Exercise 3.5.7

If $A$ is positive semidefinite, then we can write $A=\mathit{Q\Lambda }{Q}^T$, trace of $A$ equals to the sum of its eigenvalues. We have $A^{T}A=Q{\Lambda }^2{Q}^T$, so the singular values of $A$ equal to its eigenvalues (because $A$ is positive semidefinite, all its eigenvalues are larger than or equal to 0). $\parallel A{\text{∥}}_N$ equals to the sum of its singular values, which is also sum of its eigenvalues, which is the trace of $A$.

If $A=\mathit{UV}$, then we have

Where the last inequality comes from equation (13) in I.11 on page 92.