Hyperplane Through Origin is Subspace of 4-Dimensional Vector Space

Ohio State University exam problems and solutions in mathematics

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)
 
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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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