Null Space Contains Zero Vector

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_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots &  \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{m n} \\ \end{bmatrix}$

is a matrix in the matrix space $\map {\MM_\R} {m, n}$.

Then the null space of $\mathbf A$ contains the zero vector:
 * $\mathbf 0 \in \map {\mathrm N} {\mathbf A}$

where:
 * $\mathbf 0 = \mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$

Also see

 * Null Space Closed under Scalar Multiplication
 * Null Space Closed under Scalar Multiplication
 * Null Space is Subspace
 * Kernel of Linear Transformation contains Zero Vector