 2 & 2 \\ 2 &3 &2 \\ -1 & -3 & -4 \end{bmatrix} \text{ and } B=\begin{bmatrix} 1 & 2 & 2 \\ 2 &3 &2 \\ 5 & 3 & 3 \end{bmatrix}.$

Determine the null spaces of matrices $A$ and $B$.

## Proof.

### The null space of the matrix $A$

We first determine the null space of the matrix $A$.
By definition, the null space is
$\calN(A):=\{\mathbf{x}\in \R^3 \mid A\mathbf{x}=\mathbf{0}\},$ that is, the null space of $A$ consists of the solution $\mathbf{x}$ of the linear system $A\mathbf{x}=\mathbf{0}$.

To solve the system $A\mathbf{x}=\mathbf{0}$, we apply the Gauss-Jordan elimination. We reduce the augmented matrix $[A|\mathbf{0}]$ by elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 2 & 2 & 0 \\
2 &3 & 2 & 0 \\
-1 & -3 & -4 & 0
\end{array} \right] \xrightarrow{\substack{R_2-2R_1\\R_3+R_1}}
\left[\begin{array}{rrr|r}
1 & 2 & 2 & 0 \\
0 & -1 & -2 & 0 \\
0 & -1 & -2 & 0
\end{array} \right] \xrightarrow{\substack{R_1+2R_2\\R_3-5R_2}}\\
\left[\begin{array}{rrr|r}
1 & 0 & -2 & 0 \\
0 & -1 & -2 & 0 \\
0 & 0 & 0 & 0
\end{array} \right] \xrightarrow{-R_2}
\left[\begin{array}{rrr|r}
1 & 0 & -2 & 0 \\
0 & 1 & 2 & 0 \\
0 & 0 & 0 & 0
\end{array} \right].
\end{align*}
The last matrix is in reduced row echelon form, and the solutions $\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ of $A\mathbf{x}=\mathbf{0}$ satisfy
$x_1=2x_3 \text{ and } x_2=-2x_3$ and $x_3$ is a free variable.
Thus we have
$\mathbf{x}=\begin{bmatrix} 2x_3 \\ -2x_3 \\ x_3 \end{bmatrix}=x_3\begin{bmatrix} 2 \\ -2 \\ 1 \end{bmatrix}$ for any $x_3$ are solutions.

Therefore the null space of the matrix $A$ is
\begin{align*}
2 \\
-2 \\
1
\end{bmatrix} \text{ for any } x_3\in \R \right\}\\
&=\Span\left(\begin{bmatrix}
2 \\
-2 \\
1
\end{bmatrix}\right),
\end{align*}
the subspace of $\R^3$ spanned by the vector $\begin{bmatrix} 2 \\ -2 \\ 1 \end{bmatrix}$.

### The null space of the matrix $B$

The same procedure works for the matrix $B$. Thus we omit some detail below.
For the matrix $B$, the augmented matrix is reduced to
$\left[\begin{array}{rrr|r} 1 & 2 & 2 & 0 \\ 2 &3 & 2 & 0 \\ 5 & 3 & 3 & 0 \end{array} \right] \to \cdots \to \left[\begin{array}{rrr|r} 1 & 0 & 0 & 0 \\ 0 &1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right].$ (Check the computation by yourself.)
This implies that $\mathbf{x}=\mathbf{0}$ is the only solution of the system $B\mathbf{x}=\mathbf{0}$.
Therefore, the null space $\calN(B)$ of the matrix $B$ consists of just the zero vector:
$\calN(B)=\left\{\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \right\}.$

## Comment.

The dimension of the null space $\calN(A)$ is $1$, and the dimension of the null space $\calN(B)$ is $0$.
In other words, the nullity of the matrix $A$ is $1$, and the nullity of the matrix $B$ is $0$.
(Recall that the nullity of a matrix is just the dimension of the null space of the matrix.)

## Related Question.

The proof of the fact that a null space of a matrix is a subspace is give in the post The null space (the kernel) of a matrix is a subspace of $\R^n$.

### 4 Responses

1. Simon says:

there is a slight error. x_1=2x_3 and x_2=-x_3 but it should be x_2=-2x_3

• Yu says:

Dear Simon, Thank you very much for pointing out the typo. I fixed it.

• Simon says:

I am the one thanking for all these great problems and solutions.

1. 01/04/2017

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