Exercise 3.5.5

• If only one entry of a matrix $A$ is unknown, to minimize $\parallel A{\text{∥}}_S$, I’ll put 0 in the missing entry.
• To minimize $\parallel A{\text{∥}}_N$, think about $S=A^{T}A$, the eigenvalues of matrix $S$ are ${\lambda }_i$, $i=1,\dots ,r$. All ${\lambda }_i\ge 0$ because $S$ is positive semidefinite.

On the other hand, we have $\parallel A{\text{∥}}_N={\sum <!--\; FUNCTION\; APPLICATION\; -->}_i{\sigma }_i$, to minimize $\parallel A{\text{∥}}_N$ is the same as to minimize $\parallel A{\text{∥}}_N^2=\sum <!--\; FUNCTION\; APPLICATION\; -->{\sigma }_i^2+{\sum }_{i\ne j}{\sigma }_i{\sigma }_j$

Notice $\sum <!--\; FUNCTION\; APPLICATION\; -->{\sigma }_i^2=\sum {\lambda }_i=\text{trace}(A^{T}A)=\sum a_i^T{a}_i$, where $a_i$ are the columns for matrix $A$. It’s easy to see that if there’s only one entry missing in one of the $a_i$s, it’s better to make it zero to minimize the sum of ${\lambda }_i$.

I am not sure how to prove the part with ${\sigma }_i{\sigma }_j$ though.