User amin - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T01:41:19Z http://mathoverflow.net/feeds/user/27869 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/113850/auto-correlation-calculating-auto-correlation-matrix-with-given-weighted-mean [Auto Correlation] Calculating Auto Correlation Matrix with Given Weighted Mean Amin 2012-11-19T17:28:33Z 2012-11-19T20:06:52Z <p>Assume that mean of some weighted data calculated. $m=\frac{\sum_{i=1}^{n} w_i x_i}{\sum_{i=1}^{n} w_i}$</p> <p>i want to calculate $C=\sum_{k=1}^{n}w_i x_i x_i^T$</p> <p>But i dont have access to any $x_i$. </p> <p>I only got $m$ and every $w_i$.</p> <p>Is there any relation between these variables?</p> <p>Regards.</p> http://mathoverflow.net/questions/111654/eigenvalues-of-a-symmetric-positive-semi-definite-psd-matrix-after-rank-one-upd Eigenvalues of a Symmetric Positive Semi-Definite (PSD) matrix after rank one update Amin 2012-11-06T15:29:42Z 2012-11-07T04:28:55Z <p>I have a Symmetric Positive Semi-Definite matrix $A$ which i know its eigenvalue and eigenvectors. let $v$ and $u$ be a random column vector. i want to know if it is possible to have eigenvalues of matrix $A+uv^T$.</p> <p>I don't need its eigenvectors, but it is required to have the most precise eigenvalues. We know that $uv^T$ is also a rank one PSD matrix.</p> <p>Is there a close form to this problem?</p> <p>${\bf PS:}$ As you see i found the solution, but after implementation of this i can see a very small error in result, it would be appreciated if anyone know why this is happening, because we didn't used any approximation to get the result and it should be exact.</p> http://mathoverflow.net/questions/111654/eigenvalues-of-a-symmetric-positive-semi-definite-psd-matrix-after-rank-one-upd/111659#111659 Answer by Amin for Eigenvalues of a Symmetric Positive Semi-Definite (PSD) matrix after rank one update Amin 2012-11-06T16:14:57Z 2012-11-07T04:25:55Z <p>Found the solution in [1], Please let me know if i had any mistake.</p> <p>we have </p> <p>$Ax=\lambda x$ (1)</p> <p>differentiate both side we have</p> <p>$\Delta Ax+ A\Delta x= \Delta\lambda x+\lambda\Delta x$.</p> <p>multiply both side with $x^T$ and we have </p> <p>$x^T\Delta Ax+ x^TA\Delta x= x^T\Delta\lambda x+x^T\lambda\Delta x$ (2)</p> <p>from (1), by transposing both side we have $x^TA^T=\lambda x^T$. because $A$ is symmetric, we can eliminate the transpose of $A$.</p> <p>$x^TA=\lambda x^T$</p> <p>know we can eliminate last terms in both side because equality of them.</p> <p>$x^T\Delta Ax= x^T\Delta\lambda x$ </p> <p>after a simple manipulation we get</p> <p>$\frac{x^T\Delta Ax}{x^Tx}= \Delta\lambda$</p> <p>As we know $x^Tx=1$ the final result will be</p> <p>$\Delta\lambda={x^T\Delta Ax}$</p> <p>[1] Ning, H., Xu, W., Chi, Y., Gong, Y., &amp; Huang, T. S. (2010). Incremental spectral clustering by efficiently updating the eigen-system. Pattern Recognition, 43(1), 113-127.</p> http://mathoverflow.net/questions/113850/auto-correlation-calculating-auto-correlation-matrix-with-given-weighted-mean Comment by Amin Amin 2012-11-20T05:46:30Z 2012-11-20T05:46:30Z Thanks Benoit for your reply. I dont see any violation of law, but if there is one, i would delete this post for sure. let me know about that. regards. http://mathoverflow.net/questions/111654/eigenvalues-of-a-symmetric-positive-semi-definite-psd-matrix-after-rank-one-upd Comment by Amin Amin 2012-11-08T04:04:51Z 2012-11-08T04:04:51Z Thank you Federico, I appreciate you helpful tip, i hope you would give me some references about this. Regards http://mathoverflow.net/questions/111654/eigenvalues-of-a-symmetric-positive-semi-definite-psd-matrix-after-rank-one-upd/111665#111665 Comment by Amin Amin 2012-11-07T04:40:17Z 2012-11-07T04:40:17Z Dear Felix As i read the paper its about updating a diagonal matrix. im not 100% sure if i got that right but the $A$ matrix is not diagonal, its a Symmetric PSD matrix. Regards