Exercise 8.B.3

Proof. Suppose $\lambda$ is an eigenvalue of $T$ and $v\in V$ a corresponding eigenvalue. $S$ is surject, so there exists $u\in V$ such that $\mathit{Su}=v$. Then

which shows that $\lambda$ is an eigenvalue of $S^{-1}\mathit{TS}$. Hence every eigenvalue of $T$ is an eigenvalue of $S^{-1}\mathit{TS}$. We will prove these eigenvalues have the same multiplicity and it will follow that $S^{-1}\mathit{TS}$ cannot have other eigenvalues (by 8.26).

Suppose ${\lambda }_1,\dots ,{\lambda }_m$ are the distinct eigenvalues of $T$. Fix $k\in \{1,\dots ,m\}$. Let $v_1,\dots ,v_d$ be a basis of $G({\lambda }_k,T)$. There exist $u_1,\dots ,u_d\in V$ such that $S{u}_j=v_j$ for each $j=1,\dots ,d$. It easy to check that the $u$’s are linearly independent. We have

where the first equality comes from 8.11 and the second from Exercise 5 in section 5B. For each $j$, we have

where the second equality follows because $v_j\in G({\lambda }_k,T)$. This shows that $u_1,\dots ,u_d\in G({\lambda }_k,S^{-1}\mathit{TS})$. Hence

By 8.26, we must have

and

$(1)$ and $(2)$ imply that $(3)$ is only possible if $\dim <!--\; FUNCTION\; APPLICATION\; -->G({\lambda }_k,S^{-1}\mathit{TS})=\dim <!--\; FUNCTION\; APPLICATION\; -->G({\lambda }_k,T)$. Hence, their multiplicieties are the same and $S^{-1}\mathit{TS}$ cannot have other generalized eigenspaces (the ones shown here already eat up the dimension of $V$). □