Characterization of Left Null Space

Definition

Let $\mathbf A_{m \times n}$ be a matrix in the matrix space $\map {\MM_{m, n} } \R$.

Let $\map {\operatorname {N^\gets} } {\mathbf A}$ be used to denote the left null space of $\mathbf A$.

Then:

$\map {\operatorname {N^\gets} } {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf x^\intercal \mathbf A = \mathbf 0^\intercal}$

where $\mathbf X^\intercal$ is the transpose of $\mathbf X$.

Let $\mathbf x = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix} \in \R^n$.

$\ds \mathbf x \in \map {\operatorname {N^\gets} } {\mathbf A}$

$\leadstoandfrom$

$\ds \mathbf x \in \map {\operatorname N} {\mathbf A^\intercal}$

$\ds \leadstoandfrom \ \ $

$\ds \mathbf A^\intercal \mathbf x$

$=$

$\ds \mathbf 0$

Definition of Null Space

$\ds \leadstoandfrom \ \ $

$\ds \paren {\mathbf A^\intercal \mathbf x}^\intercal$

$=$

$\ds \mathbf 0^\intercal$

taking the transpose of both sides

$\ds \leadstoandfrom \ \ $

$\ds \mathbf x^\intercal \paren {\mathbf A^\intercal}^\intercal$

$=$

$\ds \mathbf 0^\intercal$

$\ds \leadstoandfrom \ \ $

$\ds \mathbf x^\intercal \mathbf A$

$=$

$\ds \mathbf 0^\intercal$

We have that $\mathbf A^\intercal \mathbf x = \mathbf 0$ is equivalent to $\mathbf x^\intercal \mathbf A = \mathbf 0^\intercal$.

This implies that $\mathbf x \in \map {\operatorname N} {\mathbf A^\intercal} \iff \mathbf x^\intercal \mathbf A = \mathbf 0^\intercal$.

Recall that:

$\mathbf x \in \map {\operatorname N} {\mathbf A^\intercal} \iff \mathbf x \in \map {\operatorname {N^\gets} } {\mathbf A}$

Hence the result, by definition of set equality.

$\blacksquare$