Two linear operators in a complex Euclidean space have a common eigenvector

Asked Mar 25 '21 at 03:52

Active Mar 25 '21 at 03:52

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The question is the following:

Let $E$ a complex Euclidean space. Let $S,T:E \rightarrow E$ linear operators with $ST=TS$. Show that $ST$ has a common eigenvector.

My idea is to use that fact that:

$ST=TS \Longleftrightarrow S$ autospaces are invariant by $T$. I know that means it exists a orthonormal base of $E$ that is made by the eigenvectors of $S$ and $T$. How can I guarantee there is a common one?

asked Mar 25 '21 at 03:52

2BDrB

3

Do you mean : "show that $S$ , $T$ have a common eigenvector" above? Does this answer your question? Does this also answer your question? – Sarvesh Ravichandran Iyer Mar 25 '21 at 04:09
1

In the textbook I took the question from it says "Show that $ST$ has a common eigenvector". I also thought it sounded weird. Is it unanswerable like this? – 2BDrB Mar 25 '21 at 04:50
Definitely sounds weird, I mean, something can be common amongst two or more things, but there's only one thing i.e. $ST$ in there. I'm sure that $S,T$ is meant there. Typo! – Sarvesh Ravichandran Iyer Mar 25 '21 at 04:51
@2BDrB Yes, the question is unanswerable unless you interpret the last $ST$ as $S,T$. – Ben Grossmann Mar 25 '21 at 14:58

Two linear operators in a complex Euclidean space have a common eigenvector

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