Algebraic and geometric multiplicities of eigenvalues of transformation Set a non zero vector $n \in \mathbb{R}^n$. Let $L$ be a linear mapping such that, $L: \mathbb{R}^n \to \mathbb{R}^n$, and $L(x) = x - 2\text{proj}_{n} (x)$

What are the algebraic and geometric multiplicities of all eigenvalues of $L$?

The issue I am having is that, I have not seen a way to get eigenvalues of a transformation, only matrices, so how does it work?
I know that $L(x) = [L]x$, where $[L]$ is the standard matrix, but working out the standard matrix is not easy either.
Hints?
 A: First we need to see what $L$ does. I can tell from looking at the formula, and having seen quite a few transformations in my time, (as well as checking explicitly by imagining $\Bbb R^2$ and $\Bbb R^3$) that this transformation mirrors across the normal hyperplane to the vector $n$.
I would therefore like to choose an orthonormal basis where $n = (0, 0, \ldots, \|n\|)^T$. In this basis, $L$ is represented by
$$
\begin{bmatrix}
1&0&0&\cdots&0\\
0&1&0&\cdots&0\\
0&0&1&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\cdots&-1
\end{bmatrix}
$$
where the diagonal has only $1$ except in the lower right corner where there is a $-1$. In this form, it's easy to see that the eigenvalues $1$ and $-1$ have algebraic multiplicities $n-1$ and $1$ respectively.
Since this is a diagonal matrix (our choice of basis vectors may be seen as a diagonalisation, if you will), the geometric and algebraic multiplicities coincide. But if you want a direct proof, the vector $(0,0,\ldots,0,1)^T$ has eigenvalue $-1$, and the linearly independent vectors
$$
(1,0,0,\ldots,0,0)^T\\
(0,1,0,\ldots,0,0)^T\\
(0,0,1,\ldots,0,0)^T\\
\vdots\\
(0,0,0,\ldots,1,0)^T
$$
all have eigenvalue $1$.
