Find the possibilities for the minimal polynomial of an operator $T: \mathbb R^5 \to \mathbb R^5$ with characteristic polynomials:
a)$p_T(x) = (x-3)^3(x-2)^2$
b)$p_T(x) = (x-1)(x-2)(x-3)(x-4)(x-5)$
c)$p_T(x) = (x-1)^m$, with $m \ge 1$
Is it possible to conclude that any of them is necessarily diagonalizable?
I don't know if I understood this exercise correctly. But this is my attempt
a)$p_{T_1} = (x-3)(x-2)$
$p_{T_2} = (x-3)(x-2)^2$
$p_{T_3} = (x-3)^2(x-2)$
$p_{T_4} = (x-3)^3(x-2)$
$p_{T_5} = (x-3)^2(x-2)^2$
$p_{T_6} = (x-3)^3(x-2)^2$
b) $p_T(x) = (x-1)(x-2)(x-3)(x-4)(x-5)$
c) $p_{T_1} = (x-1)$ . . . $p_{T_1} = (x-1)(x-1)...(x-1) = (x-1)^m $
I need the matrix to say if it's diagonalizable right? But if it is, it needs to be
a)$p_{T_1} = (x-3)(x-2)$
b) $p_T(x) = (x-1)(x-2)(x-3)(x-4)(x-5)$
c)$p_{T_1} = (x-1)$
Thanks.
Pretend the operator is in jordan canonical form with real eigen values (roots of your characteristic polynomial) and look at connection of jordan canonical to minimial polynomial (degree of a root in minimal polynomial = size of largest jordan block with the root as eigen value). This will give you clarity although what you are doing is pretty much the answer except problem c). what you did with problem a,b are correct. I am not fully clear on what is $m$ in problem $c$.
But if you are looking for minimal polynomial of diagonalizable linear operators then since the roots of characteristic polynomial is real, minimal polynomial is of degree $1$ in each eigenvalue or root of characteristic polynomial.