Exercise 6.4.23

As Hint, we denote $V_1$ and $V_2$ are the spaces with inner products $⟨\text{⋅,⋅}⟩$ and ${⟨\text{⋅,⋅}⟩}^{\prime }$. By the definition of ${⟨\text{⋅,⋅}⟩}^{\prime }$, the basis $\beta$ is orthonormal in $V_2$. So $U$ is self-adjoint on $V_2$ since it has an orthonormal basis consisting of eigenvectors. Also, we get a special positive definite, and so self-adjoint, operator $T_1$ by Exercise 6.4.22 such that

We check that $U=T_1^{-1}{U}^{\text{∗}}{T}_1$ by

for all $x$ and $y$. So we have $U=T_1^{-1}{U}^{\text{∗}}{T}_1$ and so $T_{1}U=U^{\text{∗}}{T}_1$. Pick $T_2=T_1^{-1}{U}^{\text{∗}}$ and observe that it’s self-adjoint. Pick $T_1^{\prime }=T_1^{-1}$ to be a positive definite operator by Exercise 6.4.19(c). Pick $T_2^{\prime }=U^{\text{∗}}{T}_1$ to be a self-adjoint operator. Now we have $U=T_2{T}_1=T_1^{\prime }{T}_2^{\prime }$.