# Proof review: For $A$ a non-diagonizable matrix, there exists a polynomial of degree $n-1$ s.t. $[p(A)]^ 2 = 0$

Proof review: For $A$ a non-diagonizable matrix, there exists a polynomial of degree $n-1$ s.t. $[p(A)]^ 2 = 0$.

Of course, $A$ is an $n \times n$ matrix over field $F$.

Proof: Let's look at $J$, $A$'s Jordan form. If I can find such a polynomial that will satisfy $p(J)^2 = 0$, then we're done.

$J$ isn't a diagonal matrix, therefore there exists at least one block of size that's bigger than 1. If I could annihilate the diagonal and take the matrix to the $n-1$ power, I can get a matrix of zeros except for the $n \times 1$ entry. Using this line of thinking, I've come up with the polynomial $p(t) = \sum_{i = 1}^{i = k}(t-a_i)^{n-1}-(k-1)t^{n-1}$.

Here $a_i$ are the different eigenvalues (there are $k$ of them).

This seems to work.. is this correct? What if the field is $F = \mathbb{R}$. I want a polynomial over the same field but can't figure out the trick. Any help would be appreciated.

• The proposition seems to be false; consider $\tbinom{0\ -1}{1\ 0}$ over $\Bbb{R}$. Is $F$ supposed to be algebraically closed, and $n>1$? – Servaes Nov 10 '15 at 15:19
• The example you gave seems diagonizable. – John H Nov 10 '15 at 15:21
• More generally, if there are no eigenvalues (over $F$), then $p(t)=t^{n-1}$, implying that $A$ must be nilpotent. But there are plenty of counterexamples, like the one I gave in the comment above. – Servaes Nov 10 '15 at 15:26
• Good point, I wonder if the meaning in the question was "not daigonizable over any field" or "not diagonizable over $F$"... – John H Nov 10 '15 at 15:28
• A necessary condition is that $A$ is not diagonalisable over any algebraic closure of $F$. This is different from "not diagonalisable over any field", which I also find somewhat vague. For example $\tbinom{1\ 2}{0\ 1}$ isn't diagonalisable over $\Bbb{C}$, but it is over $\Bbb{F}_2$, with some suitable interpretation. – Servaes Nov 10 '15 at 15:31

Theorem: A matrix is diagonalisable over $F$ if and only if its characteristic polynomial is a product of distinct linear factors over $F$.