Eigenvectors of a certain big upper triangular matrix - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:23:07Z http://mathoverflow.net/feeds/question/26389 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26389/eigenvectors-of-a-certain-big-upper-triangular-matrix Eigenvectors of a certain big upper triangular matrix Michael Hardy 2010-05-29T22:16:28Z 2012-05-07T12:37:58Z <p>I'm looking at this matrix:</p> <p><code>$$\begin{pmatrix} 1 &amp; 1/2 &amp; 1/8 &amp; 1/48 &amp; 1/384 &amp; \dots \\ 0 &amp; 1/2 &amp; 1/4 &amp; 1/16 &amp; 1/96 &amp; \dots \\ 0 &amp; 0 &amp; 1/8 &amp; 1/16 &amp; 1/64 &amp; \dots \\ 0 &amp; 0 &amp; 0 &amp; 1/48 &amp; 1/96 &amp; \dots \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 1/384 &amp; \dots \\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots \end{pmatrix}$$</code></p> <p>The first row contains the reciprocals of the double factorials <code>$$2, \qquad 2 \cdot 4, \qquad 2 \cdot 4 \cdot 6, \qquad 2 \cdot 4 \cdot 6 \cdot 8, \qquad \dots$$</code> Each row is a <em>shift</em> of a <em>scalar multiple</em> of the first row, and the scalar multiple is in each case itself a reciprocal of a double factorial, so that the main diagonal is the same as the first row. A consequence is that each column is proportional to the corresponding row of Pascal's triangle. E.g. the last column shown is proportional to <code>$$1, 4, 6, 4, 1.$$</code> This matrix is the matrix of coefficients in the "inversion formulas" section of <a href="http://ncatlab.org/nlab/show/trigonometric+identities+and+the+irrationality+of+pi" rel="nofollow">this rant</a> that I wrote.</p> <p>I found the first three eigenvectors: <code>$$\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ \vdots \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \\ 0 \\ \vdots \end{pmatrix}, \begin{pmatrix} 5 \\ -14 \\ 21 \\ 0 \\ 0 \\ \vdots \end{pmatrix}$$</code> Meni Rosenfeld pushed this through some software and found that up to the 40th eigenvalue, the signs of the components of the eigenvectors alternate.</p> <p>Can anything of interest be said about the eigenvectors?</p> <p>Can anything of interest be said about this matrix?</p> http://mathoverflow.net/questions/26389/eigenvectors-of-a-certain-big-upper-triangular-matrix/26431#26431 Answer by Wadim Zudilin for Eigenvectors of a certain big upper triangular matrix Wadim Zudilin 2010-05-30T07:59:30Z 2010-06-01T12:47:48Z <p>As far as I understand your construction, your matrix is $$\operatorname{diag}\biggl(1,\frac12,\frac18,\dots,\frac1{2^nn!},\dots\biggr) \cdot\exp\begin{pmatrix} 0 &amp; \frac12 &amp; 0 &amp; 0 &amp; 0 &amp; \dots \cr 0 &amp; 0 &amp; \frac12 &amp; 0 &amp; 0 &amp; \dots \cr 0 &amp; 0 &amp; 0 &amp; \frac12 &amp; 0 &amp; \dots \cr 0 &amp; 0 &amp; 0 &amp; 0 &amp; \frac12 &amp; \dots \cr \dots &amp; \dots &amp; \dots &amp; \dots &amp; \dots &amp; \dots \end{pmatrix},$$ a diagonal matrix times the exponential of a nilpotent matrix. In your question you discuss some properties of truncations of your infinite matrix, finite $n\times n$ matrices. This corresponds to the truncations of the above diagonal and nilpotent matrices. I've never seen such matrices "in work" but this one could be a nice example of understanding the alteration property of entries of its eigenvectors. In view of the other response, this could be a good point of generalising the previous results in this area.</p> http://mathoverflow.net/questions/26389/eigenvectors-of-a-certain-big-upper-triangular-matrix/26436#26436 Answer by Hans Lundmark for Eigenvectors of a certain big upper triangular matrix Hans Lundmark 2010-05-30T10:16:12Z 2010-05-30T10:16:12Z <p>Your matrix is totally nonnegative (i.e., all minors are nonnegative). This is because it can be factorized as the matrix of binomial coefficients (which is totally nonnegative by the <a href="http://qchu.wordpress.com/2009/11/17/the-lindstrom-gessel-viennot-lemma/" rel="nofollow">Karlin–McGregor–Lindström–Gessel–Viennot lemma</a>) times a diagonal matrix with positive entries $1/(2k)!!$ on the diagonal.</p> <p>For an oscillatory matrix (i.e., a totally nonnegative matrix such that some power of it is totally positive), there is a theorem by Gantmacher &amp; Krein which says that the eigenvalues are real and simple, and the eigenvector corresponding to the $k$th largest eigenvalue has $k-1$ sign changes. (Theorem 5.3 in <a href="http://www.cambridge.org/us//catalogue/catalogue.asp?isbn=0521194083" rel="nofollow">Pinkus, <em>Totally positive matrices</em></a>.)</p> <p>Unfortunately that doesn't apply here, since a power of an upper triangular matrix is upper triangular, so that some minors (below the diagonal) are always zero; hence your matrix is not oscillatory. But perhaps it is possible to use similar ideas to prove the sign changes in your case?</p> http://mathoverflow.net/questions/26389/eigenvectors-of-a-certain-big-upper-triangular-matrix/96205#96205 Answer by Dan Fodor for Eigenvectors of a certain big upper triangular matrix Dan Fodor 2012-05-07T11:32:06Z 2012-05-07T12:37:58Z <p>Might be a wild intuition , I'd say the eigenvalues are the entries of the first row , and that the eigenvector coresponding to the $nth$ eigenvalue ,$k$ is made by adjoining a column of zeroes to the eigenvector coresponding to the eigenvector coresponding to the same eigenvalue for the first $n*n$ minor of the matrix . </p> <p>Example :</p> <p>for eigenvaue $1$ we take the matrix $\begin{pmatrix} 1 \end{pmatrix}$ ,the eigenvetor corresponding to $1$ is $\begin{pmatrix} 1 \end{pmatrix}$ , so we obtain $\begin{pmatrix} 1 \cr 0 \cr 0 \cr 0 \cr 0 \cr \vdots \end{pmatrix}$ as the first eigevector . </p> <p>for eigenvaue $1/2$ we take the matrix $\begin{pmatrix} 1&amp; 1/2 \cr 0 &amp; 1/2\end{pmatrix}$ ,the eigenvetor corresponding to $1/2$ is $\begin{pmatrix} 1 \cr -1\end{pmatrix}$ , so we obtain $\begin{pmatrix} 1 \cr -1 \cr 0 \cr 0 \cr 0 \cr \vdots \end{pmatrix}$ as the second eigevector . </p> <p>The eigenvector coresponding to $1/8$ for $\begin{pmatrix} 1&amp; 1/2 &amp; 1/8 \cr 0 &amp; 1/2 &amp; 1/4 \cr 0 &amp; 0 &amp; 1/8 \end{pmatrix}$ is $\begin{pmatrix} 5 \cr -14 \cr 21 \cr \end{pmatrix}$, you get the ideea .Also , the eigenvectors span the entire space , ie if a possibly infinite (but convergent) sum of eigenvectors is $\vec 0$ then the coefficients of those vectors are $0$ . </p> <p>Here is an explicit formula for the eigenvectors :first select $M_n$ , the $n*n$ truncation of the matrix and calculate $M_n - I*v_n$ , the nt'h eigenvalue . Example : for n=3 , we obtain \begin{pmatrix} 7/8 &amp; 1/2 &amp; 1/8 \cr 0 &amp; 3/8 &amp; 1/4 \cr 0 &amp; 0 &amp; 0 \end{pmatrix} . Now let $S$ be the $(n-1)*(n-1)$ truncation of that , ie \begin{pmatrix} 7/8 &amp; 1/2 \cr 0 &amp; 3/8 &amp; \end{pmatrix} Calculate $S^{-1}$ = \begin{pmatrix} 8/7 &amp; -32/21 \cr 0 &amp; 8/3 &amp; \end{pmatrix} , now multiply $S^{-1}$ with the truncation of the last column of $M_n$ , \begin{pmatrix} 1/8 \cr 1/4 \end{pmatrix} You obtain \begin{pmatrix} -5/21 \cr 2/3 \cr \end{pmatrix} . Concatenating $-1$ to that , you obtain \begin{pmatrix} -5/21 \cr 2/3 \cr -1 \end{pmatrix} , the third eigenvector ,or the nt'h eigenvector in the general case .</p>