Consider a representation $(K^n, J_n(\lambda))$ of an algebra $K[T]$ where $K$ is a field, $\lambda \in \mathbb{R}$ and $J_n$ denotes the $n \times n$ Jordan block.
How can we see that $$End_{K[T]}(K^n,J_n(\lambda)) \cong K[T]/(T^n)?$$
We know that every $f \in End_{K[T]}(K^n,J_n(\lambda))$ must fulfill the property $f(Tx) = Tf(x)$ and if we write $f$ in its Jordan form $J$ then this is equivalent to $JJ_n(\lambda) = J_n(\lambda)J$, which in turn implies that $J$ must be a polynomial of $J_n(\lambda)$. Now, since $(J_n(\lambda)-\lambda I_n)^n = 0$, we then have
$$End_{K[T]}(K^n,J_n(\lambda)) \cong K[T]/(T-\lambda)^n.$$
Is this correct? If so, then the statement should follow from the fact that $K[T]/(T^n) \cong K[T]/(T-\lambda)^n$? Any feedback would be appreciated, thanks!