Exercise 6.5.21

Since $A$ is unitarily equivalent to $B$, we write $B=P^{\text{∗}}\mathit{AP}$ for some unitary matrix $P$.

1.

Compute

$$\mathrm{tr}<!--\; FUNCTION\; APPLICATION\; -->(B^{\text{∗}}B)=\mathrm{tr}<!--\; FUNCTION\; APPLICATION\; -->({(P^{\text{∗}}\mathit{AP})}^{\text{∗}}(P^{\text{∗}}\mathit{AP}))=\mathrm{tr}<!--\; FUNCTION\; APPLICATION\; -->(P^{\text{∗}}{A}^{\text{∗}}\mathit{AP})=\mathrm{tr}<!--\; FUNCTION\; APPLICATION\; -->(A^{\text{∗}}A).$$

2.

We compute the trace of $A^{\text{∗}}A$ and get

$$\mathrm{tr}<!--\; FUNCTION\; APPLICATION\; -->(A^{\text{∗}}A)=\sum _{i=1}^n{(A^{\text{∗}}A)}_{\mathit{ii}}$$

$$=\sum _{i=1}^n\sum _{k=1}^n{(A^{\text{∗}})}_{\mathit{ik}}{A}_{\mathit{ki}}=\sum _{i,j}|A_{\mathit{ij}}{\text{|}}^2.$$

Use the result in the previous argument we get the conclusion.

3.

By the previous argument, they are not unitarily equivalent since

$$|1{\text{|}}^2+|2{\text{|}}^2+|2{\text{|}}^2+|i{\text{|}}^2=10$$

is not equal to

$$|i{\text{|}}^2+|4{\text{|}}^2+|1{\text{|}}^2+|1{\text{|}}^2=19.$$