Finding eigenvalues and eigenspaces for the matrix A A is a matrix that reflects vectors x $\in$ $\mathbb{R}^3$ about a fixed plane P in $\mathbb{R}^3$. Find all eigenvalues and eigenspaces of A.  Does A admit an eigenbasis for $\mathbb{R}^3$? Why or why not?
Another part of this question: Let a = $\begin{bmatrix}\alpha & \beta & \gamma \end{bmatrix}^T$ $\in$ $\mathbb{R}^3$ be nonzero and fixed.  Consider the linear transformation T : $\mathbb{R}^3 \to \mathbb{R}^3$ given by T(x) = a $\times$ x. Find all real eigenvalues of T and all real eigenspaces of T.  Does T admit an eigenbasis for $\mathbb{R}^3$? Why or why not?
for the first part I don't know how I would find an eigenvalue for a matrix I don't know, I would assume A would look like a reflection matrix but I've looked online and I can't find anything that tells me what a 3$\times$3 reflection matrix would look like.
for the second part, I know that a would look like $\begin{bmatrix}\alpha\\\ \beta \\\ \gamma \end{bmatrix}$, but wouldn't a $\times$ x look like $\begin{bmatrix}\beta x_3 - \gamma x_2 & \gamma x_1 - \alpha x_3 & \alpha x_2 - \beta x_1 \end{bmatrix}$? And if so how would you even find the eigenvalues and eigenspaces of that?
Any help is appreciated, thanks!
 A: The key to doing this problem efficiently is to understand the geometric interpretation of an eigenvalue. Recall that for a linear transformation $T:\Bbb R^n \to \Bbb R^n$, we say that $x \in \Bbb R^n$ is an eigenvector of $T$ associated with the eigenvalue $\lambda$ if it is a non-zero vector for which $T(x) = \lambda x$.
For the first problem, our linear transformation is $T_A(x) = Ax$. It is given that $T$ takes a vector and reflects it across the plane $P$. Suppose that $v_1$ is a non-zero vector perpendicular to $P$.


*

*I claim that $v_1$ must be an eigenvector of $T_A$. Do you see why? What is the eigenvalue associated with $v_1$?

*Take any two linearly independent vectors $v_2,v_3$ that are orthogonal to $v_1$ (that is, two vectors from the plane $P$).  I claim that these are also eigenvectors. Do you see why? What are the associated eigenvalues?

*$v_1,v_2,v_3$ form a basis for $\Bbb R^3$, and each vector is an eigenvector. So, $A$ admits an eigenbasis for $\Bbb R^3$.


For the second problem, our linear transformation is $T(x) = \alpha \times x$.


*

*By the properties of the cross product, $T(x)$ must be orthogonal to $x$.  There is exactly one real number $\lambda$ for which it is possible to have $x \neq 0$ and $T(x) = \lambda x$, what is this $\lambda$? If $x$ is an eigenvector associated with this $\lambda$, what must be true about $x$?

*From the above, we have found out that $T$ has only one real eigenvalue, and the eigenspace associated with that eigenvalue is $1$-dimensional. Since we have no more eigenvectors with which to construct an eigenbasis, we conclude that $T$ does not admit an eigenbasis for $\Bbb R^3$.



As I explained above, using the specific entries for the matrices associated with these transformations turns out not to be a "nice" approach. If you are interested in seeing what these matrices might look like, then you should know that the reflection matrix from the first part is the Householer transformation $A = I - 2vv^T$, and the cross-product matrix for the second part is $[\alpha]_{\times}$, as is explained in this section of the cross-product wiki page.
