# The Null Space (the Kernel) of a Matrix is a Subspace of $\R^n$

## Problem 121

Let $A$ be an $m \times n$ real matrix. Then the null space $\calN(A)$ of $A$ is defined by
$\calN(A)=\{ \mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}_m\}.$ That is, the null space is the set of solutions to the homogeneous system $A\mathbf{x}=\mathbf{0}_m$.

Prove that the null space $\calN(A)$ is a subspace of the vector space $\R^n$.
(Note that the null space is also called the kernel of $A$.)

## Proof.

### Subspace criteria

We check the following criteria for $\calN(A)$ to be a subspace of $\R^n$.

(a) The zero vector $\mathbf{0}_n \in \R^n$ is in $\calN(A)$.
(b) If $\mathbf{x}, \mathbf{y} \in \calN(A)$, then $\mathbf{x}+\mathbf{y}\in \calN(A)$.
(c) If $\mathbf{x} \in \calN(A)$ and $c\in \R$, then $c\mathbf{x} \in \calN(A)$.

### Check (a)

Since we have $A\mathbf{0}_n=\mathbf{0}_m$, the zero vector $\mathbf{0}_n \in \R^n$ is in $\calN(A)$. Thus (a) is met.

### Check (b)

For (b), take arbitrary vectors $\mathbf{x}, \mathbf{y} \in \calN(A)$.
By the definition of $\calN(A)$, we have
$A\mathbf{x}=\mathbf{0}_m \text{ and } A\mathbf{y}=\mathbf{0}_m. \tag{*}$ We want to show that $\mathbf{x}+ \mathbf{y} \in \calN(A)$. Thus we need to show $A(\mathbf{x}+\mathbf{y})=\mathbf{0}$.

This equality can be proved as follows.
We have
\begin{align*}
A(\mathbf{x}+\mathbf{y}) &= A\mathbf{x}+A\mathbf{y}\\
&=\mathbf{0}+\mathbf{0}=\mathbf{0}.
\end{align*}
Here the second equality follows from (*).
Thus $\mathbf{x}+\mathbf{y}\in \calN(A)$ and (b) is satisfied.

### Check (c)

To check (c), take arbitrary $\mathbf{x} \in \calN(A)$ and a scalar $c \in \R$.
Since $\mathbf{x} \in \calN(A)$, we have
$A\mathbf{x}=\mathbf{0}_m.$

Multiplying by the scalar $c$, we have
$cA\mathbf{x}=c\mathbf{0}_m.$ This is equivalent to
$A(c\mathbf{x})=\mathbf{0}_m.$ This equality implies that $c\mathbf{x} \in \calN(A)$, hence (c) is met.

Thus we checked the conditions (a), (b), and (c) and conclude that the null space $\calN(A)$ is a subspace of $\R^n$.

## Related Question.

In the following problem, we determine the null space of a matrix explicitly.

### 4 Responses

1. 01/04/2017

[…] 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 $R^n$. […]

2. 01/12/2017

[…] the null space of an $mtimes n$ matrix is a subspace of the vector space $R^n$. (See the post The null space (the kernel) of a matrix is a subspace of $R^n$.) Since we showed that $W$ is the null space of $1times 3$ matrix $A$, we conclude that $W$ is a […]

3. 01/23/2017

[…] Bmathbf{x}=mathbf{0}}\ &=calN(B), end{align*} the null space of the matrix $B$. Since any null space is a vector space, this shows that $V$ is a subspace of […]

4. 10/02/2017

[…] Observe that the conditions [mathbf{a}^{trans}mathbf{x}=0, mathbf{b}^{trans}mathbf{x}=0, text{ and } mathbf{c}^{trans}mathbf{x}=0] can be combined into the following matrix equation [Amathbf{x}=mathbf{0},] where [A=begin{bmatrix} 1 & 0 & 1 & 0 \ 1 &1 & 0 & 0 \ 0 & 1 & -1 & 0 end{bmatrix}] and $mathbf{0}$ is the three dimensional zero vector. Note that the rows of the matrix $A$ are $mathbf{a}^{trans}$, $mathbf{b}^{trans}$, and $mathbf{c}^{trans}$. It follows that the subset $V$ is the null space $calN(A)$ of the matrix $A$. Being the null space, $V=calN(A)$ is a subspace of $R^4$. (See the post “The Null Space (the Kernel) of a Matrix is a Subspace of $R^n$“.) […]

##### If Vectors are Linearly Dependent, then What Happens When We Add One More Vectors?

Suppose that $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_r$ are linearly dependent $n$-dimensional real vectors. For any vector $\mathbf{v}_{r+1} \in \R^n$, determine whether...

Close