linear operator on inner product space with $(B,A)=\operatorname{tr}(AB^\dagger)$

Question

In the linear space of $n \times n$ matrices on $\mathbb{C}$ consider: $$(B,A)=tr(AB^\dagger)$$

1. Prove that it's a Hilbert product (inner product)
2. Show a orthogonal base for the space
3. if we have a linear operator so defined: $$ \mathcal H: B \to H^\dagger BH$$ then get $\mathcal H^\dagger$
4. If H is a unitary matrix, get the norm of $\mathcal H$

1) I've tried to use the definition of inner product:

i) $(B,B)=tr(BB^\dagger)= \sum_{i=1}^n b_{ii} \overline{b_{ii}}=\sum_{i=1}^n |b_{ii}|^2 >0 $

ii) $(B,A+C)=tr((A+C)B^\dagger)= \sum_{i=1}^n (a_{ii}+c_{ii}) \overline{b_{ii}}=(B,A)+(B,C) $

iii) the same with scalar: $(B,\lambda A)= \lambda(B,A)$

i) $\overline{(B,A)}=tr \overline{(AB^\dagger)}=tr(\overline{A}B)=(A,B)$

2) I think that an orthogonal base could be: $\{E_i\}_{i=1}^n$ with $$E_1= diag(1, 0, 0, ...)$$ $$E_2=diag(0, 1, 0, ...)$$ etc, so that: $(E_i, E_j)=0$ $\forall i \neq j$

3) Here I have the biggest troubles. I'm looking for the self adjoint operator of $ \mathcal H $: $$(\mathcal H B,A)=(B,\mathcal H ^\dagger A)$$ $$(H^\dagger BH, A)= \sum_{i=1}^n a_{ii} \overline{\overline{h_{ii}}h_{ii}b_{ii}}=\sum_{i=1}^n a_{ii} \overline{b_{ii}} |h_{ii}|^2$$ And I think that $\mathcal H^\dagger$ could be: $$\mathcal H^\dagger: A \to HAH^\dagger$$

4) Being $H$ unitary it preserves the distances (hence the norm). So we have: $$||\mathcal H B||=||B||$$ The norm of $\mathcal H$ should be at most 1?

Answer 1

Regarding 1: you have incorrectly computed the inner product. Note that $$ \operatorname{tr}(AB^\dagger) = \sum_{i=1}^n \left(\sum_{j=1}^n a_{ij}\overline{b_{ij}}\right) = \sum_{i,j = 1}^n a_{ij}\overline{b_{ij}} $$ This should make your work going forward easier.

Regarding 2: your set does not give a full base of the space. To get a full basis, use the set $\{E_{ij}\}_{i,j = 1}^n$, where $E_{ij}$ is the matrix whose $i,j$ entry is $1$ and whose other entries are $0$. It is useful to note that $E_{ij} = e_ie_j^\dagger$, where $e_1,\dots,e_n$ is the canonical basis of $\Bbb C^n$.

Regarding 3, consider the following: $$ (\mathcal H(B),A) = (H^\dagger BH,A) = \operatorname{tr}(A[H^\dagger BH]^\dagger) = \operatorname{tr}(AH^\dagger B^\dagger H) = \operatorname{tr}(HAH^\dagger B^\dagger)\\ = \operatorname{tr}([HAH^\dagger] B^\dagger) = (B,HAH^\dagger) $$

Regarding 4: Remember that in this context, the definition of the norm is $$ \|B\|^2 = (B,B) = \operatorname{tr}(BB^\dagger) $$