Null Space Closed under Vector Addition

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 and each the column matrix $\mathbf {x}_{n \times 1}$ is interpreted as a vector in $\R^n$.

Then the null space is closed under vector addition:


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

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

By the definition of null space:

Next, observe that:

The dimensions are 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