# 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 $\mathbf X$.

## Proof

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

 $\ds \mathbf x \in \map {\operatorname {N^\gets} } {\mathbf A}$ $\leadstoandfrom$ $\ds \mathbf x \in \map {\operatorname N} {\mathbf A^\intercal}$ Definition of Left Null Space $\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 \ \$ $\ds \mathbf x^\intercal \paren {\mathbf A^\intercal}^\intercal$ $=$ $\ds \mathbf 0^\intercal$ Transpose of Matrix Product $\ds \leadstoandfrom \ \$ $\ds \mathbf x^\intercal \mathbf A$ $=$ $\ds \mathbf 0^\intercal$ Transpose of Transpose of Matrix

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.

$\blacksquare$