Null Space Closed under Scalar Multiplication

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 where each column is an element of a real vector space.

Then $\operatorname{N}\left({ \mathbf{A} }\right)$ is closed under scalar multiplication:


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

Proof
Let $\mathbf{v} \in \operatorname{N}\left({ \mathbf{A} }\right)$,$\lambda \in \R$.

By the definition of null space:

Observe that:

Hence the result, by the definition of null space.

Also see

 * Null Space Contains Zero Vector
 * Null Space Closed under Vector Addition
 * Null Space is Subspace