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Given a norm $N$ over ${\bf M}_n(\mathbb C)$, it is a natural question to find the best constant $C_N$ such that $$N([A,B])\le C_N N(A)N(B),\qquad\forall A,B\in{\bf M}_n(\mathbb C).$$

The answer is known at least in the following cases:

  • the operator norm $\|A\|_2=\sup\frac{\|Ax\|_2}{\|x\|_2}$ where the norm over $\mathbb C^n$ is the standard Hermitian $\|x\|_2^2=\sum_j|x_j|^2$. Then $$\|[A,B]\|_2\le2\|A\|_2\|B\|_2$$ is optimal for $n\ge2$.
  • the Frobenius norm $\|A\|^2_F=\sum_{i,j}|a_{ij}|^2$. Then a theorem by Böttcher & Wentzel (2008) tells us that $$\|[A,B]\|_F\le\sqrt2\|A\|_F\|B\|_F,$$ and again this is optimal.

I have a third norm in mind, yet of a different nature: the numerical radius $$r(A)=\sup_{x\ne0}\frac{|x^*Ax|}{\|x\|^2}.$$ This is the smallest radius of a disk $D(0;r)$ containing the numerical range (or Hausdorffian) of the matrix. What is the optimal constant $C_{nr}$ such that $r([A,B])\le C_{nr}r(A)r(B)$ for all $A,B$ in ${\bf M}_n(\mathbb C)$ ?

Let me point out that $r$ is not submultiplicative. We have at best $r(MN)\le 4r(M)r(N)$, which gives by the triangle inequality $r([A,B])\le8r(A)r(B)$, but this is certainly not optimal. However, it is a super-stable norm, in the sense that $r(M^k)\le r(M)^k$ for every $k\ge1$.

This question naturally extends to $n$-commutators, in the spirit of my previous question Standard polynomials applied to matrices .

Edit. See below Piotr Migdal's answer and my adaptation of it. It gives $C_{nr}=4$.

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    $\begingroup$ For random gaussian matrices, some experimental figures: $C_2 = 2.8409$, $C_3=2.4264$, $C_4=2.3061$, etc., that conjecture of mine is probably wrong because of numerical issues. For uniformly random matrices, it seems $C_2 = 1.367$, $C_3=0.8188$, etc.; but since I cannot trust these results, this brings up the question of how to reliably compute / estimate the numerical radius? $\endgroup$
    – Suvrit
    Feb 29, 2012 at 3:16
  • $\begingroup$ One has ${\rm Tr}[A,B]=0$. If $n=2$, we may apply to $[A,B]$ the following formula, valid whenever ${\rm Tr}M=0$ : $$4r(M)^2=\|M\|_F^2+2|\det M|.$$ $\endgroup$
    – Denis Serre
    Mar 6, 2012 at 17:14

3 Answers 3

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I got $$r([A,B])\leq 4\sqrt{2} r(A) r(B).$$

It is lower than $8$ but still higher than the conjuncture $C_{nr}=4$.

I used the following facts:

  • For normal (i.e. $X^\*X=XX^\*$) matrices we have $r(X)=\sigma_1(X)$ (the largest singular value of X).

  • $\sigma_1(XY-YX)\leq 2 \sigma_1(X)\sigma_1(Y)$

  • Also, note that for $X$ and $Y$ hermitian (i.e. $X^\*= X$ and $Y^\* = Y$ ) we have

$$r(X+iY) \geq \max\left(\sigma_1(X),\sigma_1(Y)\right),$$ $$r(X+iY) \leq \sqrt{\sigma_1^2(X) + \sigma_1^2(Y)}.$$

Lets decompose $A$ and $B$ in their hermitian and antihermitian parts, $$A = A_h + i A_a, \quad B= B_h+i B_a.$$

Then $$r^2([A,B]) \leq \left( \sigma_1([A_h,B_h]-[A_a,B_a])\right)^2 + \left( \sigma_1([A_h,B_a]+[A_a,B_h])\right)^2$$ $$\leq\left( 2\sigma_1(A_h)\sigma_1(B_h)+2\sigma_1(A_a)\sigma_1(B_a) \right)^2 + \left( 2\sigma_1(A_h)\sigma_1(B_a)+2\sigma_1(A_a)\sigma_1(B_h) \right)^2$$ $$=4 ( \sigma_1^2(A_h) + \sigma_1^2(A_a) )( \sigma_1^2(B_h) + \sigma_1^2(B_a) ) +16 \sigma_1(A_h) \sigma_1(A_a) \sigma_1(B_h) \sigma_1(B_a)$$ $$\leq 8( \sigma_1^2(A_h) + \sigma_1^2(A_a) )( \sigma_1^2(B_h) + \sigma_1^2(B_a) )$$ $$\leq 32 r^2(A)r^2(B).$$

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    $\begingroup$ nice idea to use the cartesian decomposition here. $\endgroup$
    – Suvrit
    Mar 8, 2012 at 0:46
  • $\begingroup$ @Piotr. Oops, sorry! Your result seems to be valid in every dimension. let me see it in details. $\endgroup$
    – Denis Serre
    Mar 8, 2012 at 6:12
  • $\begingroup$ This could lead to a better bound: Let $k$ be the best constant such that $\sigma_1([X,Y])\le k\sigma_1(X)\sigma_1(Y)$ for every Hermitian matrices $X,Y$. Then your calculation gives a bound with constant $2k\sqrt2$. Does anyone know the constant $k$ ? If we drop the restriction that $X$ and $Y$ are Hermitian, then the best constant is $2$. $\endgroup$
    – Denis Serre
    Mar 8, 2012 at 8:02
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    $\begingroup$ @Denis Even for Hermitian $X$ and $Y$ we get the same constant, $k=2$. Just take $X=[[1,0],[0,-1]]$ and $Y=[[0,1],[1,0]]$. $\endgroup$
    – Piotr Migdal
    Mar 8, 2012 at 8:18
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    $\begingroup$ @Piotr. See below, I have adapted your proof to get the optimal constant $4$. $\endgroup$
    – Denis Serre
    Mar 8, 2012 at 12:37
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Here is a partial result.

Claim. $4 \le C_{nr} \le 8$.

Proof. The upper bound has already been shown by the OP. The lower-bound follows by \begin{equation*} A = \begin{bmatrix} 0 & 1\\\\ 0 & 0 \end{bmatrix},\qquad B = \begin{bmatrix} 0 & 0\\\\ -1 & 0 \end{bmatrix} \end{equation*} for which $$\frac{r([A,B])}{r(A)r(B)} = \frac{1}{\frac 12\times\frac 12}=4.$$ (Note: Slightly more generally, the $1$ in the above matrices can be replaced by an nonzero scalar).

Based on some experiments mentioned in my comments to Denis, I am led to the following attractive conjecture.

Conjecture: $C_{nr}=4$.

Additional Remarks.

Define $$X := \begin{bmatrix} 0 & 1\\\\ 0 & 0 \end{bmatrix},$$

and let $A$, $B$ be arbitrary. Then, it is easy to see that we have the commutator inequality:

\begin{equation*} r(X \otimes [A,B]) \le 4 r(X \otimes A) r(X \otimes B), \end{equation*} where $\otimes$ denotes the Kronecker product.

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  • $\begingroup$ @Suvrit. See my partial answer below. $\endgroup$
    – Denis Serre
    Mar 7, 2012 at 14:28
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The answer by Piotr Migdal can be modified to give the accurate inequality $$r([A,B])\le4r(A)r(B),\qquad\forall A,B\in{\bf M}_n(\mathbb C).$$ The only new argument is that for every matrix $M$, there exists an angle $\theta$ such that $r(M)=\|{\rm Re}(e^{-i\theta}M)\|_2.$

Actually, we do have $$r(M)=\sup_\alpha\|{\rm Re}(e^{-i\alpha}M)\|_2.$$ Hereabove, the real part is defined as ${\rm Re} N=\frac12(N+\bar N^T)$. Notice that I employ the notation $\|\cdot\|$ (operator norm) which coincides with $\sigma_1$.

Let us apply this to $M=[A,B]$. With $\theta$ as above, let us decompose $e^{-i\theta}A=A_{\theta h}+iA_{\theta a}$. Then let us proceed as Piotr did: $$r([A,B])=\|{\rm Re}[e^{-i\theta}A,B]\| = \|i[A_{\theta h},B_a]+i[A_{\theta a},B_h]\|\le2(\|A_{\theta h}\|\cdot\|B_a\|+\|A_{\theta a}\|\cdot\|B_h\|),$$ which gives $$r([A,B])\le4r(e^{-i\theta}A)r(B).$$ Hence the result.

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  • $\begingroup$ @Denis: Maybe it is very simple, but why the equality with the angle $\alpha$ does hold? $\endgroup$
    – Piotr Migdal
    Mar 8, 2012 at 14:57
  • $\begingroup$ @Piotr. You take the supremum of a continuous function of $\alpha\in[0,2\pi]$. It is achieved at some $\theta$. $\endgroup$
    – Denis Serre
    Mar 8, 2012 at 17:20
  • $\begingroup$ @Denis Actually, I was concerned with $r(M)=\sup_a \Vert\Re(e^{i\alpha}M)\Vert_2$. But now I see it: $r(M) = \sup_\alpha \sup_{|x|=1} \Re (x^\* e^{i\alpha}M x)$ and $\Vert (e^{i \alpha}A + e^{-i \alpha}A^\*)/2 \Vert_2 = \sup_{|x|=1} \Re(x^* e^{i \alpha}A x)$. Nice trick to remove the complex part anyway. $\endgroup$
    – Piotr Migdal
    Mar 9, 2012 at 12:42
  • $\begingroup$ And one remark: for a commutator ($X$ and $Y$ are Hermitian), the real part is with $i$ as $(i[X,Y])^\* = -i [Y,X] = i [X,Y]$, whereas $[X,Y]^\* = [Y,X] = - [X,Y]$. $\endgroup$
    – Piotr Migdal
    Mar 9, 2012 at 17:21
  • $\begingroup$ @Piotr. Yes! This is always puzzling. Only skew-symmetric matrices form a Lie algebra. $\endgroup$
    – Denis Serre
    Mar 9, 2012 at 19:29

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