# Characterization of Left Null Space

## Definition

Let $\mathbf A_{m \times n}$ be a matrix in the matrix space $\mathcal M_{m, n} \left({\R}\right)$.

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

Then:

$\operatorname {N^\gets} \left({\mathbf A}\right) = \left\{{\mathbf x \in \R^n: \mathbf x^\intercal \mathbf A = \mathbf 0^\intercal}\right\}$

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

## Proof

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

 $\displaystyle \mathbf x \in \operatorname {N^\gets} \left({\mathbf A}\right)$ $\iff$ $\displaystyle \mathbf x \in \operatorname {N}\left({\mathbf A^\intercal}\right)$ definition of left null space $\displaystyle \iff \ \$ $\displaystyle \mathbf A^\intercal \mathbf x$ $=$ $\displaystyle \mathbf 0$ definition of null space $\displaystyle \iff \ \$ $\displaystyle \left({\mathbf A^\intercal \mathbf x}\right)^\intercal$ $=$ $\displaystyle \mathbf 0^\intercal$ taking the transpose of both sides $\displaystyle \iff \ \$ $\displaystyle \mathbf x^\intercal \left({\mathbf A^\intercal}\right)^\intercal$ $=$ $\displaystyle \mathbf 0^\intercal$ Transpose of Matrix Product $\displaystyle \iff \ \$ $\displaystyle \mathbf x^\intercal \mathbf A$ $=$ $\displaystyle \mathbf 0^\intercal$ 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 $\mathbf x \in \operatorname{N} \left({\mathbf A^\intercal}\right) \iff \mathbf x^\intercal \mathbf A = \mathbf 0^\intercal$.

Recall $\mathbf x \in \operatorname{N} \left({\mathbf A^\intercal}\right) \iff \mathbf x \in \operatorname {N^\gets} \left({\mathbf A}\right)$

Hence the result, by definition of set equality.

$\blacksquare$