Null Space Contains Zero Vector

From ProofWiki
Jump to navigation Jump to search

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:

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


Proof 1

\(\ds \mathbf A \mathbf 0\) \(=\) \(\ds \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{m n} \\ \end {bmatrix} \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \mathbf 0\)

The order is 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

Retrieved from "https://proofwiki.org/w/index.php?title=Null_Space_Contains_Zero_Vector&oldid=512014"