# 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 $\R^3$.

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## 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 $\R^3$.

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