Null Space is Subspace

Theorem

Let:

$\operatorname{N} \left({\mathbf A}\right) = \left\{{\mathbf x \in \R^n: \mathbf {Ax} = \mathbf 0}\right\}$

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 $\operatorname{N} \left({\mathbf A}\right)$ is a linear subspace of $\R^n$.

Proof

$\operatorname{N} \left({\mathbf A}\right) \subseteq \R^n$, by construction.

We have:

• $\mathbf 0 \in \operatorname{N} \left({\mathbf A}\right)$, from Null Space Contains Zero Vector

• $\forall \mathbf v, \mathbf w \in \operatorname{N} \left({\mathbf A}\right): \mathbf v + \mathbf w \in \operatorname{N} \left({\mathbf A}\right)$, from Null Space Closed under Vector Addition

• $\forall\mathbf v \in \operatorname{N} \left({\mathbf A}\right), \lambda \in \R: \lambda \mathbf v \in \operatorname{N} \left({\mathbf A}\right)$, 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.