Generalized eigenvalue space of $P(\lambda)$ and $\lambda$ Suppose $A$ is a linear transformation on a finite dimensional vector space, we know that the  eigenvalue of $P(A)$ is exactly $P(\lambda)$, where $\lambda$ is the eigenvalue of $A$, $P$ is a polynomial, my question is if the generalized eigenvalue space of $P(\lambda)$ and $\lambda$ are the same? If not, is it 
 A: To ensure that all eigenvalues of $P[A]$ correspond to an eigenvalue of $A$, one must first of all assume that the characteristic polynomial of $A$ splits into linear factors, i.e., the spectrum of $A$ contains the all eigenvalues that exist in an algebraic closure of the field. Then one has

For every eigenvalue $\lambda$ of a linear operator$~A$ and for any polynomial$~P$, the generalised eigenspace of$~A$ for$~\lambda$ is contained in the generalised eigenspace of$~P[A]$ for its eigenvalue$~P[\lambda]$. The generalised eigenspace of $P[A]$ for some eigenvalue$~\mu$ is the direct sum of the generalised eigenspaces of$~A$ for those eigenvalues$~\lambda$ with $P[\lambda]=\mu$.

Note in particular that there is no equality of generalised eigenspaces in general.
The proof is simple: that generalised eigenspace of$~A$ consists by definition of the vectors in the kernel of $\def\id{\,\mathrm{id}}(A-\lambda\id)^k$ for some $k\in\Bbb N$, while the mentioned generalised eigenspace of$~P[A]$ consists of the vectors in the kernel of $(P[A]-P[\lambda]\id)^k$ for some $k\in\Bbb N$. But since $\lambda$ is clearly a root of $P-P[\lambda]$, the latter polynomial is divisible by $X-\lambda$, say $P-P[\lambda]=(X-\lambda)Q$, and then $(P-P[\lambda])^k=(X-\lambda)^kQ^k$ is a multiple of $(X-\lambda)^k$, and any vector annihilated by $(X-\lambda)^k[A]=(A-\lambda\id)^k$ is also annihilated by $(P-P[\lambda])^k[A]=(P[A]-P[\lambda]\id)^k$, which is the desired inclusion.
By hypothesis, the (direct) sum of the generalised eigenspaces for all eigenvalues$~\lambda$ of$~A$ fills the whole space. But each one is contained in the generalised eigenspace for the corresponding eigenvalue$~P[\lambda]$ of$~P[A]$, so the sum of those generalised eigenspaces for all occurring values$~P[\lambda]$ also fills the whole space, and those values constitute the entire spectrum of$~P[A]$. Obviously the generalised eigenspace for$~\mu$ is given by the mentioned direct sum.
