If $S,T$ are linear operators on $V$ over $\mathbb{C}$ and $ST = TS$ show there exist one common eigenvalue of $S$ and $T$

Question

I think I have to show that $Sx = \lambda_S x$ and $Ty = \lambda_T y$ with $\lambda_S = \lambda_T$? I tried doing $TSx = T\lambda_S x = STx \implies S(Tx)= \lambda_S Tx$

and $STy = S\lambda_Ty = TSy \implies T(Sy) = \lambda_T Sy$

but this isn't leading anywhere. Can I get a hint or something?

Both the above comment and answer below provide counterexamples to your statement. — Sangchul Lee, Jul 14 '20 at 03:32

score 1 · Answer 1 · answered Jul 14 '20 at 03:16

1

It's not true, as you can see if $V$ is one-dimensional, in which case all linear operators commute with each other.

answered Jul 14 '20 at 03:16

Robert Shore

23,332

Assume over $\mathbb{C}$ – Lemon Jul 14 '20 at 03:18
4

It should be a 'common eigenvector' and in that case mse has an answer already here https://math.stackexchange.com/questions/1227031/do-commuting-matrices-share-the-same-eigenvectors – ShBh Jul 14 '20 at 03:19
You are right. I completely misread the original question, that's why I had so much trouble with this question. – Lemon Jul 14 '20 at 03:52

If $S,T$ are linear operators on $V$ over $\mathbb{C}$ and $ST = TS$ show there exist one common eigenvalue of $S$ and $T$

1 Answers1