Exercise 5.B.7(b) in “Linear Algebra Done Right” 4th edition by Sheldon Axler

Question

The following exercise is part (b) of exercise number 7 in Sheldon Axler's Linear Algebra Done Right, 4th edition:

Suppose $V$ is finite-dimensional and $S, T \in \mathcal{L}(V)$. Prove that if at least one of $S, T$ is invertible, then the minimal polynomial of $ST$ equals the minimal polynomial of $TS$.

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This is what I have attempted so far:

Let's assume $S$ is invertible. Let $p$ be the minimal polynomial of $ST$ and let $q$ be the minimal polynomial of $TS$. This implies that

$$
p(ST) = 0 \ \ \text{and} \ \ q(TS)= 0
$$

Using the fact that $p(TS) = S^{-1}p(ST)S$ (easy to show), we have that $p(TS) = 0$. This implies that $p$ is a polynomial multiple of $q$. Thus, there exists a nonzero (since $p$ is monic) $r \in \mathcal{P}(\mathbb{F})$ such that

$$
p = rq
$$

Also, $q(TS) = S^{-1}q(ST)S$. Since $q(TS)=0$, we use the invertibility of $S$ to get that $q(ST) = 0$. Hence, $q$ is a polynomial multiple of $p$. Thus, there exists a nonzero (since $q$ is monic) $s \in \mathcal{P}(\mathbb{F})$ such that

$$
q = sp
$$

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Question.

I've been stuck here for a while. I need to somehow show both $r$ and $s$ equal $1$, but I can only show that $rs=1$, which doesn't really help. I tried so show $p \neq q$ would lead to a contradiction, but I wasn't able to. Any suggestions or hints?

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A note on notation:

• $\mathbb{F}$ is either the set of real or complex numbers.

• $\mathcal{P}(\mathbb{F})$ is the infinite-dimensional vector space of
  polynomials.

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Similar questions, but different focus:

• When are the minimal polynomials of $ST$ and $TS$ not equal for linear operators $T$ and $S$?
• If $m_1,m_2$ are the minimal polynomials of $ST$ and $TS$ prove $m_2(x)=x^im_1(x)$ where $i=-1,0$ or $1$

Answer 1

You seem to have essentially solved the problem, but a slightly more straight-forward way to think about it is the following:

If $A$ is a linear map over an $\mathbb F$-vector space, then $m_A$, the minimal polynomial for $A$, is the monic polynomial characterised by the fact that if $f(A)=0$ then $m_A$ divides $f$. That is $I_A: = \{f(t)\in \mathbb F[t]: f(A)=0\}$, the set of polynomials which $A$ satisfies, is precisely the set of polynomials which are divisible by $m_A$.

Now the statement of the problem is symmetric in $S$ and $T$, so we may as well assume that $S$ is invertible. But then $ST = S(TS)S^{-1}$ so that if $p(ST)=0$ then $Sp(TS)S^{-1}=0$, and hence $p(TS)=0$. On the other hand, since $TS = S^{-1}(ST)S$, if $q(TS)=0$ then $S^{-1}q(ST)S=0$ so that $q(ST)=0$. It follows that $p(ST)=0$ if and only if $p(TS)=0$, that is $I_{ST} = I_{TS}$. But then it follows immediately that $ST$ and $TS$ has the same minimal polynomial.