Null Space Closed under Scalar Multiplication

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

Then $\map {\mathrm N} {\mathbf A}$ is closed under scalar multiplication:


 * $\forall \mathbf v \in \map {\mathrm N} {\mathbf A} ,\forall \lambda \in \R: \lambda \mathbf v \in \map {\mathrm N} {\mathbf A}$

Proof
Let $\mathbf v \in \map {\mathrm N} {\mathbf A}$, $\lambda \in \R$.

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