Question

This question has been bugging me for at least a month.

Suppose \(S,T\in\mathcal{L}(V)\), where \(V\) is a finite dimensional vector space over \(\mathbb{C}\) or \(\mathbb{R}\). Prove that \(ST\) and \(TS\) have the same eigenvalues.

Equivalently, suppose \(S\) and \(T\) are two square matrices with entries in \(\mathbb{C}\) or \(\mathbb{R}\), prove that \(ST\) and \(TS\) have the same eigenvalues.

Answer 1

In case anyone is wondering, this question is directly from Linear Algebra Done Right. A determinant free proof is therefore preferred.

Hi ganeshie8!

Answer 2

Hey!

If \(\rm{\lambda}\) is an eigenvalue of \(A\), then we have :
\[A\rm{x}=\lambda \rm{x}\]
for some vector \(\rm{x}\)

Answer 3

If \(\rm{\lambda}\) is an eigenvalue of \(ST\), then we have :
\[(ST)\rm{x}=\lambda \rm{x}\]
for some vector \(\rm{x}\)

Answer 4

Since matrix multiplication is associative, we can rearrange it as :

\[S(T\rm{x})=\lambda \rm{x}\]

Answer 5

Left multiply \(T\) both sides and it becomes easy to conclude...

Answer 6

.......\[
\begin{align*}
STx&=\lambda x \\
S(Tx)&=\lambda x\\
TS(Tx)&=\lambda (Tx)
\end{align*}\]I feel ashamed...

Answer 7

Happens haha ;)
Interesting to see that Tx = x if S and T commute !

Answer 8

Why if S and T commute Tx=x?

Answer 9

ST = TS

so they are same, they have same eigenvalues, same eigenvectors

Answer 10

In our proof, we showed this :

If \((\lambda, x)\) is an (eigenvalue, eigenvector) pair of ST,
then \((\lambda, Tx)\) will be an (eigenvalue, eigenvector) pair of TS.

Answer 11

When ST = TS, these are same matrices, eigenvectors and everything must be same for ST and TS.

so, \(x = Tx\)

Answer 12

But by reversing the proof we can conclude that x=Sx? But then S=T=Identity matrix?

Answer 13

\[
\begin{align*}
STx&=\lambda x\\
TS(Tx)&=\lambda Tx\\
ST(Tx)&=\lambda Tx
\end{align*}
\]
Combining line 1 and line 3 we can say that both \(x\) and \(Tx\) are eigenvectors of the same eigenvalue, but can we say that \(x=Tx\)?

Answer 14

@ganeshie8 plz don't leave lol

Answer 15

We cannot say x = Tx if S and T don't commute

Answer 16

But can we say x=Tx if T and S commute. I am sure that we can say Tx and x are two eigenvectors corresponding to the same eigenvalue but x=Tx?

Answer 17

I think I see the issue. My earlier statement looks silly haha. We may massage it a bit and say \(x \propto Tx\) if all the eigenvalues are distinct.

Answer 18

Lol. Yes if all eigenvalues are distinct what you said after massaging is true. If eigenvalues are not distinct it is really hard to say anything about it. Non-distinct eigenvalues are generally hard to dealt with comparatively.

Answer 19

Let me close this and open another linear algebra question.