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)
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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
- 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
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