# Linear Algebra - Reflection in a hyperplane

We have a matrix $A$: $$A = \dfrac{1}{7} \cdot \begin{pmatrix} 5 & -4 & -2 & 2 \\ 4 & -1 & -4 & 4 \\ -2 & -4 & 5 & -2 \\ 2 & 4 & 2 & 5 \end{pmatrix}$$

The map $f_a : \mathbb{R}^4 \to \mathbb{R}^4$ is a reflection in the hyperplane $H \subset \mathbb{R}^4$. Determine $H$.

I don't quite how to find the hyperplane $H$. I think the root of my problem is that I don't really understand what is meant mathematically with "reflection in a hyperplane". Can anyone clear up what is asked and a strategy to find the hyperplane?

• I'm guessing you mean $f_a$ to denote multiplication by $A$, but you left this out. To find the hyperplane $H$ I would compute the fixed points of that mapping (fixed by reflection implies point lying on $H$). – hardmath Oct 26 '15 at 23:55
• In two dimensions, reflection in a hyperplane is reflection w/r to some line, in three it's w/r to some plane, &c. – amd Oct 27 '15 at 0:16
• Double-check the matrix $A$. A reflection should have determinant $-1$, but this one’s determinant is $-\frac9{49}$. – amd Oct 28 '15 at 8:27

A reflection about the hyperplane $H$ will fix vectors in $H$ and reflect other vectors of $\mathbb{R}^4$ across $H$. Since vectors in $H$ are fixed by $f_A$, they are eigenvectors of $f_A$ with eigenvalue 1. To find $H$, you must find the eigenspace of $f_A$ associated to the eigenvalue 1.
An example in a smaller dimension is the matrix \begin{equation*} A = \left(\begin{matrix}-1 & 0 \\ 0 & 1\end{matrix}\right), \end{equation*} which represents reflection in $\mathbb{R}^2$ about the $y$ axis. The eigenvectors of this $A$ with eigenvalue 1 are vectors which lie on the $y$ axis.