Let $A$ and $B$ be $4\times4$ traceless matrices with Hilbert-Schmidt norms summing up to $1/4$, i.e. $$\text{Tr}\left[ A\right]=\text{Tr}\left[ B\right] = 0,\qquad\text{Tr}\left[ A^\dagger A + B^\dagger B \right ] = \frac{1}{4}.$$ Let C be their Kronecker sum, i.e. $$C = A \otimes I_4 + I_4 \otimes B,$$ where $I_4$ is the 4d identity matrix and $\otimes$ is the Kronecker product.

Is it true that the sum of its two largest singular values squared is bounded from above by $1/2$, i.e. $$\sigma_1^2(C)+\sigma_2^2(C)\leq\frac{1}{2}\qquad?$$

It is the simplest non-trivial case (and of my interest on its own) of a more general conjecture that for $A$ and $B$ traceless $n\times n$ matrices for their Kronecker sum $C = A \otimes I_{n} + I_{n} \otimes B$ we have $$\sum_{i=1}^{n/2}\sigma_i^2(C)\leq\frac{1}{2}\text{Tr}\left[C^\dagger C\right].$$ For even $n$ it is saturated for $A_{1,2}=1/\sqrt{n}$ (and zero for every other entry).


  • For normal $A$ and $B$ the inequality holds; the proof is in arXiv:0711.2613. However, the relation $\lambda_{i,j}(C) = \lambda_i(A)+\lambda_j(B)$ significantly simplifies the problem.
  • The left hand side is a singular norm between Schatten and Ky Fan norm, $\Vert X \Vert_{(p,k)}=\sqrt[p]{\sum_{i=1}^k \sigma_i^p}$. So the relation can viewed as a norm inequality $\Vert C \Vert_{(2,n/2)}\leq \frac{1}{\sqrt{2}}\Vert C \Vert_{(2,n^2)}$ for a special class of $C$.
  • The tracelessness of both $A$ and $B$ is necessary (otherwise numerical tests show that the inequality does not hold). Consequently, standard tools (which does not take trace into account) like $k$th Ky Fan norm (here $k=2$) do not suffice. For example: $$||C^\dagger C||^{k=2} \leq ||A^\dagger A\otimes I_4||^{k=2} + ||I_4\otimes B^\dagger B||^{k=2} + || A^\dagger \otimes B + A \otimes B^\dagger||^{k=2} $$ $$\leq 2 \sigma_1^2(A)+2 \sigma_1^2(B)+2\sigma_1(A)\sigma_1(B)+2\sigma_1(A)\sigma_2(B) \leq 0.793,$$ which isn't that strong as we already have $\text{Tr}[C^\dagger C]=1$.
  • $\begingroup$ ok, i deleted my answer which had a wrong counterexample because of using 3x3 instead of 4x4 matrices. $\endgroup$ – Suvrit Feb 29 '12 at 19:38

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