Exercise 7.D.3

Proof. By the Polar Decomposition (7.45), there exists an isometry $S\in L(V)$ such that $T^{\text{∗}}=S\sqrt{T{T}^{\text{∗}}}$. Taking the adjoint of each side, we get

where the last equality follows because $\sqrt{T{T}^{\text{∗}}}$ is self-adjoint. This yields the desired result, because $S^{\text{∗}}$ is also an isometry. □