# 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 $B$.

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## 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*}

\calN(A)&=\left\{\mathbf{x}\in \R^3 \quad \middle| \quad \mathbf{x}=x_3\begin{bmatrix}

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 $\R^n$.

Add to solve later

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

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

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