# Equation of Linear Operators

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.

• $(STS^{-1}-\lambda I)u$ 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. Jan 5 at 7:43
• Ah good catch. So there is definitely something missing here... Jan 5 at 7:47
• For the first part, you also need to show that all of the eigenvalues of $STS^{-1}$ are also eigenvalues of $T$. Jan 5 at 8:22
• 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. Jan 12 at 21:55
• @user8675309 okay great this has helped a lot. I appreciate it! Jan 13 at 2:31