Null Space Closed under Vector Addition

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

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}$, $\mathbf x_{n \times 1} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$ and $\mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$ are matrices


 * the column matrix $\mathbf x_{n \times 1}$ is interpreted as a vector in the real Euclidean space $\R^n$.

Then $\map {\mathrm N} {\mathbf A}$ is closed under vector addition:


 * $\forall \mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}: \mathbf v + \mathbf w \in \map {\mathrm N} {\mathbf A}$

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

By the definition of null space:

Next, observe that:

The order is correct, by hypothesis.

Hence the result, by the definition of null space.

Also see

 * Null Space Contains Zero Vector
 * Null Space Closed under Scalar Multiplication
 * Null Space is Subspace