Given a matricial expression, the eigenvalues are restricted to the polynomial solutions?

Question

Given a matricial expression, like $A^2-4I=0$, I want to know if it is true that the eigenvalues of $A$ are restricted to the solutions of the polynomial $x^2-4=0$, so $x \in \{-2,2\}$.

With Cayley-Hamilton, I know that with the caracteristical polynomial $x^2-4=0$, I would get that the eigenvalues are $-2$ and $2$. I know it is not true given that $A^2-4I=0$, because I could have $A=2I$, and the eigenvalues are only $2$, but $A$ could also be $-2I$ and the eigenvalues are $-2$. I want to show that no other number could be eigenvalue, so $x \in \{-2,2\}$.

What I am trying to prove looks like the other way of the Cayley-Hamilton theorem, I guess it is true, but I am unable to prove.

Answer 1

Do you know what the minimal polynomial of a matrix is? If you do then it's very simple: the eigenvalues are exactly the roots of the minimal polynomial which we denote that $m$. And by definition of minimal polynomial we know that if a polynomial $f$ (over the same field) satisfies $f(A)=0$ then $m|f$, and hence the eigenvalues must be roots of $f$. ($f$ might have other roots as well, but the eigenvalues are always roots of $f$)

Answer 2

If you have a polynomial $p$ that annihilates the matrix $A$, then every eigenvalue of $A$ must be a root of $p$. Let $\mathbf v$ be an eigenvector of $A$ with eigenvalue $\lambda$. Clearly, $A^k\mathbf v=\lambda^k\mathbf v$, and so $$p(A)\mathbf v = p(\lambda)\mathbf v=0,$$ but $\mathbf v\ne 0$, therefore $p(\lambda)=0$. The converse does not hold, as you’ve already shown: the eigenvalues of $A$ must be roots of $p$, but not all roots of $p$ need be eigenvalues of $A$.