Exercise 6.4.21

As Hint, we check whether $\mathit{UT}$ is self-adjoint with respect to the inner product ${⟨x,y⟩}^{\prime }$ or not. Denote $F$ to be the operator $\mathit{UT}$ with respect to the new inner product. Compute that

for all $x$ and $y$. This means that $\mathit{UT}$ is self-adjoint with respect to the new inner product. And so there’s some orthonormal basis consisting of eigenvectors of $\mathit{UT}$ and all the eigenvalue is real by the Lemma before Theorem 6.17. And these two properties is independent of the choice of the inner product. On the other hand, $T^{-1}$ is positive definite by Exercise 6.4.19(c). So the function ${⟨x,y⟩}^{″}:=⟨T^{-1}(x),y⟩$ is also a inner product by the previous exercise. Denote $F^{\prime }$ to be the operator $\mathit{TU}$ with respect to this new inner product. Similarly, we have

for all $x$ and $y$. By the same argument we get the conclusion.