14,480 reputation
32979
bio website suvrit.de
location MIT
age
visits member for 4 years, 8 months
seen 3 hours ago

Researcher in Optimization and Machine Learning.

Hobbyist in Inequalities, Matrix Analysis, Polynomials, etc.

I'm here mostly to learn mathematics!


3h
comment Positive roots of a polynomial
For $n=3$ and also for $n=4$ it seems there is just one change of sign in the consecutive terms, so Descartes' rule of signs should kick in and yield at most one positive root. Perhaps one can even inductively show that there is just one change of sign, and a small additional argument shows that there is at least one positive root.
9h
comment How to prove this determinant is positive?
@GHfromMO Great! I too got quite some headaches with this excellent problem!
11h
comment How to prove this determinant is positive?
@user23765: I see the difference now. In my answer, I assume a "sufficient" condition, and frequently it is satisfied. In the cases where it is not (the $\pm a + \log(-1)$ case in your bullet point 3), things still work out. So it remains to guarantee that these are the only possible cases.
11h
comment How to prove this determinant is positive?
$M \in \mathbb{C}^{2n\times 2n}$ and seems to either have the structure mentioned in my answer, or have the structure in bullet point 3---though latter structure seems to happen due to "non principal matrix logarithm"...
12h
revised How to prove this determinant is positive?
fixed a typo, and changed flow a tiny bit.
17h
comment How to prove this determinant is positive?
@user23765: it seems that one can always have complex $M$ that satisfies the skew-Hamiltonian-like structure mentioned in my answer. Since only existence of $M$ is needed but not its actual construction, should it really matter?
17h
comment On closest unitary matrix
If $A=UP$ where $U$ is unitary and $P$ is positive definite, then we know that $\|A-U\| \le \|A-V\|$ for any unitary matrix $V$ (the norm above is any unitarily invariant norm). This may help extend the proof to your case, perhaps by using a unitary dilation of $B$...
18h
revised How to prove this determinant is positive?
expanded the answer....
1d
comment How to prove this determinant is positive?
@GHfromMO: thanks for the clarification! (my prev. comment was written out of sequence with your comment though)
1d
comment How to prove this determinant is positive?
Ok, I finally got the argument I think. Since each $e^{A_i}$ lies in $O(n,n)$ and because we can make the product of these exponentials approach any element in $O(n,n)$, we should eventually have a counterexample for large enough $N$.
1d
comment How to prove this determinant is positive?
Do we need to have zero trace for $\log T$?
1d
awarded  Cleanup
1d
answered How to prove this determinant is positive?
1d
comment How to prove this determinant is positive?
The case $N=1$ is trivial because $e^{A_1}$ is positive definite, while the case $N=2$ is also trivial because $\lambda(e^{A_1}e^{A_2}) > 0$.
1d
comment How to prove this determinant is positive?
The bipartite case above is special, in particular as also others noticed (and you too, I think), $\det(I+P_1\cdots P_N)$ can be easily negative for positive definite $P_i$, each of which has determinant $1$ (just like each $e^{A_i}$ above); seems like some kind of "stable polynomials" are involved here...
2d
comment How to prove this determinant is positive?
This is a really nice problem; has already given me plenty of numerical headaches with my mistaken counterexamples!
Apr
30
reviewed Leave Closed Disprove this Piece of Jensen's Inquality “Black Magic”
Apr
29
reviewed Leave Open Programming workbooks in C++ and Research Math
Apr
29
reviewed Close is there a collected works of J.P. Lagrange?
Apr
29
reviewed Leave Open Area of a plane surface that gives a lot of theoretical problems