Prove that $T$ and $S^{-1}TS$ have the same eigenvalues

Question

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.

Answer 1

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 $\lambda$.

Answer 2

Another approach:

$det(A-\lambda I_n)=det(S^{-1})\cdot det(A-\lambda I_n)\cdot det(S)=det(S^{-1}(A-\lambda I_n)S)=det(S^{-1}AS- S^{-1}\lambda S)=det(S^{1}AS-\lambda I_n) $

Answer 3

Set $U=S^{-1}TS$, so that $SU=TS$. If $v$ is an eigenvector for $U$ relative to $\lambda$, then $$ TSv=SUv=S(\lambda v)=\lambda Sv, $$ which means that $Sv$ is an eigenvector of $T$, after noting that $v\ne0$ implies $Sv\ne0$.

Similarly, from $US^{-1}=S^{-1}T$ we can deduce that if $w$ is an eigenvector for $T$, $S^{-1}w$ is an eigenvector for $U$ with respect to the same eigenvalue.

Note that, setting $E_U(\lambda)=\{v:Uv=\lambda v\}$ and $E_T(\lambda)=\{w:Tw=\lambda v\}$, we can define \begin{align} &f\colon E_U(\lambda)\to E_T(\lambda),\qquad v\mapsto Sv\\ &g\colon E_T(\lambda)\to E_U(\lambda),\qquad w\mapsto S^{-1}w \end{align} which are linear maps inverse to one another. As a consequence, the operators $T$ and $U$ have the same eigenvalues, with the same geometric multiplicity, when $V$ is finite dimensional.

The fact that eigenvalues of $T$ and $U$ are the same follows also, when $V$ is finite dimensional, from considering that the two operators have the same characteristic polynomial. This has the consequence that also the algebraic multiplicity is the same.

Answer 4

Sorry for bumping up an old problem, but I just encountered this problem while reviewing for final exam, and here is my attempt. I think it is the most straightforward because it follows from theorem 8.29 of Axler that states operators on complex vector spaces have block diagonal matrix representation with upper triangular blocks, with eigenvalues on the diagonal.

Proof: By theorem 8.29, there exists some basis in which matrix of $T$, denoted as $A$, is block diagonal as described above. Thus, $A=P^{-1}TP$ for some invertible matrix $P$. Now since $S$ is invertible, we can also write $A=(P^{-1}S)S^{-1}TS(S^{-1}P)=(S^{-1}P)^{-1}(S^{-1}TS)(S^{-1}P)$. Hence, $T$ and $S^{-1}TS$ have the same block diagonal matrix representation with upper triangular blocks, with eigenvalues on the diagonal as Axler has shown in 8.29. Therefore, we conclude that these two operators have same eigenvalues with same multiplicities by just reading off the diagonal entries. Done!