Null Space is Subspace

Theorem

Let:

$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf {A x} = \mathbf 0}$

be the null space of $\mathbf A$, where:

$\mathbf A_{m \times n} = \begin {bmatrix}

a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end {bmatrix}$, $\mathbf x_{n \times 1} = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$, $\mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$

are matrices.

Then $\map {\mathrm N} {\mathbf A}$ is a linear subspace of $\R^n$.

Proof

$\map {\mathrm N} {\mathbf A} \subseteq \R^n$, by construction.

We have:

$\mathbf 0 \in \map {\mathrm N} {\mathbf A}$, from Null Space Contains Zero Vector

$\forall \mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}: \mathbf v + \mathbf w \in \map {\mathrm N} {\mathbf A}$, from Null Space Closed under Vector Addition

$\forall \mathbf v \in \map {\mathrm N} {\mathbf A}, \lambda \in \R: \lambda \mathbf v \in \map {\mathrm N} {\mathbf A}$, from Null Space Closed under Scalar Multiplication

The result follows from Vector Subspace of Real Vector Space.

$\blacksquare$

Sources

• For a video presentation of the contents of this page, visit the Khan Academy.