Non diagonalizable matrix Suppose $A \in M^{F}_{n \times n}$ is non diagonalizable.
Is there a polynomial $P(t) \in F[t]$ of degree $n-1$ such that $P(A)^2=0$ when:


*

*$F=\mathbb{R}$

*$F=\mathbb{C}$


How do I approach this question?
 A: The answer is in the comments already made, but perhaps some expansion would not be amiss. 
$A$ has $n$ eigenvalues (in $\bf C$). If they are all distinct, then $A$ is diagonalizable, so we may assume there is a repeated eigenvalue, hence at most $n-1$ distinct eigenvalues. 
Let $Q$ be the characteristic polynomial of $A$. The ring of polynomials with coefficients in $F$ is a unique factorization domain, so $Q=R^2S$ for some polynomials $R$ and $S$ with coefficients in $F$, where $S$ has no repeated factors. We can't have $R$ of degree zero, since that would mean $Q$ has no repeated factors, which we have already ruled out. Now $P=RS$ is the polynomial you're looking for. 
A: For $F=\mathbb R$, the solutions doesn't work because repeated eignevalues are not the only thing that stands in the way of diagonalizability.  Indeed, you cannot diagonalize unless the eigenvalues are real.
For example the rotation matrices $R_{\theta}=\pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta}$ with $\theta \neq 0,\pi$ give a counterexample to the claim.

In general, if suppose that $A$ is an $n\times n$ real matrix with unique eigenvalues, some of which are complex, and let $D=SAS^{-1}$ be a diagonalization of $A$ over $\mathbb{C}$ (there can't be one over $\mathbb{R}$ because there are complex eigenvalues).  Let $\lambda_1, \ldots, \lambda_n$ the the diagonal entries of $D$ (the eigenvalues of $A$).  Then for any polynomial $P$, we have $P(D)$ is diagonal with entries $P(\lambda_1), \ldots, P(\lambda_n)$.  Because $P(D)$ is conjugate to $P(A)$, one vanishes if and only if the other does.
So suppose that $P^2(A)=P^2(D)=0$.  Then each $\lambda_i$ would be a root of $P^2$, and hence a root of $P$.  Therefore, $P$ would have to have degree at least $n$.
