Bounding the commutator [A,B] in terms of the numerical radius - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:46:57Z http://mathoverflow.net/feeds/question/89762 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89762/bounding-the-commutator-a-b-in-terms-of-the-numerical-radius Bounding the commutator [A,B] in terms of the numerical radius Denis Serre 2012-02-28T13:44:13Z 2012-03-12T10:25:58Z <p>Given a norm $N$ over ${\bf M}_n(\mathbb C)$, it is a natural question to find the best constant $C_N$ such that $$N([A,B])\le C_N N(A)N(B),\qquad\forall A,B\in{\bf M}_n(\mathbb C).$$</p> <p>The answer is known at least in the following cases:</p> <ul> <li>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 $n\ge2$.</li> <li>the Frobenius norm $\|A\|^2_F=\sum_{i,j}|a_{ij}|^2$. Then a theorem by Böttcher &amp; Wentzel (2008) tells us that $$\|[A,B]\|_F\le\sqrt2\|A\|_F\|B\|_F,$$ and again this is optimal.</li> </ul> <blockquote> <p>I have a third norm in mind, yet of a different nature: the <em>numerical radius</em> $$r(A)=\sup_{x\ne0}\frac{|x^*Ax|}{\|x\|^2}.$$ This is the smallest radius of a disk $D(0;r)$ containing the <em>numerical range</em> (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 ${\bf M}_n(\mathbb C)$ ?</p> </blockquote> <p>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 <em>super-stable</em> norm, in the sense that $r(M^k)\le r(M)^k$ for every $k\ge1$.</p> <p>This question naturally extends to $n$-commutators, in the spirit of my previous question <a href="http://mathoverflow.net/questions/38698" rel="nofollow">http://mathoverflow.net/questions/38698</a> .</p> <p><strong>Edit</strong>. See below Piotr Migdal's answer and my adaptation of it. It gives $C_{nr}=4$.</p> http://mathoverflow.net/questions/89762/bounding-the-commutator-a-b-in-terms-of-the-numerical-radius/89773#89773 Answer by S. Sra for Bounding the commutator [A,B] in terms of the numerical radius S. Sra 2012-02-28T15:34:10Z 2012-02-29T04:49:50Z <p>Here is a partial result. </p> <p><strong>Claim.</strong> $4 \le C_{nr} \le 8$.</p> <p><strong>Proof.</strong> The upper bound has already been shown by the OP. The lower-bound follows by \begin{equation*} A = \begin{bmatrix} 0 &amp; 1\\ 0 &amp; 0 \end{bmatrix},\qquad B = \begin{bmatrix} 0 &amp; 0\\ -1 &amp; 0 \end{bmatrix} \end{equation*} for which $$\frac{r([A,B])}{r(A)r(B)} = \frac{1}{\frac 12\times\frac 12}=4.$$ (<em>Note:</em> Slightly more generally, the $1$ in the above matrices can be replaced by an nonzero scalar).</p> <p>Based on some experiments mentioned in my comments to Denis, I am led to the following attractive conjecture.</p> <p><strong>Conjecture:</strong> $C_{nr}=4$.</p> <hr> <p><strong>Additional Remarks.</strong> </p> <p>Define $$X := \begin{bmatrix} 0 &amp; 1\\ 0 &amp; 0 \end{bmatrix},$$</p> <p>and let $A$, $B$ be arbitrary. Then, it is easy to see that we have the commutator inequality:</p> <p>\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.</p> http://mathoverflow.net/questions/89762/bounding-the-commutator-a-b-in-terms-of-the-numerical-radius/90458#90458 Answer by Denis Serre for Bounding the commutator [A,B] in terms of the numerical radius Denis Serre 2012-03-07T14:27:52Z 2012-03-07T14:27:52Z <p>When $n=2$, I have found that $$r([A,B])\le4r(A)r(B),$$ where the constant $4$ is optimal. This uses a characterization of the extremal points of the unit ball associated with the numerical radius (MO <a href="http://mathoverflow.net/questions/90290" rel="nofollow">question</a>). See the <a href="http://www.umpa.ens-lyon.fr/~serre/DPF/extBnr.pdf" rel="nofollow">proof</a>.</p> http://mathoverflow.net/questions/89762/bounding-the-commutator-a-b-in-terms-of-the-numerical-radius/90499#90499 Answer by Piotr Migdal for Bounding the commutator [A,B] in terms of the numerical radius Piotr Migdal 2012-03-07T19:50:26Z 2012-03-08T06:31:15Z <p>I got $$r([A,B])\leq 4\sqrt{2} r(A) r(B).$$</p> <p>It is lower than $8$ but still higher than the conjuncture $C_{nr}=4$. </p> <p>I used the following facts: </p> <ul> <li><p>For normal (i.e. $X^*X=XX^*$) matrices we have $r(X)=\sigma_1(X)$ (the largest singular value of X).</p></li> <li><p>$\sigma_1(XY-YX)\leq 2 \sigma_1(X)\sigma_1(Y)$</p></li> <li><p>Also, note that for $X$ and $Y$ hermitian (i.e. $X^*= X$ and $Y^* = Y$ ) we have</p></li> </ul> <p>$$r(X+iY) \geq \max\left(\sigma_1(X),\sigma_1(Y)\right),$$ $$r(X+iY) \leq \sqrt{\sigma_1^2(X) + \sigma_1^2(Y)}.$$</p> <p>Lets decompose $A$ and $B$ in their hermitian and antihermitian parts, $$A = A_h + i A_a, \quad B= B_h+i B_a.$$</p> <p>Then $$r^2([A,B]) \leq \left( \sigma_1([A_h,B_h]-[A_a,B_a])\right)^2 + \left( \sigma_1([A_h,B_a]+[A_a,B_h])\right)^2$$ $$\leq\left( 2\sigma_1(A_h)\sigma_1(B_h)+2\sigma_1(A_a)\sigma_1(B_a) \right)^2 + \left( 2\sigma_1(A_h)\sigma_1(B_a)+2\sigma_1(A_a)\sigma_1(B_h) \right)^2$$ $$=4 ( \sigma_1^2(A_h) + \sigma_1^2(A_a) )( \sigma_1^2(B_h) + \sigma_1^2(B_a) ) +16 \sigma_1(A_h) \sigma_1(A_a) \sigma_1(B_h) \sigma_1(B_a)$$ $$\leq 8( \sigma_1^2(A_h) + \sigma_1^2(A_a) )( \sigma_1^2(B_h) + \sigma_1^2(B_a) )$$ $$\leq 32 r^2(A)r^2(B).$$</p> http://mathoverflow.net/questions/89762/bounding-the-commutator-a-b-in-terms-of-the-numerical-radius/90571#90571 Answer by Denis Serre for Bounding the commutator [A,B] in terms of the numerical radius Denis Serre 2012-03-08T12:36:22Z 2012-03-12T10:25:58Z <p>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 $r(M)=\|{\rm Re}(e^{-i\theta}M)\|_2.$ </p> <p>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 $\sigma_1$.</p> <p>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.</p>