# Hyperplane Through Origin is Subspace of 4-Dimensional Vector Space

## Problem 371

Let $S$ be the subset of $\R^4$ consisting of vectors $\begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix}$ satisfying
$2x+3y+5z+7w=0.$ Then prove that the set $S$ is a subspace of $\R^4$.

(Linear Algebra Exam Problem, The Ohio State University)

## Proof.

First, in set theoretical notation, the definition of $S$ can be written as
$S=\left\{\, \begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix}\in \R^4 \quad \middle| \quad 2x+3y+5z+7w=0 \,\right\}.$

Let $A=\begin{bmatrix} 2 & 3 & 5 & 7 \end{bmatrix}$ be the $1 \times 4$ matrix. Then the defining equation $2x+3y+5z+7w=0$ can be written as
$A\mathbf{x}=0,$ where
$\mathbf{x}=\begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix}.$ It follows that the set $S$ is the null space of $A$, that is, $S=\calN(A)$.
Since every null space is a subspace, we see that $S$ is also a subspace of $\R^4$.

## Linear Algebra Midterm Exam 2 Problems and Solutions

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