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Let $V$ be a finite-dimensional vector space over $\mathbb{C}$, and let $S$ and $T$ be linear operators on $V$.

(a) Prove that, if $S$ is invertible, then $T$ and $STS^{-1}$ have the same set of eigenvalues.

(b) Prove that, if $ST-TS = S$, then $S$ is not invertible. Hint: Consider the eigenvalue of $T$ with the largest real part.

Would appreciate a proof check for (a) and some help for (b).

(a) Since $V$ is a complex vector space, $T$ has at least 1 eigenvalue, say $\lambda$. Let $u$ be an eigenvector of $T$ corresponding to $\lambda$. Now, consider the image of $Su$ under $(STS^{-1})$. We have,

\begin{align*} STS^{-1}(Su) = STu = S\lambda u = \lambda Su, \end{align*} so $Su$ is an eigenvector of $STS^{-1}$ with eigenvalue $\lambda$.

(b) I'm not quite sure how the hint comes into play here. My first idea was to assume to get a contradiction that $S$ was invertible. Then $S^{-1}$ exists. By the same reason in (a), $T$ has an eigenvalue $\lambda$ and corresponding eigenvector $u$. Now consider the image $S^{-1}u$ under $ST-TS$. We have,

\begin{align*} (ST-TS)S^{-1}u = STS^{-1} u-Tu = (STS^{-1}-\lambda I)u\end{align*}.

However, this is also equal to $Su$, which cannot be 0 (again by the same reason in (a)) but $(STS^{-1}-\lambda I)$ must be 0 since $T$ and $STS^{-1}$ share eigenvalues, a contradiction.

This seems like it is correct to me, but my solution doesn't use the hint for part (b). Am I missing something?

Update: One can prove an even stronger version of (a), that being that $T$ and $STS^{-1}$ has the same characteristic polynomial as $T$ hence the same multiset of eigenvalues.

Proof: Starting with the characteristic polynomial of $STS^{-1}$, we see, \begin{align*} \det(STS^{-1}-\lambda I) = \det(S(T-\lambda I)S^{-1}) = \det(S)\det(T-\lambda I)\det(S^{-1})= \det(T-\lambda I), \end{align*} as desired.

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    $\begingroup$ $(STS^{-1}-\lambda I)u$ need not be $0$. $\lambda$ is an eigen value of $STS^{-1}$ but $u$ need not be an eigen vector corresponding to this eigen value. $\endgroup$ Jan 5 at 7:43
  • $\begingroup$ Ah good catch. So there is definitely something missing here... $\endgroup$ Jan 5 at 7:47
  • $\begingroup$ For the first part, you also need to show that all of the eigenvalues of $STS^{-1}$ are also eigenvalues of $T$. $\endgroup$
    – Seeker
    Jan 5 at 8:22
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    $\begingroup$ Yes. Aside from some parenthesis nits that's the typical argument for (a)-- best to put it in the OP with an 'update' note. So re-consider $STS^{-1}=T+I $$\implies \det(T-\lambda I)=\det(STS^{-1}-\lambda I)=\det(T+I-\lambda I)=\det(T+(1-\lambda) I)$. My approach is to look at the trace. Your book's hint is if $\lambda'$ is an eigenvalue of $T$ such that $a'=Re(\lambda')$ is the max real part amongst all eigenvalues, then the Left Hand Side (LHS) is $\neq 0$ for any $\lambda\mapsto z$ when $Re(z)\gt a'$ by definition, but the RHS $=0$ at $\lambda\mapsto\lambda' +1$ which is a contradiction. $\endgroup$ Jan 12 at 21:55
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    $\begingroup$ @user8675309 okay great this has helped a lot. I appreciate it! $\endgroup$ Jan 13 at 2:31

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