# 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)

Add to solve later

Sponsored Links

## 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

- True of False Problems and Solutions: True or False problems of vector spaces and linear transformations
- Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces”
- Problem 2 and its solution: Determine whether trigonometry functions $\sin^2(x), \cos^2(x), 1$ are linearly independent or dependent
- Problem 3 and its solution: Orthonormal basis of null space and row space
- Problem 4 and its solution: Basis of span in vector space of polynomials of degree 2 or less
- Problem 5 and its solution: Determine value of linear transformation from $R^3$ to $R^2$
- Problem 6 and its solution: Rank and nullity of linear transformation from $R^3$ to $R^2$
- Problem 7 and its solution: Find matrix representation of linear transformation from $R^2$ to $R^2$
- Problem 8 and its solution (current problem): Hyperplane through origin is subspace of 4-dimensional vector space

Add to solve later

Sponsored Links

## 4 Responses

[…] Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space […]

[…] Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space […]

[…] Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space […]