For a non diagnolizable Matrix $A_{n\times n}$ exists a non zero polynomial $p(t)$ of degree $< n$ s.t. $(p(A))^2=0$

Question

True or False: For a non diagnolizable Matrix $A_{n \times n}$ exists a non zero polynomial $p(t)$ of degree $< n$ such that $(p(A))^2=0$

1. Over $\Bbb C$
2. Over $\Bbb R$

I don't know about $\Bbb R$ but over $\Bbb C$ I think it's false since the minimal polynomial should coincide woth the characteristic polynomial, but I'm not sure about this, I feel I'm missing the point of the question which would make this obvious.

Edit: fixed question to be about non diagnolizable matrix

Answer 1

Let $r(\lambda)$ be the characteristic polynomial of the matrix $A=A_{n\times n}$ over $\Bbb C$, which is non-diagonalizable over $\Bbb C$. Since $A$ is non-diagonalizable, its Jordan normal form is not diagonal, so there exists a multiple root $\lambda_0$ of the polynomial $r(\lambda)$. Let $r(\lambda)=(\lambda-\lambda_0)p(\lambda)$. Then $\deg p(\lambda)=n-1<n$, but $r(\lambda)|p(\lambda)^2$. Since by Hamilton-Cayley theorem $r(A)=0$, $p(A)^2=0$ too.

When the matrix $A$ is over $\Bbb R$, but non-diagonalizable over $\Bbb C$ the proof is similar. Indeed, since the characteristic polynomial $r(\lambda)$ of the matrix $A$ has real coefficients, all its non-real roots can be split into pairs of conjugate numbers. Since $A$ is a non-diagonalizable, there exists a multiple root $\lambda_0$ of the polynomial $r(\lambda)$. If $\lambda_0$ is real put $p(\lambda)= r(\lambda)/ (\lambda-\lambda_0)$, otherwise put $p(\lambda)= r(\lambda)/((\lambda-\lambda_0)(\lambda-\overline{\lambda_0}))$. Remark that in the latter case $\lambda_0\ne\overline{\lambda_0}$, so both $\lambda_0$ and $\overline{\lambda_0}$ are roots of the polynomial p(\lambda). Then $\deg p(\lambda)<n$, but $r(\lambda)|p(\lambda)^2$. Since by Hamilton-Cayley theorem $r(A)=0$, $p(A)^2=0$ too.

For a matrix $A$ over $\Bbb R$ which is non-diagonalizable over $\Bbb R$ the claim is non-necessarily true. Indeed, consider any $2\times 2$ real matrix $A$ whose characteristic polynomial $r(\lambda)$ has no real roots (for instance, we can pick the matrix from Dietrich Burde’s answer, for which $r(\lambda)=\lambda^2+1$). The matrix $A$ is non-diagonalizable over $\Bbb R$, because otherwise its Jordan normal form would be diagonal with real entries, which implies all roots of $r(\lambda)$ are real, a contradiction. If $p(A)^2=0$
and $\deg p<2$ then $p(\lambda)^2$ is divisible by the minimal polynomial $m(\lambda)$ of the matrix $A$, which coincides with $r(\lambda)$, because the roots of $r(\lambda)$ are distinct. But $p(\lambda)^2$ can have at most one distinct (complex) root, a contradiction.

Answer 2

Here is a counterexample for $n=2$, with $$ A=\begin{pmatrix} 0 & 1 \cr -1 & 0 \end{pmatrix} $$ The polynomial $p$ must have degree $<2$, so it is given by $p(x)=rx+s$ for scalars $r,s$, not both zero. Since $A^2=-I$ we have $$ p(A)^2=(rA+sI)^2=2rsA+(s^2-r^2)I $$ This cannot be zero except for $r=s=0$.

This can be generalized to all $n\ge 2$.