Let $C \in \mathbb{R}^{4 \times 4}$ satisfy $C^3 + 6C = 5C^2$. Prove that $C$ is diagonalizable over the real numbers

Question

Question:

Let $C \in \mathbb{R}^{4 \times 4}$ satisfy $C^3 + 6C = 5C^2$. Prove that $C$ is diagonalizable over the real numbers.

My Attempt:

If $C$ satisfies $C^3 + 6C = 5C^2$ then $C$ is a root of the polynomial $p(x) = x^3 - 5x^2 + 6x = x(x-2)(x-3)$. The minimal polynomial $m(x)$ for $C$ then must divide $p(x)$. Hence

$$ m(x) \in \{x, x-2, x-3, x(x-2), x(x-3), (x-2)(x-3), x(x-2)(x-3) \}. $$

In any of these instances, the minimal polynomial for $C$ splits into linear factors over $\mathbb{R}$, and each of the possibilities for $m(x)$ has distinct roots. This implies that the matrix $C$ will always have distinct eigenvalues, and that $C$ can be diagonalized.

Follow up question:

How does the value of $n$ in $\mathbb{R}^{n \times n}$ affect the proof of this solution? For instance, if $C$ were a 5 by 5 matrix instead, would the proof still hold?

Answer 1

Your solution is correct. Even more concisely, I think you could just say that since $p$ splits into distinct linear factors over $\mathbb{R}$, any divisor of $p$ does too (no need to explicitly list out all the possibilities for $m$).

Indeed the value of $n$ does not matter! The same argument works for $C \in \mathbb{R}^{n \times n}$ for any $n$.

Answer 2

As long as an annihilating polynomial $a(x)=x(x-1)(x-2)$ splits into linear factors, the minimal polynomial too contains linear factors and hence the matrix $C$ is diagonalizable for all $n$.