Let the Jordan block of size $n$ and parameter $\lambda$ be $J_n(\lambda)$ $$J_n(\lambda) = \begin{pmatrix} \lambda & 1&&&& \\ & \lambda& 1 && \\ & & \ddots & \ddots & \\ & & & \lambda&1 \\ & & & &\lambda \end{pmatrix}$$
Show that the dimension of the centralizer of $J_n(\lambda)$ is $n$.
Through working this out explicitly for small $n$, I've worked that the centralizer for $J_n(\lambda)$, $C$, is of the form $$\begin{pmatrix} a & b & \ddots &m & n\\ & a& b & \ddots & m\\ & & \ddots & \ddots & \ddots \\ & & & a&b \\ & & & &a \end{pmatrix}$$ where $C$ is upper triangular where the entries in each diagonal are equal to one another. However I'm unable to formulate a proof for this. I think splitting up $J_n(\lambda)$ to $\lambda I_n + N$ might be useful, since if C centralizes $N$, and it centralizes $I_n$, then consequently $J_n(\lambda)$ as well. But I'm stuck as to proving it