most recent 30 from http://mathoverflow.net 2013-05-21T16:38:07Z http://mathoverflow.net/feeds/user/15182 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90844/a-holder-like-inequality Zae Kwong 2012-03-10T20:53:08Z 2012-05-29T15:01:18Z

<p>If $0&lt; a_1\le a_2\le \cdots \le a_n\le a_{n+1}$ and $p>1$, is it true that $$\left(\frac{n+1}{n}\right)^{1-\frac{1}{p}}\left(\frac{\sum_{i=1}^{n+1}a_i^p}{\sum_{i=1}^{n}a_i^p}\right)^{\frac{1}{p}}\ge \frac{\sum_{i=1}^{n+1}a_i}{\sum_{i=1}^{n}a_i}?$$ The numerator and denominator looks like Hölder's inequality.</p>

http://mathoverflow.net/questions/70171/is-there-a-relation-between-pdp-and-pdp Zae Kwong 2011-07-12T20:00:00Z 2011-10-25T16:17:55Z

<p>Considering in the complex fields. Let $P$ be a nonsingular matrix, $P^* $ be its conjugate transpose, is there a relation between $P^*|D|P$ and $|P^*DP|$, where $D$ is a diagonal matrix? In particular, is it true </p> <p>$P^* |D|P \ge |P^*DP|$ in the sense of Lowner order, or is there an order for eigenvalues?</p> <p>Here $|A|=(A^*A)^{1/2}$, the absolute value of a complex matrix.</p> <p><strong>Edit</strong> As I know from Suvrit's answer, there is no relation like $P^* |D|P \ge |P^* DP|$ in the sense of Lowner order. So my question becomes, is the $i$th largest eigenvalue of $P^* |D|P$ larger than that of $|P^*DP|$?</p>

http://mathoverflow.net/questions/72887/existence-of-a-symmetric-matrix Zae Kwong 2011-08-14T21:03:32Z 2011-08-15T02:46:13Z

<p>In the real field. Given a diagonal matrix $D$ and a symmetric matrix $A$. For every skew symmetric matrix $S$, is there always a symmetric matrix $H$ such that $-\operatorname{trace}(DSAS)=\operatorname{trace}(DHAH)$? </p> <p>If $A$ is also diagonal, this can be easily seen true. </p>

http://mathoverflow.net/questions/71678/arithmetic-geometric-mean-of-positive-matrices Zae Kwong 2011-07-30T19:06:07Z 2011-08-01T18:09:05Z

<p>Let $A,B$ be positive definite (Hermitian) matrices. Define the Arithmetic-geometric means of positive matrices by $A_0=A, G_0=B$, $A_{n+1}=\frac{A_n+G_n}{2}, G_{n+1}=A_n\natural G_n$, where $A_n\natural G_n$ means the geometric mean defined in <a href="http://www.isid.ac.in/~statmath/eprints/2011/isid201102.pdf" rel="nofollow">http://www.isid.ac.in/~statmath/eprints/2011/isid201102.pdf</a></p> <p>Will ${A_n}$ and ${G_n}$ converge to the same matrix? </p>

http://mathoverflow.net/questions/71433/a-matrix-eigenvalue-problem Zae Kwong 2011-07-27T20:11:32Z 2011-07-27T20:57:15Z

<p>This question is related to <a href="http://mathoverflow.net/questions/70689/ask-some-matrix-eigenvalue-inequalities" rel="nofollow">http://mathoverflow.net/questions/70689/ask-some-matrix-eigenvalue-inequalities</a></p> <p>Let $\begin{bmatrix} A&amp; B \\ B^* &amp;A \end{bmatrix}$ be positive semidefinite. Is it true $\lambda_i^{1/2}(B^*B)\le \lambda_i(A)$? Here, $λ_i(⋅)$ means the ith largest eigenvalue of ⋅. </p>

http://mathoverflow.net/questions/90844/a-holder-like-inequality Zae Kwong 2012-03-10T22:36:22Z 2012-03-10T22:36:22Z

If it is true, then I have something to say, but....

http://mathoverflow.net/questions/90844/a-holder-like-inequality/90853#90853 Zae Kwong 2012-03-10T22:36:00Z 2012-03-10T22:36:00Z

Clever argument.

http://mathoverflow.net/questions/71678/arithmetic-geometric-mean-of-positive-matrices Zae Kwong 2011-09-12T12:42:52Z 2011-09-12T12:42:52Z

Hi Wadim:Inspired from this paper <a href="http://lab.rockefeller.edu/cohenje/PDFs/140CohenNussbaumArithmeticGeometricMeansPositiveMatricesMath.pdf" rel="nofollow">lab.rockefeller.edu/cohenje/PDFs/&hellip;</a> Instead of entrywise definition, I would like to see a natural definition.

http://mathoverflow.net/questions/73096/does-the-determinant-equality-hold Zae Kwong 2011-08-17T20:20:02Z 2011-08-17T20:20:02Z

Not homework, for sure.

http://mathoverflow.net/questions/72887/existence-of-a-symmetric-matrix Zae Kwong 2011-08-15T18:11:53Z 2011-08-15T18:11:53Z

It is a very specific problem, it does not have interesting background.

http://mathoverflow.net/questions/71678/arithmetic-geometric-mean-of-positive-matrices/71767#71767 Zae Kwong 2011-08-01T18:58:18Z 2011-08-01T18:58:18Z

I don't think there is &quot;no loss of generality&quot; to assume $A=I, B=D$. $(P^*AP)^{1/2}\ne P^*A^{1/2}P$ generally.

http://mathoverflow.net/questions/70937/the-largest-entry-of-inverse-m-matrix Zae Kwong 2011-07-22T12:44:24Z 2011-07-22T12:44:24Z

Paul's comment is exactly what I what to express. If $A=diag(1/2,1/2)$, then its inverse is $diag(2,2)$, not what I want.

http://mathoverflow.net/questions/70937/the-largest-entry-of-inverse-m-matrix Zae Kwong 2011-07-21T22:22:20Z 2011-07-21T22:22:20Z

GH: I edited that to make it clear.

http://mathoverflow.net/questions/70171/is-there-a-relation-between-pdp-and-pdp/70184#70184 Zae Kwong 2011-07-13T17:12:51Z 2011-07-13T17:12:51Z

I think there is a relation between eigenvalues, i.e. the $k$th largest eigenvlaue of $P^* |D|P$ is no less than the $k$th largest eigenvalue of $|P^*DP|