5
$\begingroup$

Given $M\in M_n({\mathbb R})$ and $\ell\in{0,\ldots,n-1}$, we define $$d_\ell(M)=\sum_{j=1}^nm_{j,j+\ell},$$ where the indices are understood mod $n$. In particular, $d_0$ is the trace operator.

Let $A\in M_n({\mathbb R})$ be given. We define a map $\Delta: O_n({\mathbb R})\rightarrow{\mathbb R}^{n-1}$ by $$\Delta(R)=(d_1(R^TAR),\ldots,d_{n-1}(R^TAR)).$$ Mind that we omit $d_0(R^TAR)$, because we know in advance that it equals the trace of $A$.

Question. Does it exist an orthogonal matrix $R$ such that $\Delta(R)=(0,\ldots,0)$ ?

The requested property ressembles one for which the answer is known to be positive: find $R$ orthogonal such that the diagonal $R^TAR$ is constant (thus equal to $\frac{1}{n}{\rm Tr}A$). Both properties consist of $n-1$ linear constraints, and both are consistent with the fact that the mean value of $R^TAR$ over $SO_n$ is $(\frac{1}{n}{\rm Tr}A) I_n$. Thus the answer would certainly be positive if the stronger following statement is true.

Statement. The image of $SO_n$ under $\Delta$ is convex. True or False ?

This statement looks ambitious, since $\Delta$ is not linear, and $SO_n$ is not a convex set. But an optimistic mathematicien will say that it ressembles the Toeplitz-Hausdorff theorem about the convexity of the image of the complex unit sphere under the quadratic map $x\mapsto x^*Mx$. Note that the T-H thm is used to find an $R^TAR$ with constant diagonal.

$\endgroup$
6
  • $\begingroup$ Just to understand better your question: if we work with complex matrices, $A$ is unitarily equivalent to a lower triangular matrix, so the question becomes trivial, right? is there any relation with your problem? $\endgroup$ Sep 29, 2010 at 15:13
  • $\begingroup$ No, it is not trivial. Let me emphasize the indices are understood mod n. Thus the functions $d_\ell$ are circular sums. There are $n$ terms in each sum. $\endgroup$ Sep 29, 2010 at 15:37
  • $\begingroup$ I see. Not trivial. $\endgroup$ Sep 29, 2010 at 16:15
  • 3
    $\begingroup$ For this question, I would have suggested to go and see a certain very recent book on matrices... but will not, as it's your book ;-) $\endgroup$ Sep 29, 2010 at 16:20
  • $\begingroup$ If you took $R$ to be unitary, then I think it is equivalent to the other problem, by diagonalizing the cyclic permutation matrix. $\endgroup$
    – Ian Agol
    Sep 30, 2010 at 5:52

1 Answer 1

2
$\begingroup$

Believe me, I didn't know the answer when I asked the question. But now I do. It is No. Here is a counterexample, a $3\times3$ matrix $A$ for which $\Delta_A$ does not vanish over the orthogonal group.

The matrix is that of a rotation of angle $2\pi/3$ around some axis. For instance $A$ can be taken as the matrix of the permutation $[1,2,3]$. Its orbit under orthogonal conjugation is the set of all rotations of angle $2\pi/3$. So let $B$ be such a rotation, and let $v=(a,b,c)$ be the unitary vector about which the rotation takes place. Then $d_1(B)$ and $d_2(B)$ are $$\frac{3}{2}(ab+bc+ca)\pm\frac{\sqrt3}{2}(a+b+c).$$ Thus $d_1=d_2=0$ means $ab+bc+ca=0$ and $a+b+c=0$, which are incompatible with $a^2+b^2+c^2=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.