Prove that the trace norm is dual to the spectral norm

Question

Suppose $A\in L(X,Y)$, $\|\cdot\|$ denotes the spectral norm, i.e., the largest eigenvalue of $\sqrt{A^*A}$, and $\|\cdot\|_{\rm tr}:={\rm tr}\sqrt{A^*A}$ denotes trace norm. So I would like to prove the statement that $$\|A\|_{\rm tr}=\max\{|{\rm tr}(A^*B)|: B\in L(X,Y), \|B\|=1 \}$$

I know from Lemma 9.5 in Nielsen and Chuang that $$|{\rm tr}(AU)|\le {\rm tr}|A|=\|A\|_1$$ for all unitaries $U$, and that equality is achieved by $U=V^\dagger$ where $A=|A|V$ is the polar decomposition of $A$.

I think my question is if $B$ is not a unitary but has norm 1, can we have that $$|{\rm tr}(AB)|> {\rm tr} |A|\ge |{\rm tr}(AU)|$$ for any unitary? And if yes, why the maximum is still achieved by a unitary?

Answer 1

There are different ways to prove what you want to prove, including the solution tsgeorgios has suggested, but for the sake of gaining greater intuition I would suggest starting with the recognition that the trace norm of any matrix is equal to the sum of its singular values.

Once you have this, the inequality you are trying to prove follows pretty easily. In particular, consider a singular value decomposition $$ A = \sum_k s_k |\psi_k\rangle \langle \phi_k|. $$ For any choice of $B$ we have, by the triangle inequality and a simple property of the spectral norm, that $$ |\operatorname{Tr}(A^{\ast} B)| = \biggl| \sum_{k} s_k \langle \psi_k | B | \phi_k\rangle \biggr| \leq \sum_k s_k |\langle \psi_k | B | \phi_k\rangle| \leq \sum_k s_k \|B\| = \| A \|_{\text{tr}} \| B \|. $$

Answer 2

We still have $ \big| \langle B, A \rangle \big| = \big|\text{Tr}(AB^{\dagger}) \big| \leq \text{Tr}|A| $ for any operator $B$ with operator norm $ ||B|| \leq 1 $.

First observe that $ ||B|| \leq 1 $ implies that for any positive semi-definite operator $ Q = \sum_i \lambda_i |\psi_i \rangle\langle \psi_i| \in L(X, X) $: $$ \text{Tr}(Q) = \sum_i \lambda_i \langle \psi_i|\psi_i \rangle \geq \sum_i \lambda_i \langle \psi_i|B^{\dagger}B|\psi_i \rangle = \sum_i \lambda_i \text{Tr} \big( B|\psi_i \rangle \langle \psi_i|B^{\dagger} \big) = \text{Tr}(B Q B^{\dagger}) $$

Now a similar proof to Nielsen and Chuang lemma 9.5 is possible: $$ \text{Tr}(AB^{\dagger}) = \text{Tr}(V|A|B^{\dagger}) = \text{Tr}(\big(V|A|^{1/2}\big) \big(|A|^{1/2} B^{\dagger}\big)) \leq \sqrt{\text{Tr}(|A|^{1/2} V^{\dagger} V |A|^{1/2}) \cdot \text{Tr}(B |A| B^{\dagger})} \leq \sqrt{\text{Tr}|A| \cdot \text{Tr}|A|} = \text{Tr}|A| $$

where we wrote $ A = V |A| $ for an isometry $ V \in L(X, Y) $ based on the SVD decomposition, the first inequality is the Cauchy–Schwarz and the second is our previous 'observation' with $ Q = |A| $.

Together with equality for $ B = V $ we have that $$ max\{ \big| \langle B, A \rangle \big|, B\in L(X,Y), ||B|| \leq 1\} = \text{Tr}|A| $$