I'm having troble proving this:
Let $A\in M_n(R). $ If $n=2$ and $A$ has exactly one eigenvalue $\lambda$ then $A$ is similar to $\begin{bmatrix} \lambda & 1\\ 0 & \lambda \\ \end{bmatrix}$ or to $\begin{bmatrix} \lambda & 0\\ 0 & \lambda \\ \end{bmatrix}$
I did prove that if the geometric multiplicity of $\lambda$ is $2$, then $A$ is similar to $\begin{bmatrix} \lambda & 0\\ 0 & \lambda \\ \end{bmatrix}$, but I'm struggling with the case when $A$ is not diagonalizable. Any thoughts?
Let $x$ be a eigenvector for $\lambda$ and $y$ be any other vector such that $(x,y)$ is a basis. Then, by changing basis you get : $$A \sim \begin{bmatrix}\lambda & a\\0&\mu\end{bmatrix}$$
for some $a,\mu\in\mathbb R$. Since $A$ has only eigenvalue $\lambda$, we have $\mu = \lambda$. By rescaling $y$, we can set $a= 1$.