2
$\begingroup$

Suppose $A\in L(X,Y)$. $||\cdot||$ denotes spectral norm and denotes the largest singular value of a matrix, i.e. the largest eigenvalue of $\sqrt{A^*A}$.

$||\cdot||_{tr}$ denotes trace norm. We have that $$||A||_{tr}=tr\sqrt{A^*A}$$ So I would like to prove the statement that $$||A||_{tr}=\max\{|tr(A^*B)|: B\in L(X,Y), ||B||=1 \}$$

I know that from Nielsen and Chuang lemma 9.5 that

$$|tr(AU)|\le tr |A|$$ and equality is achieved by a unitary.

We have by definition that $|A|=\sqrt{A^*A}$. So $||A||_{tr}=tr|A|$.

I think my question is if $B$ is not a unitary but has norm 1, can we have that

$$|tr(AB)|> tr |A|\ge |tr(AU)|$$ for any unitary? And if yes, why the maximum is still achieved by a unitary?

$\endgroup$
2
  • 1
    $\begingroup$ Quick question: If $A\in L(X,Y)$ and $B\in L(X,Y)$, we can't compose them unless $Y=X$. Should it be $B\in L(Y,X)$, or $tr(A^*B)$, maybe? $\endgroup$
    – Sam Jaques
    Oct 28 '20 at 9:50
  • $\begingroup$ Surely $tr|A|\geq|tr(AB)|$? $\endgroup$
    – DaftWullie
    Oct 28 '20 at 10:22
5
$\begingroup$

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 $$ |\operatorname{Tr}(A^{\ast} B)| = \biggl| \sum_{k} s_k \langle \psi_k | B | \phi_k\rangle \biggr| \leq \sum_k s_k |\langle \psi_k | B | \phi_k\rangle| \leq \sum_k s_k \|B\| = \| A \|_{\text{tr}} \| B \|. $$

$\endgroup$
3
  • $\begingroup$ Thank you, that's an insightful approach too, together with your book helps to understand the dual norms better! $\endgroup$
    – user777
    Oct 28 '20 at 19:05
  • $\begingroup$ Why is $|\langle \psi_k | B | \phi_k\rangle| \leq ||B||$? $\endgroup$
    – BlackHat18
    Oct 28 '20 at 19:31
  • 1
    $\begingroup$ @BlackHat18: start with Cauchy-Schwarz, and then I suppose it's pretty much down to the definition of the spectral norm. $\endgroup$ Oct 28 '20 at 20:59
3
$\begingroup$

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 $$ max\{ \big| \langle B, A \rangle \big|, B\in L(X,Y), ||B|| \leq 1\} = \text{Tr}|A| $$

$\endgroup$
1
  • $\begingroup$ Right! It is a good idea to adapt the Nielsen and Chuang's proof! $\endgroup$
    – user777
    Oct 28 '20 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.