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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.

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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 $\lambda$.

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  • $\begingroup$ So then I can try to show that $S^{-1}(TS) = \lambda (TS)$? What I tried to do is say $S^{-1}(T(v)S) = S^-1 \lambda S = \lambda S^{-1} S = \lambda.$ Is that ok to do?? $\endgroup$ – Soaps Jul 26 '14 at 22:53
  • $\begingroup$ First, use S^{-1} to get it to display as $S^{-1}$. To show that $S^{-1}v$ is an eigenvector of $S^{-1}TS$, you need to show that $(S^{-1}TS)(S^{-1}v) = \lambda(S^{-1}v)$. Can you do this? Remember that if $v$ is an eigenvalue of $T$, then $Tv = \lambda v$. $\endgroup$ – JimmyK4542 Jul 26 '14 at 22:55
  • $\begingroup$ Yeah I can show that, so that would sufficient to say that $\lambda$ is an eigenvalue of $S^{-1}TS$ with corresponding eigenvector $S^{-1}v$? $\endgroup$ – Soaps Jul 26 '14 at 23:01
  • $\begingroup$ Yes, that is exactly what the problem wants you to prove. $\endgroup$ – JimmyK4542 Jul 26 '14 at 23:02
  • $\begingroup$ Thank you so much that was really helpful! $\endgroup$ – Soaps Jul 26 '14 at 23:03
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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) $

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  • $\begingroup$ This is cleanest, but this problem is from Linear Algebra Done Right, so no determinants allowed! $\endgroup$ – galois Apr 14 '16 at 4:56
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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.

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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!

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  • $\begingroup$ I have a question. How come A = P^-1TP? Thank you $\endgroup$ – Mangostino Oct 4 '20 at 16:50
  • $\begingroup$ @Mangostino If T is written as a matrix in the standard basis, you can use a change of basis matrix P to transform it to A, which is block diagonal. $\endgroup$ – Macrophage Oct 4 '20 at 16:53
  • $\begingroup$ Oh okey. Thanks a lot $\endgroup$ – Mangostino Oct 4 '20 at 16:54

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