User bart - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T06:10:01Z http://mathoverflow.net/feeds/user/14039 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93337/bandwidth-reduction-of-multiple-matrices/93395#93395 Answer by Bart for Bandwidth reduction of multiple matrices Bart 2012-04-07T08:25:00Z 2012-04-07T08:25:00Z <p>If you use the typical reorderings (like reverse Cuthill-McKee ordering), $P^T A P$ will have a smaller bandwidth (in general, not tridiagonal though). Since $P$ is a permutation matrix, all $P^T D_i P$ will remain diagonal too.</p> <p>Bart</p> http://mathoverflow.net/questions/73028/numerical-linear-algebra-how-to-compute-bta-1b-efficiently/73847#73847 Answer by Bart for Numerical linear algebra: how to compute $b^TA^{-1}b$ efficiently Bart 2011-08-27T14:19:03Z 2011-08-27T14:19:03Z <p>This is possibly an answer from a practical point of view: If you use the CG method for solving $x=A^{-1}b$ then $b^T A^{-1}b$ can be obtained along the way. However, it has been shown that computing $b^T A^{-1}b$ during the iteration can converge faster than first solving for $x$ and then multiplying $b^T x$. See "Z. Strakos and P. Tichy, On efficient numerical approximation of the bilinear form c*A-1b , SIAM Journal on Scientific Computing (SISC), 33, 2011, pp. 565-587" and the references therein for the positive definite case.</p> http://mathoverflow.net/questions/73551/smoothness-along-rays-sufficient-for-global-smoothness Smoothness along rays sufficient for global smoothness Bart 2011-08-24T09:20:48Z 2011-08-24T09:20:48Z <p>Hi,</p> <p>Suppose I have a function $f:\mathbb{R}^d \to \mathbb{R}$ and I know that $f$ is smooth ($C^\infty$) along each ray $t \mapsto f(td)$ on $t \in [-\epsilon, \epsilon]$ and all directions $d \in \mathbb{R}^d$.</p> <p>Is smoothness along these rays sufficient for $f$ to be smooth around $0$ as a multivariate function (all partial derivatives exist)?</p> <p>Thanks.</p> http://mathoverflow.net/questions/60938/is-any-n-1-times-n-1-submatrix-of-an-n-times-n-vandermonde-matrix-invert/60939#60939 Answer by Bart for Is any $(n-1)\times (n-1)$ submatrix of an $n \times n$ Vandermonde matrix invertible? Bart 2011-04-07T13:42:15Z 2011-04-07T13:42:15Z <p>No. Take $n=2$ and $\alpha_1=0, \alpha_2=1$. Then the Vandermonde matrix $\begin{bmatrix} 1 &amp; 0 \\ 1 &amp; 1 \end{bmatrix}$ is invertible, but the upper-right submatrix is not.</p> http://mathoverflow.net/questions/60298/explanation-of-y-x-exp-triangle-for-a-lie-group/60311#60311 Answer by Bart for Explanation of $y = x \exp(\triangle)$ for a Lie Group Bart 2011-04-01T17:11:59Z 2011-04-01T17:11:59Z <p>The answer of Theo basically says it all what the exponential is concerned, but I maybe can shed some light regarding the optimization perspective. </p> <p>Let $M$ be your Lie group and suppose it is a subgroup of $\textrm{GL}_n$. Now, if you want to solve $\min f(x) \quad \textrm{s.t. x \in M}$ by an iterative method, one usually uses a so-called smooth retraction map $R_x: T_x M \to M.$ to replace your current approximation $x$ to $x_+ := R_x(\Delta)$. Think for example of doing a line-search $R_x(-t \textrm{ grad}_x f)$ for $t>0$ where $\textrm{ grad}_x f$ is the Riemannian gradient of $f$.</p> <p>(The retraction map has to fulfill some properties, in order for this to work, likes smoothness and being a first-order approximation of the geodesics (see below).)</p> <p>Due to the left (or right) action of a Lie group on itself by multiplication, the exponential mapping at the identity $\exp$ can be transported to get a retraction at $x$ as $x\exp$. As Theo already explained, $\exp$ does not need to be a global diffeomorphism, but that is not needed for optimization, since we perform updates locally.</p> <p>Another typical choice for $R_x$ are the geodesics in $x$. For some metrics, the $\exp$ coincides with the geodesics (for instance, bi-invariant metrics), but not always. One can also use cheaper alternatives for $R_x$ if your are only concerned with optimization. A nice reference for this retraction-based optimization on manifolds (and so, also Lie groups) is <a href="http://press.princeton.edu/titles/8586.html" rel="nofollow">http://press.princeton.edu/titles/8586.html</a> .</p> http://mathoverflow.net/questions/59894/find-the-point-on-the-stiefel-manifold-that-is-closest-to-a-matrix/60221#60221 Answer by Bart for Find the point on the Stiefel Manifold that is closest to a matrix Bart 2011-03-31T18:29:42Z 2011-03-31T18:29:42Z <p>Are you sure you need the distance function $\|XX^T-YY^T\|_2$? If not, the solution to $\min \{\|E\|: E \in \mathbb{R}^{m \times n}, (X+E)^T(X+E) = I \}$ for the 2-norm and the Frobenius norm is the polar decomposition; see, e.g., section 4 in <a href="http://www.maths.man.ac.uk/~nareports/narep161.pdf" rel="nofollow">http://www.maths.man.ac.uk/~nareports/narep161.pdf</a> .</p> <p>Maybe you do want $\|XX^T-YY^T\|_2$ as distance function because of the equivalence by the orthogonal group. In that case you maybe want to rephrase your problem on the Grassmann manifold on linear subspaces. In fact, the Stiefel manifold with the orthogonal group factored out is exactly the Grassmann manifold. There is a nice paper about this: <a href="http://dx.doi.org/10.1137/S0895479895290954" rel="nofollow">http://dx.doi.org/10.1137/S0895479895290954</a> .</p> http://mathoverflow.net/questions/60168/orbits-of-semi-algebraic-actions Orbits of semi-algebraic actions Bart 2011-03-31T09:24:40Z 2011-03-31T09:24:40Z <p>Hello all,</p> <p>I recently came across the following Theorem in Gibson (Singular points of smooth mappings, 1979). Since I haven't seen this result somewhere else and this reference is not so widespread, I was wondering if it is correct, known or trivial?</p> <p>Theorem B4 in Gibson79: Let $\phi: G \times M \to M$ be a smooth action of a Lie group $G$ on a smooth manifold $M$. Suppose that the action is semi-algebraic (i.e., the graph of $\phi$ is a semi-algebraic set). Then all the orbits are smooth submanifolds of $M$.</p> <p>(Smooth means $C^\infty$; smooth submanifold means a differential manifold that is embedded.)</p> <p>The proof goes as follows. The orbit at $x \in M$ is semi-algebraic by Tarski-Seidenberg. Every non-void semi-algebraic has at least a one neighbourhood where it is a submanifold in $M$. Since orbits are homogeneous by the action of $G$, this neigbourhood extends to the whole orbit.</p> <p>Thanks!</p> http://mathoverflow.net/questions/60126/do-surjections-exist-which-are-not-submersions-on-a-set-of-measure-non-zero Do surjections exist which are not submersions on a set of measure non-zero. Bart 2011-03-30T21:28:12Z 2011-03-30T23:54:27Z <p>Let $f: M \to N$ be a smooth maps between smooth manifolds. Then $f$ is a submersion (by definition) if its differential is also surjective. Now suppose $f$ is surjective. Is it possible that the surjective map $f$ fails to be a submersion on a set in $N$ of measure non-zero? If so, what is such a map?</p> <p>Suppose the manifolds $M$ and $N$ are non-compact. Does this change the previous answer?</p> <p>Thanks!</p> http://mathoverflow.net/questions/73551/smoothness-along-rays-sufficient-for-global-smoothness Comment by Bart Bart 2011-08-24T14:19:46Z 2011-08-24T14:19:46Z Apparently, in the published version there is the condition that the k-th partial derivative of $f \circ u$ needs to be smaller than $C k! / r^k$ for some $r&gt;0$. http://mathoverflow.net/questions/73551/smoothness-along-rays-sufficient-for-global-smoothness Comment by Bart Bart 2011-08-24T13:35:46Z 2011-08-24T13:35:46Z Thanks Willie. Looking at the Boman paper, demanding smoothness along only analytic curves is indeed not sufficient. It turns out that when $f \circ u$ is real analytic for every real analytic curve $u$, that $f$ is real analytic. See &quot;An Ontology of Directional Regularity Implying Joint Regularity&quot; published in Real Analysis Exchange, available at <a href="http://www.math.wustl.edu/~sk/joint.pdf" rel="nofollow">math.wustl.edu/~sk/joint.pdf</a> . http://mathoverflow.net/questions/73551/smoothness-along-rays-sufficient-for-global-smoothness Comment by Bart Bart 2011-08-24T12:15:55Z 2011-08-24T12:15:55Z That is indeed a nice counterexample. Would you happen to have a reference of this result of Jan Boman? In addition, would smoothness along all real analytic curves be sufficient too? http://mathoverflow.net/questions/55480/geodesics-for-a-homogeneous-space/55482#55482 Comment by Bart Bart 2011-04-04T08:34:07Z 2011-04-04T08:34:07Z @Jos&#233;: Thanks for the explanation! (Maybe you can specify in the answer that you consider the $G$-invariant metric.) http://mathoverflow.net/questions/55480/geodesics-for-a-homogeneous-space/55482#55482 Comment by Bart Bart 2011-04-03T16:35:50Z 2011-04-03T16:35:50Z @Jos&#233;, maybe I am missing something here, but I don't see what the positivity of the OP (orthogonal product?) has to do with $H$ being a subgroup of the orthogonal group? Any $G/H$ with $H$ a closed Lie subgroup of $G$ is a homogeneous space. For sure, one can put a positive OP on $G/H$. http://mathoverflow.net/questions/55480/geodesics-for-a-homogeneous-space/55482#55482 Comment by Bart Bart 2011-04-03T09:25:04Z 2011-04-03T09:25:04Z I don't think you can find a reductive split for any Riemannian homogeneous space since $H$ does not need to be compact. It only needs to be closed in $G$.