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.
I am unable to follow your argument; what do you mean by 'If I could annihilite the diagonal', for example?
My first thoughts would be to use the following theorem:
Theorem: A matrix is diagonalisable over $F$ if and only if its characteristic polynomial is a product of distinct linear factors over $F$.