General question first: upper/lower bound a sum of Kronecker products by its components. More specifically, how is $$ \Vert\sum_{\alpha}S_{\alpha}\otimes B_{\alpha}\Vert$$ bounded by the operator norms of $d_{S}\times d_{S}$ dimensional matrices $\{ S_{\alpha}\}$ and $d_{B}\times d_{B}$ dimensional matrices $\{ B_{\alpha}\}$? The operator norm $\Vert\cdot\Vert$ denotes the (usual) max abs singular value norm. The symbol $\otimes$ denotes the Kronecker or tensor product of matrices. Do we have a Cauchy-like inequality?

More specific question:

We have a "basis" set $\{ S_{\alpha} \}$ of $d_{S}^{2}-1$ Hermitian and traceless matrices: they satisfy $$\text{Tr}(S_{\alpha}S_{\beta}^\dagger)=d_{S}\delta_{\alpha\beta}$$ where $\delta_{\alpha\beta}$ is the Kronecker symbol and are all normalized with respect the operator norm: $\Vert S_{\alpha}\Vert=1$. In short, they would be an orthonormal basis for the Hilbert-Schmidt inner product if the identity matrix was included and they were normalized properly. For $d_s=2$ the $\{S_\alpha\}$ can be taken to be the Pauli matrices. The matrices $B_{\alpha}$ are also Hermitian and traceless but otherwise arbitrary. I would like to bound $\Vert\sum_{\alpha}S_{\alpha}\otimes B_{\alpha}\Vert$ from up and down by a factor of $\max_{\alpha}\Vert B_{\alpha}\Vert$. The best I have done so far is rather miserable:$$ \frac{1}{d_S}\max_{\alpha}{\left\Vert B_{\alpha}\right\Vert }\leq\Vert\sum_{\alpha}S_{\alpha}\otimes B_{\alpha}\Vert\leq(d_S^{2}-1)\max_{\alpha}{\left\Vert B_{\alpha}\right\Vert }$$

The closest reference that I have found on this material is the work by Chansangiam et al [ScienceAsia 35, 106 (2009)] which I am still going through. This question is related to a question in physics for bounding fidelity of evolutions in open quantum systems. In short I want to bound the maximum energy scale in a combined system with that of its components.

  • $\begingroup$ What happens when all the $B_\alpha$ are equal? Your upper-bound seems to be a good one then. $\endgroup$
    – Suvrit
    Feb 8, 2011 at 5:46
  • $\begingroup$ @Suvrit: In that case I think the upper bound would be $\sqrt{d_s^2-1}=O(d_s)$ and would be far from tight. The $S_\alpha$ are as spread out as they can be. $\endgroup$ Feb 8, 2011 at 15:32
  • $\begingroup$ By the way a lower bound is probably more challenging. $\endgroup$ Feb 8, 2011 at 15:33

1 Answer 1


I do not know if this helps, but here are some upper bounds, standard in operator space theory. The first inequality, attributed to Haagerup, is an analog of the Cauchy-Schwarz inequality in your setting:

$$\Vert\sum_{\alpha}S_{\alpha}\otimes B_{\alpha}\Vert\leq \Vert\sum_{\alpha}S_{\alpha}\otimes \overline{S_{\alpha}}\Vert^{1/2} \Vert\sum_{\alpha}B_{\alpha}\otimes \overline{B_{\alpha}}\Vert^{1/2}.$$

Here for a matrix $A = (A_{i,j})$, $\overline{A}$ denotes the matrix $(\overline{A_{i,j}})$. The expressions appearing on the right-hand side of this inequality are the norms of $(S_\alpha)$ and $(B_\alpha)$ in the operator Hilbert space OH. For a proof, see for example page 123 in Pisier's Introduction to Operator Space Theory.

Another inequality (no longer symmetric) that reduces to the usual Cauchy-Schwarz inequality when the matrices are of size $1$ is the following (and the same with the role of S and B reversed): $$\Vert\sum_{\alpha}S_{\alpha}\otimes B_{\alpha}\Vert\leq \Vert\sum_{\alpha}S_{\alpha}S_{\alpha}^*\Vert^{1/2} \Vert\sum_{\alpha}B_{\alpha}^* B_{\alpha}\Vert^{1/2}.$$

Now the terms appearing on the left are, in the language of operator spaces, the row (resp. column) norm of $(S_\alpha)$ (resp. $(B_\alpha)$). The row (resp. column) norm of $S=(S_\alpha)$ is just the norm of the matrix $ROW(S)$ (resp. $COLUMN(S)$) obtained, in a block-decomposition, by putting $S_\alpha$'s on the first row (resp. column) and $0$'s on the other rows (resp. columns).

This last inequality is very easy to prove, and more generally we have $\Vert\sum_i a_i b_i\Vert\leq \Vert\sum_i a_i a_i^*\Vert^{1/2} \Vert\sum_i b_i^* b_i\Vert^{1/2}$ for any matrices $a_i$ and $b_i$. Indeed, the LHS of this inequality is $\Vert ROW(a) COLUMN(b)\Vert$, and its RHS is $\Vert ROW(a)\Vert \Vert COLUMN(b)\Vert$. This inequality is thus just expressing that the operator norm is sub-multiplicative.

Edit (for a lower bound, without the typo this time). In the case when the $S_\alpha$'s form an orthonormal family for the scalar product $\langle A,B\rangle = Tr(B^* A)/d_S$, you get the following lower bound:

$$\Vert\sum_{\alpha}S_{\alpha}\otimes B_{\alpha}\Vert\geq \max(\Vert\sum_{\alpha}B_{\alpha}^* B_\alpha\Vert^{1/2}, \Vert \sum_{\alpha}B_{\alpha} B_\alpha^*\Vert^{1/2}).$$

This is because $\sum_{\alpha}B_{\alpha}^* B_\alpha$ is $1/d_S Tr \otimes id$ applied to $X^*X$, where $X=\sum_{\alpha}S_{\alpha}\otimes B_{\alpha}$. And since $1/d_s Tr$ is a state, $1/d_S Tr \otimes id$ is of norm $1$ from $M_{d_S} \otimes M_{d_B}$ to $M_{d_B}$.

In the specific situation of your problem, here are the bounds one actually gets. Your assumptions on the $S_\alpha$'s imply that they all are unitary, and orthonormal for $\langle A,B\rangle = Tr(B^* A)/d_S$. Therefore you have that $$\Vert\sum_{\alpha}S_{\alpha}\otimes \overline{S_{\alpha}} \Vert = \Vert\sum_{\alpha}S_{\alpha}^*S_{\alpha}\Vert= \Vert\sum_{\alpha}S_{\alpha}S_{\alpha}^*\Vert = N$$ where $N$ is the number of terms in the sum (you want to take $N=d_S^2-1$).

You therefore get $$ \max(\Vert\sum_{\alpha}B_{\alpha}^* B_\alpha\Vert^{1/2}, \Vert \sum_{\alpha}B_{\alpha} B_\alpha^*\Vert^{1/2}) \leq \Vert\sum_{\alpha}S_{\alpha}\otimes B_{\alpha} \Vert $$ and $$\Vert\sum_{\alpha}S_{\alpha}\otimes B_{\alpha} \Vert \leq \sqrt N \min(\Vert\sum_{\alpha}B_{\alpha}^* B_\alpha\Vert^{1/2}, \Vert \sum_{\alpha}B_{\alpha} B_\alpha^*\Vert^{1/2}).$$

The lower bound is tight (take all the $B_\alpha$'s but one equal to zero). The upper bound too (take $B_\alpha = \overline{S_{\alpha}}$), and implies the one you gave in your question.

  • $\begingroup$ Thanks. I have to check these out... but any ideas about a lower bound esp. for the special problem where S_\alpha are some sort of basis? $\endgroup$ Feb 8, 2011 at 15:35
  • $\begingroup$ A last comment: if you really want to compare with $max_\alpha\Vert B_\alpha\Vert$, the best constants one gets is $1$ for the lower bound and $N = d_S^2-1$ for the upper bound, which are both tight. $\endgroup$ Feb 8, 2011 at 16:21
  • $\begingroup$ I think the upper bound will scale with $\sqrt{d_s^2-1}$ actually, as the $S_\alpha$ are well spread out in a sense but I have no proof... How do you get the lower bound of 1, by the way? $\endgroup$ Feb 8, 2011 at 17:46
  • $\begingroup$ I do not understand your first sentence (but notice that for any unitary matrix $S_1,\dots S_N$, if $B_\alpha = \overline{S_\alpha}$ you get $\Vert \sum_[\alpha=1}^N S_\alpha \otimes B_\alpha \Vert =N$). For your question, the lower bound $1$ follows from my answer and the inequality $\sum B_\alpha^* B_\alpha \geq B_i^* B_i$, which implies $\Vert \sum B_\alpha^* B_\alpha \Vert \geq \Vert B_i^* B_i\Vert = \|B_i\|^2$ for all $i$. $\endgroup$ Feb 8, 2011 at 17:55
  • $\begingroup$ Thanks Mikael for the update. Consider a case where $B_\alpha$ are all the same $B$. Then the upper bound must be larger than $\Vert \sum S_\alpha\Vert\Vert B\Vert$ which despite having $d_s^2-1$ term, does not scale with $d_s^2-1$... Or I have no clue! Funny, the way I got my lower bound also used a similar method (we call it partial trace in physics) but I arrived at a different bound! I still need to check things more carefully. $\endgroup$ Feb 8, 2011 at 23:18

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