11. Solution: It follows from 7.45 that $T=S\sqrt{T^*T}$ for an isometry $S\in\ca L(V)$. Note that $\sqrt{T^*T}$ is self-adjoint and $S^{-1}=S^*$ (7.42 (e), (f)), we have

\[

TT^*=S\sqrt{T^*T}(S\sqrt{T^*T})^*=S\sqrt{T^*T}\sqrt{T^*T}S^*=S(T^*T)S^{-1}.

\]It follows from Problem 15 of Exercise 5A that $TT^*$ and $T^*T$ have the same eigenvalues. Moreover, each eigenvalue has the same multiplicity.

Since $TT^*$ and $T^*T$ are positive operators, see Problem 4 of Exercise 7C, they have nonnegative eigenvalues. Also the singular value of $T$ are the nonnegative square roots of the eigenvalues of $T^*T$ while the singular value of $T$ are the nonnegative square roots of the eigenvalues of $TT^*$, see 7.52. We conclude that $T$ and $T^*$ have the same singular values. Moreover, each eigenvalue has the same multiplicity.

See: Wu Jinyang’s Comment. Basically, matrices or operators related in the way of Problem 15 of Exercise 5A can be considered as the “same”. They have almost the same structure. It is called they are similar to each other.

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