Jordan canonical form of $p(\alpha)$ in terms of that of $\alpha$ Let $\alpha$ be a linear transformation defined in a finite-dimensional vector space $V$ over a field $F$. If polynomials $p(x)\in F[x]$ are such that for all eigenvalues $\lambda$ of $\alpha$, $p'(\lambda)$ (the formal derivative of $p(x)$ evaluated at $\lambda$) $\neq 0$, then what can we say about the Jordan Form of $p(\alpha)$ with respect to that of $\alpha$?
 A: If $\alpha$ has repeated eigenvalues then $f(\alpha)$ also has repeated eigenvalues and hence its derivative will be zero. So by hypothesis $\alpha$ has distinct eigenvalues and hence is diagonalisable. $f(\alpha)$ has no repeated eigenvalues since its derivative is non zero. Hence it is also diagonalisable. Hence its Jordan form will be diagonal where the diagonal entries are f(a) where “a” is eigenvalue of $\alpha$.
A: I did this:
Lemma: For all $v\in\ V$, $k_v\in\ \mathbb{Z}^+$ s.t. $m_{\alpha,v}(x)=(x-\lambda_v)^{k_v}$. We have $\left< v \right>_{p(\alpha)} = \left< v \right>_{p(\alpha)}$ and $m_{p(\alpha)}(x)=(x-p(\lambda_v))^{k_v}$.
Prerequisite:
Suppose $m_\alpha(x)$ exists and equals to some $\Pi_i (x-\lambda_i)^{n_i}$. We can decompose $V$ into $\bigoplus_i ker(\alpha-\lambda_i\ id_V)^{n_i}$, and the minimal polynomial of $\alpha$ restricted to each kernel are exactly the $(x-\lambda_i)^{n_i}$ (the product of their minimal polynomials is exactly that of $\alpha$). By Primary Cyclic Decomposition, we can further decompose each kernel labeled by $i$ to some $\bigoplus_s \left< v_{i,s} \right>_{\alpha}$. We have the minimal polynomial of each $v_{i,s}$ divides that of the kernel it belongs to, thus $\exists k_{i,s}<n_i,\ m_{\alpha, v_{i,s}}(x)=(x-\lambda_i)^{k_{i,s}}$. We can see that $$\mathcal{B}_{i,s}=\{ (\alpha-\lambda_{i,s}id_V)^{k_{i,s}-1}(v_{i,s}),\ (\alpha-\lambda_{i,s}id_V)^{k_{i,s}-2}(v_{i,s}),\ ...\ , (\alpha-\lambda_{i,s}id_V)(v_{i,s}),\ v_{i,s}\}$$ is an ordered basis of $\left< v_{i,s} \right>_{\alpha}$, and the matrix representation of $\alpha$ restricted to $\left< v_{i,s} \right>_{\alpha}$ is exactly the Jordan matrix $J_{k_{i,s}}(\lambda_i)$. This is how we derive the Jordan Canonical Form.
Then:
By Lemma, for $p(\alpha)$, we can decompose $V$ by the same set of $\{v_{i,s}\}_{i,s}$. Also, the minimal polynomials of $p(\alpha)$ restricted to $\left< v_{i,s} \right>_{p(\alpha)}$ is that of $\alpha$, but replacing $\lambda_{i}$ with $p(\lambda_{i})$. The same derivation follows and we have that the Jordan Form of $p(\alpha)$ is that of $\alpha$, replacing the diagonal $\lambda_{i}$ with $p(\lambda_{i})$.
