Null Space Contains Zero Vector

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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}$

is a matrix in the matrix space $\mathbf M_{m, n} \left({\R}\right)$.


Then the null space of $\mathbf A$ contains the zero vector:

$\mathbf 0 \in \operatorname{N}\left({\mathbf A}\right)$

where:

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


Proof 1

\(\displaystyle \mathbf{A0}\) \(=\) \(\displaystyle \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end{bmatrix}\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}\)
\(\displaystyle \) \(=\) \(\displaystyle \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}\)
\(\displaystyle \) \(=\) \(\displaystyle \mathbf 0\)

The dimensions are correct by hypothesis.

The result follows by the definition of null space.

$\blacksquare$


Proof 2

From Matrix Product as Linear Transformation, $\mathbf {Ax} = \mathbf 0$ defines a linear transformation from $\R^m$ to $\R^n$.

The result then follows from Linear Transformation Maps Zero Vector to Zero Vector.


Also see


Sources