Show that $V=\bigg\{\begin{bmatrix}\mathbf{x}\\A\mathbf{x}\\\end{bmatrix}: x\in\mathbb{R}^n\bigg\}\subset\mathbb{R}^{n+1}$ is a hyperplane.

Question

Problem:

Let $A$ be a $m \times n$ matrix.

1. Show that $V=\bigg\{\begin{bmatrix}\mathbf{x}\\A\mathbf{x}\\\end{bmatrix}: \mathbf{x}\in\mathbb{R}^n\bigg\}\subset\mathbb{R}^{m+n}$ is a subspace of $\mathbb{R}^{m+n}$ (solved)
2. When $m=1$, show that $V\subset\mathbb{R}^{n+1}$ is a hyperplane by finding a nonzero vector $\mathbf{b}\in\mathbb{R}^{n+1}$ so that $V=\{\mathbf{z}\in\mathbb{R}^{n+1}:\mathbf{b}\cdot \mathbf{z}=0\}$ (unsure)

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Attempt:

1. Show that $V=\bigg\{\begin{bmatrix}\mathbf{x}\\A\mathbf{x}\\\end{bmatrix}: \mathbf{x}\in\mathbb{R}^n\bigg\}\subset\mathbb{R}^{m+n}$ is a subspace of $\mathbb{R}^{m+n}$ (solved)

   • Trivially, $\begin{bmatrix}\mathbf{0}\\A\mathbf{0}\end{bmatrix} = \mathbf{0}\in\mathbb{R}^{m+n}$
   • Choose $c\in\mathbb{R}$ and $\mathbf{v}\in V$. Then, $\begin{bmatrix}c\mathbf{v}\\Ac\mathbf{v}\end{bmatrix} = c\begin{bmatrix}\mathbf{v}\\A\mathbf{v}\end{bmatrix}$
   • Choose $\mathbf{v}, \mathbf{v'}\in V$. Then $\begin{bmatrix}\mathbf{v}\\A\mathbf{v}\end{bmatrix}+\begin{bmatrix}\mathbf{v'}\\A\mathbf{v'}\end{bmatrix} = \begin{bmatrix}\mathbf{v+v'}\\A(\mathbf{v'}+\mathbf{v})\end{bmatrix}$
2. When $m=1$, show that $V\subset\mathbb{R}^{n+1}$ is a hyperplane by finding a nonzero vector $\mathbf{b}\in\mathbb{R}^{n+1}$ so that $V=\{\mathbf{z}\in\mathbb{R}^{n+1}:\mathbf{b}\cdot \mathbf{z}=0\}$ (unsure)
   • We can choose a vector $\begin{bmatrix}A^T\\-1\end{bmatrix}$ ($A^T$ is a $n\times1$ matrix) so that $$\begin{bmatrix}A \hspace{2mm} (-1)\end{bmatrix}\cdot \begin{bmatrix}\mathbf{x}\\A\mathbf{x}\end{bmatrix} = A\mathbf{x} + (-1)(A\mathbf{x}) = 0$$.

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I'm fairly certain part 1 is correct but feel free to correct me if I'm wrong.

Questions:

1. Is my answer sufficient for question 2? If not, what are they looking for?

Answer 1

Yes, you are on the right track, I would only suggest a different notation and a little more details in the proof: use a lower case letter, such as $\mathbf a$, for the $n\times 1$ matrix $A^T$, then let $\mathbf{b}=\left[\begin{array}{c} \mathbf{a} \\-1\end{array}\right]$. Now show that for $\mathbf{x}\in {\mathbb R}^{n+1}$, if $\mathbf{x}\cdot \mathbf{b}=0$, then $\mathbf{x}\in V$, and if $\mathbf{x}\in V$, then $\mathbf{x}\cdot \mathbf{b}=0$.