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Let $A$ and $B$ be two matrices. How can I prove that $$ \text{tr}\left(A^T B\right) \le \|A\|_{op} \|B\|_1$$

where $\|\cdot\|_{op}$ is the largest singular value of $A$, and $\|B\|_1$ is the sum of all singular values of $B$? I can see this is true if $A=B$ by using SVD, but don't know how to prove this more general statement.

Yuki Kawabata
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  • The trace of $A^T B$ is the sum of the eigenvalues of $A^T B$ which is less than or equal to the sum of the singular values of $A^T B$. Can you show that the $i$th singular value of $A^T B$ is less than or equal to the largest singular value of $A$ times the $i$th singular value of $B$? – Eric Apr 24 '19 at 09:05
  • @Eric I don't know how to show that... – Yuki Kawabata Apr 24 '19 at 18:56
  • This can be easily proved by using the characterisation that $|B|1=\max{Q^TQ=I}\operatorname{tr}(BQ)$. – user1551 Aug 26 '20 at 13:44

1 Answers1

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Presumably $A$ and $B$ are real. By padding them with zeroes, we may assume that they are square matrices. Let $A=USV^T$ be a singular value decomposition and let $D$ be any orthogonal diagonal matrix (i.e. a diagonal matrix whose diagonal entries are $\pm1$) such that $U^TBVD$ has a nonnegative diagonal. Then \begin{aligned} \operatorname{tr}(A^TB) &=\operatorname{tr}(VSU^TB)\\ &=\operatorname{tr}(SU^TBV)\\ &\le\operatorname{tr}(SU^TBVD)\\ &\le\sigma_1(S)\operatorname{tr}(U^TBVD)\\ &=\sigma_1(S)\operatorname{tr}(BVDU^T)\\ &\le\sigma_1(S)\max_{Q^TQ=I}\operatorname{tr}(BQ)\\ &=\sigma_1(S)\|B\|_1.\\ \end{aligned} The inequality in question also follows directly from von Neumann's trace inequality $\operatorname{tr}(A^TB)\le\sum_i\sigma_i(A)\sigma_i(B)$: $$ \operatorname{tr}(A^TB) \le\sum_i\sigma_i(A)\sigma_i(B) \le\sigma_1(A)\sum_i\sigma_i(B) =\sigma_1(A)\|B\|_1. $$

user1551
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