I need help in Linear Algebra Done Right, section 5.A, problem 15a:
Suppose $T \in L(V)$. Suppose $S \in L(V)$ is invertible.
(a) Prove that $T$ and $S^{-1}TS$ have the same eigenvalues.
Note: In this chapter, all vector spaces are assumed to be finite-dimensional over $\mathbb R$ or $\mathbb C$.
I've found that if $x$ is an eigenvector of $T$ corresponding to eigenvalue $\lambda$, $S^{-1}x$ is an eigenvector corresponding to eigenvalue $\lambda$, so all eigenvalues of $T$ are eigenvalues of $S^{-1}TS$, but I need help proving in the other direction.
My Proof:
Because $Tx=\lambda x$, $S^{-1}TS(S^{-1}x)=S^{-1}Tx=S^{-1}\lambda x=\lambda S^{-1}x$. Since $S^{-1}$ is invertible, $S^{-1}x\neq 0$. So $S^{-1}x$ is an eigenvector of $S^{-1}TS$.
Could you help me?
Best Answer
You have proven that any eigenvalue of $T$ is an eigenvalue of $S^{-1}TS$. It remains to go the other way.
Hint: $x\mapsto S^{-1}x$ successfully took you one way. It would make sense that $y \mapsto Sy$ takes you the other way. (You may need to rewrite $T$ a little.)