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 $A$.)

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.

Intersection of Two Null Spaces is Contained in Null Space of Sum of Two Matrices
Let $A$ and $B$ be $n\times n$ matrices. Then prove that
\[\calN(A)\cap \calN(B) \subset \calN(A+B),\]
where $\calN(A)$ is the null space (kernel) of the matrix $A$.
Definition.
Recall that the null space (or kernel) of an $n \times n$ matrix […]

Dimension of Null Spaces of Similar Matrices are the Same
Suppose that $n\times n$ matrices $A$ and $B$ are similar.
Then show that the nullity of $A$ is equal to the nullity of $B$.
In other words, the dimension of the null space (kernel) $\calN(A)$ of $A$ is the same as the dimension of the null space $\calN(B)$ of […]

Prove a Given Subset is a Subspace and Find a Basis and Dimension
Let
\[A=\begin{bmatrix}
4 & 1\\
3& 2
\end{bmatrix}\]
and consider the following subset $V$ of the 2-dimensional vector space $\R^2$.
\[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\]
(a) Prove that the subset $V$ is a subspace of $\R^2$.
(b) Find a basis for […]

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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.)
Solution.
The null space $\calN(A)$ of the matrix $A$ is by […]

Row Equivalent Matrix, Bases for the Null Space, Range, and Row Space of a Matrix
Let \[A=\begin{bmatrix}
1 & 1 & 2 \\
2 &2 &4 \\
2 & 3 & 5
\end{bmatrix}.\]
(a) Find a matrix $B$ in reduced row echelon form such that $B$ is row equivalent to the matrix $A$.
(b) Find a basis for the null space of $A$.
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Determine Null Spaces of Two Matrices
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 […]

Quiz 6. Determine Vectors in Null Space, Range / Find a Basis of Null Space
(a) Let $A=\begin{bmatrix}
1 & 2 & 1 \\
3 &6 &4
\end{bmatrix}$ and let
\[\mathbf{a}=\begin{bmatrix}
-3 \\
1 \\
1
\end{bmatrix}, \qquad \mathbf{b}=\begin{bmatrix}
-2 \\
1 \\
0
\end{bmatrix}, \qquad \mathbf{c}=\begin{bmatrix}
1 \\
1 […]

[…] 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 $R^n$. […]

[…] 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 […]

[…] 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 […]

[…] 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 $R^n$“.) […]

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...

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[…] 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 $R^n$. […]

[…] 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 […]

[…] 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 […]

[…] 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 $R^n$“.) […]