Take $\mathbb{R^3}$ with the standard inner product. Let $a,b \in \mathbb{R^3}$ so that $\langle a,b\rangle = 2$
$L: \mathbb{R^3} \rightarrow \mathbb{R^3}:x \longmapsto x - \langle x,a\rangle b$
Let $c\in\mathbb R^3$ such that $c$ is orthogonal to both $a$ and $b$. Then $\{a,b,c\}$ is a basis of $\mathbb R^3$ and the matrix of $L$ with respect to this basis is$$\begin{bmatrix}1&0&0\\\lVert a\rVert^2&-1&0\\0&0&1\end{bmatrix}.$$
I know the eigenvalues are $X_1 = 1$ and $X_2 = -1$ with algebraic multiplicity $2$ and $1$, but someone told me that saying that $X_1$ occurs twice does not fully proof that it's not diagonazible. I don't know what else I need to show.
Thanks in advance.
That matrix is diagonalizable. You can check that $(0,0,1)$ and $(2,\|a\|^2,0)$ are eigenvectors with eigenvalue $1$, and that $(0,1,0)$ is an eigenvector with eigenvalue $-1$.