# Eigenvalues of a polynomial of a matrix

I have two polynomials, $p$ and $q$, and a square matrix $A$. I also have that $q(A)$ is invertible. I want to prove the following statement.

Suppose that $\mu$ is an eigenvalue of $p(A)q(A)^{-1}$. Then $A$ has an eigenvalue $\lambda$ such that $\mu =\frac{p(\lambda)}{q(\lambda)}$.

I have gotten this far:

Since $q(A)$ and $p(A)$ commute, $p(A)q(A)^{-1} = q(A)^{-1}p(A)$. If $\mu$ is an eigenvalue of $p(A)q(A)^{-1}$, then I can write

$$p(A)q(A)^{-1}v = \mu v$$ for some $v \neq 0$. Hence, $q(A)^{-1}p(A)v = \mu v$, which would then imply $r(A)v = 0$ where $r(x) = p(x) -\mu q(x)$ is another polynomial. So $(0,v)$ is an eigenpair for $r(A)$. Not sure how to carry this forward. I thought about minimal polynomials but I don't think they are of any use here.

Maybe I could also use the fact that I can find two matrices in Jordan form, $P$ and $Q$, and an invertible matrix $T$, such that $p(A) = TPT^{-1}$ and $q(A) = TQT^{-1}$ thanks to $p(A)$ and $q(A)$ commuting. This gives right to the fact that

$$QPw = PQw = \mu w$$ where $w = T^{-1}v$. The fact that $(\mu,w)$ is an eigenpair for both $QP$ and $PQ$ seems noteworthy but I don't see how it is useful here.

I would very much appreciate seeing different kinds of answers; some based on results for finite dimensional spaces and some based on operator theory or whatever abstraction level is appropriate.

• Do you mean $q (A^{-1})$ or $[q (A)]^{-1}$? – user May 23 '18 at 18:43
• @user The second one. – Calculon May 23 '18 at 18:44
• How do you know that $p(A)$ and $q(A)$ commute? – Sean Roberson May 23 '18 at 19:19
• @SeanRoberson $A$ commutes with itself and all its powers. – Calculon May 23 '18 at 19:20

## 2 Answers

Assume you are working over the complex numbers. The spectrum $\sigma(A)$ consists of the eigenvalues of $A$. The spectral mapping theorem states that $$\sigma(p(A))= p(\sigma(A)).$$ For the given problem, $\mu$ is an eigenvalue of $p(A)q(A)^{-1}$ iff $\mu I-p(A)q(A)^{-1}$ is singular, which is true iff $\mu q(A)-p(A)$ is singular because $q(A)$ is assumed to be invertible. This in turn is equivalent to knowing that $\mu q(\lambda)-p(\lambda)$ maps one of the eigenvalues $\lambda$ of $A$ to $0$, by the Spectral Mapping Theorem. So $\mu q(\lambda)-p(\lambda)=0$ for some eigenvalue $\lambda$ of $A$. Because $q(A)$ is invertible, then $q(\lambda)\ne 0$ for every eigenvalue of $A$. Hence, $$\mu = \frac{p(\lambda)}{q(\lambda)}$$ for some $\lambda\in\sigma(A)$.

Let $B=q(A)$. Then $B^{-1}=h(B)$ for some polynomial $h$. (See a proof here.)

Therefore, $q(A)^{-1}=h(q(A))$ and $p(A)q(A)^{-1}=f(A)$ for $f(x)=p(x)h(q(x))$, also a polynomial.

The result is now well known.

You need to assume that the scalar field is algebraically closed or at least that $A$ is triangularizable.

• Thank you. Your first remark follows from Cayley-Hamilton, right? – Calculon May 24 '18 at 9:49
• @Calculon, yes, but not necessarily. See math.stackexchange.com/a/2785699/589. I've edited my answer. – lhf May 24 '18 at 11:45