# Find a Matrix so that a Given Subset is the Null Space of the Matrix, hence it’s a Subspace

## Problem 252

Let $W$ be the subset of $\R^3$ defined by
$W=\left \{ \mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\in \R^3 \quad \middle| \quad 5x_1-2x_2+x_3=0 \right \}.$ Exhibit a $1\times 3$ matrix $A$ such that $W=\calN(A)$, the null space of $A$.
Conclude that the subset $W$ is a subspace of $\R^3$.

## Solution.

Note that the defining equation $5x_1-2x_2+x_3=0$ can be written as
$\begin{bmatrix} 5 & -2 & 1 \\ \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=0.$

Hence if we put $A=\begin{bmatrix} 5 & -2 & 1 \\ \end{bmatrix}$, then the defining equation becomes
$A\mathbf{x}=0.$ Hence, we have
$W=\{x \in \R^3 \mid A\mathbf{x}=0\},$ which is exactly the null space of the $1\times 3$ matrix $A$.
Hence we have proved that $W=\calN(A)$.

In general, the null space of an $m\times n$ matrix is a subspace of the vector space $\R^n$.
(See the post The null space (the kernel) of a matrix is a subspace of $\R^n$.)
Since we showed that $W$ is the null space of $1\times 3$ matrix $A$, we conclude that $W$ is a subspace of $\R^3$.

Let $A$ be an $n\times n$ invertible matrix. Prove that the inverse matrix of $A$ is uniques.