Find the minimal polynomial of a matrix $A$ that satisfies $A^3 = 25 A$

Question

Let $A$ be a matrix that satisfies $A^3 = 25 A$. Find the form of its minimal polynomial. Is the matrix diagonalizable?

I said that its characteristic polynomial is $$P_A(x) = x^3 - 25x = x(x-5)(x+5)$$ so its minimal polynomial polynomial is $m_A(x) = x(x-5)(x+5)$, but i'm not sure if that's correct.

Answer 1

You cannot conclude that $t^3-25t$ is the characteristic polynomial; in fact that's impossible unless $A$ is $3\times 3$, which is not given.

It's easy to give examples (diagonal matrices) showing that the minimal polynomial can be any non-trivial polynomial that divides $t^3-25t$; since you have a factorization of $t^3-25t$ you can enumerate the seven possibilities if you want.

Something I figured out a few days ago:

(Over any field) Suppose $p$ is a polynomial that factors $p(t)=\prod_{j=1}^n(t-\lambda_j)$, where $\lambda_j\ne\lambda_k$ for $j\ne k$. If $A$ is a square matrix and $p(A)=0$ then $A$ is diagonalizable.

(Of course this is very familiar if $p$ is the characteristic polynomial.)

Proof: The familiar partial-fractions expansion shows that there exist scalars $\alpha_j$ such that$$\frac1{p(t)}=\sum_{j=1}^n\frac{\alpha_j}{t-\lambda_j}.$$(At least that looks familiar from calculus. It was just a few days ago I finally saw a purely algebraic proof; see below)

That says $$\sum\alpha_jp_j(t)=1,$$where $p_j$ is the polynomial such that $$p(t)=(t-\lambda_j)p_j(t).$$So for any vector $x$ we have $$x=\sum x_j,$$where $x_j=\alpha_jp_j(x)$.

It follows that $$(A-\lambda_jI)x=0,$$or $Ax_j=\lambda_jx_j$. So each $x_j$ is an eigenvector of $A$, unless of course $x_j=0$. In any case, each $x_j$ lies in an eigenspace. So the eigenspaces span, hence $A$ is diagonalizable.

Proof of that partial-fractions thing: Since $p_1,\dots,p_n$ have no common factor and the polynomials form a PID there exist polynomials $q_j$ with $$1=\sum q_jp_j.$$There exist polynomials $r_j$ and scalars $\alpha_j$ such that $$q_j(t)=(t-\lambda_j)r_j(t)+\alpha_j.$$ Since $(t-\lambda_j)p_j(t)=p(t)$ this shows that $$1=p(t)\sum r_j(t)+\sum\alpha_jp_j(t),$$or$$1-p(t)\sum r_j(t)=\sum\alpha_jp_j(t).$$

If $\sum r_j\ne0$ then the left side above has degree strictly larger than the right side. So $\sum r_j=0$, hence $\sum\alpha_jp_j=1$.

Exercise Suppose just that $p(t)=\prod(t-\lambda_j)^{n_j}$ and $\deg(d)<\deg(p)$, and show that $$\frac{d(t)}{p(t)}=\sum\frac{d_j(t)}{(t-\lambda_j)^{n_j}}$$with $\deg(d_j)<n_j$.