2
$\begingroup$

I'm wondering whether there is certain relationship between the largest eigenvalue of a positive matrix(every element is positive, not neccesarily positive definite) $A$, $\rho(A)$ and that of $A∘A^T$, $\rho( A∘A^T)$, where $∘$ denotes hadamard product.

Here's a result I find for many numerical cases. I create a matrix of size $n$ whose elements are uniformly drawn from $[0,M]$, as $n$ gets large (>20), $\rho(A)\rightarrow 2M\rho( A∘A^T)$.

I've read some papers on the bound of eigenvalue of $A∘B$, yet none of them mention the special case of $A∘A^T$. I'm wondering whether there's a theory about this and moreover, whether this result could be extended to general linear operators, such as integral operators $T(f(x))=\int k(x,y)f(y)dy$ and $T(f(x))=\int k(x,y)k(y,x)f(y)dy$

Any reference is appreciated. hanks in advance!

$\endgroup$
5
  • $\begingroup$ Yes, in my simulation it's the "size" of an average element(sorry for the abuse of terminology, I'm not quite sure myself). And according to this paper link Bounds on eigenvalues of the Hadamard product, it may have something to do with the largest diagonal element of $A$ $\endgroup$ – Sylvan Jun 17 '14 at 17:49
  • 2
    $\begingroup$ We can simply define $f(A):=\rho (A)/\rho(A\circ A^t)$. $\endgroup$ – Christian Remling Jun 17 '14 at 17:53
  • $\begingroup$ Ahh.. that's true, then my question is how does $ρ(A)/ρ(A∘At)$ relates to the $A$'s norm, element size, etc.. $\endgroup$ – Sylvan Jun 17 '14 at 17:56
  • $\begingroup$ Is your question about all matrices or random matrices? The answers could be quite different. $\endgroup$ – Noah Stein Jun 17 '14 at 21:47
  • $\begingroup$ I'm asking about general matrices, yet I'll also appreciate it if I can know something about the random case. $\endgroup$ – Sylvan Jun 17 '14 at 21:53
3
$\begingroup$

The following information may be useful, though probably you already know it.

  1. $\rho(A\circ A^T) \le \rho(A)\rho(A^T)=\rho^2(A)$
  2. The other direction of course fails easily; though it is interesting to note that $\rho(A) \le \rho( (A+A^T)/2)$
  3. Section 5.7 of Topics in matrix analysis by R. A. Horn, C. J. Johnson contains a wealth of material about the Hadamard product, especially for nonnegative matrices.
$\endgroup$
1
  • 1
    $\begingroup$ Thank you for the book recommendation, I'll give it a proper read! $\endgroup$ – Sylvan Jun 18 '14 at 4:53
2
$\begingroup$

If I understand correctly the question, the answer is that no reasonable such function exists. Take the matrix that is zero everywhere except that $A_{i,i+1}=1$, $i=1,\ldots,n-1$, and $A_{n,1}=1$. Then $\rho(A)=1$ but $\rho(A\circ A^T)=0$.

$\endgroup$
4
  • $\begingroup$ Thank you for this example, yet I'm only concerned about positive matrices here, I should have wrote it as strictly positive. Indeed this is an immature conjecture, yet from my numerical result, I believe we could say something about it under proper circumstances. $\endgroup$ – Sylvan Jun 17 '14 at 20:52
  • 2
    $\begingroup$ Make all the zero entries equal to 10^{-23} (that's a particularly nice small number). Same problem, only quantitative. Until you restrict what you mean by f, the only reasonable answer it seems is the one by Christian Remling in comments. $\endgroup$ – ofer zeitouni Jun 17 '14 at 21:09
  • $\begingroup$ If $f(A)$ is something like $1/\min(A_{ii})$, is this relation still possible? Sorry for the way I presented this question, I've made it more open now. $\endgroup$ – Sylvan Jun 17 '14 at 21:33
  • $\begingroup$ Your result is an a.s. statement, and i believe in it. $\endgroup$ – Italo Cipriano Jun 28 '14 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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