Exercise 3.5.6

$U^{\text{∗}}=U{\mathrm{\Sigma }}^{\frac{1}{2}}=U{\mathrm{\Sigma }}^{\frac{1}{2}}I$, which is the SVD of matrix $U^{\text{∗}}$, and its singular values are the entries in ${\mathrm{\Sigma }}^{\frac{1}{2}}$, thus we have its eigenvalues as entries in $\mathrm{\Sigma }$. $\parallel U^{\text{∗}}{\text{∥}}_F^2={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^r{\lambda }_i$, i.e. the sum of entries in matrix $\mathrm{\Sigma }$.

However, on the other side, $\parallel A{\text{∥}}_N$ is the sum of its singular values, which are the entries of $\mathrm{\Sigma }$ according to its SVD.

This proves that $\parallel U^{\text{∗}}{\text{∥}}_F^2=\parallel A{\text{∥}}_N$. Similarly we have $\parallel V^{\text{∗}}{\text{∥}}_F^2=\parallel A{\text{∥}}_N$.