Suppose $T \in L(V)$. Suppose $S \in L(V)$ is invertible.
Prove that $T$ and $S^{-1}TS$ have the same eigenvalues.
What is the relationship between the eigenvectors of T and the eigenvectos of $S^{-1}TS$?
I started by saying suppose $\lambda$ is an eigenvalue of T such that $T(v) = \lambda v$
Show $S^{-1}TS\,(v) = T(v) = \lambda v$
I have no idea how to proceed with the rest of the problem though, help?
Update: Can anyone help me figure out how the eigenvectors of these two relate? All I can get from this is that just because they have the same eigenvalues doesn't mean the eigenvectors are the same.
Best Answer
Hint: If $v$ is an eigenvector of $T$, then $v$ is not necessarily an eigenvector of $S^{-1}TS$. So instead, try to show that $S^{-1}v$ is an eigenvector of $S^{-1}TS$ with the same eigenvalue $\lambda$.