Exercise 3.5.1

Let $A_0=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill a\hfill & \hfill b\hfill \end{array}\right]$, let ${\lambda }_1,{\lambda }_2$ be the eigenvalues of $A_0^T{A}_0=\left[\begin{array}{cc}\hfill 1+a^2\hfill & \hfill 2+\mathit{ab}\hfill \\ \hfill 2+\mathit{ab}\hfill & \hfill 4+b^2\hfill \end{array}\right]$, so we have ${\lambda }_1+{\lambda }_2=1+a^2+4+b^2=5+a^2+b^2$ and ${\lambda }_1{\lambda }_2=(1+a^2)(4+b^2)-{(2+\mathit{ab})}^2={(2a-b)}^2$.
The singular values of $A_0$ are thus

This equals to ${(a+2)}^2+{(b-1)}^2$ when $2a-b>0$ and ${(a-2)}^2+{(b+1)}^2$ when $2a-b<=0$

They achieve minimum of 5 when $a=b=0$, so

$A_0=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ minimizes $\parallel A_0{\text{∥}}_N$

Let $B_0=\left[\begin{array}{cc}\hfill 1\hfill & \hfill a\hfill \\ \hfill b\hfill & \hfill 4\hfill \end{array}\right]$, let ${\lambda }_1,{\lambda }_2$ be the eigenvalues of $B_0^T{B}_0=\left[\begin{array}{cc}\hfill 1+b^2\hfill & \hfill a+4b\hfill \\ \hfill a+4b\hfill & \hfill 16+a^2\hfill \end{array}\right]$, so we have ${\lambda }_1+{\lambda }_2=1+b^2+16+a^2=17+a^2+b^2$ and ${\lambda }_1{\lambda }_2=(1+b^2)(16+a^2)-{(a+4b)}^2={(\mathit{ab}-4)}^2$.

This equals to ${(a+b)}^2+9$ when $\mathit{ab}>4$ and $(a-b)^2 + 25 $ when $\mathit{ab}<=4$. The second achieves minimal of 25 when $a=b=0$. We want to check whether there are $a,b$ that can make the first one less than 25 when $\mathit{ab}>4$, so we check ${(a+b)}^2+9<25$, and we find it requires ${(b-2)}^2<0$ or ${(b+2)}^2<0$, which is impossible. But on the boundary, we see that $a=b=2$ or $a=b=-2$ also achieves the same minimum 25.

So $B_0=\left[\begin{array}{cc}\hfill 1\hfill & \hfill a\hfill \\ \hfill a\hfill & \hfill 4\hfill \end{array}\right]$ minimizes $\parallel B_0{\text{∥}}_N$, where $a=0,2,-2$.

Let $C_0=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill a\hfill \end{array}\right]$, let ${\lambda }_1,{\lambda }_2$ be the eigenvalues of $C_0^T{C}_0=\left[\begin{array}{cc}\hfill 10\hfill & \hfill 2+3a\hfill \\ \hfill 2+3a\hfill & \hfill 4+a^2\hfill \end{array}\right]$, so we have ${\lambda }_1+{\lambda }_2=14+a^2$ and ${\lambda }_1{\lambda }_2=10(4+a^2)-{(2+3a)}^2={(a-6)}^2$.

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Take derivative w.r.t. $a$, we find $a=1$, so $C_0=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 1\hfill \end{array}\right]$ minimizes $\parallel C_0{\text{∥}}_N$