Timeline for Eigenvalues and eigenvectors of the matrix with entries $\dbinom{n+1}{2j-i}$ for $i, j = 1, 2, \ldots, n$
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Aug 21, 2018 at 11:05 | comment | added | Francois Ziegler | How explicit a formula do you need for the eigenvectors? As $B$ is diagonalizable, there is always this... | |
| Aug 20, 2018 at 17:22 | comment | added | darij grinberg | Then the claim by @MTyson that $V^{-1}BV$ is upper-triangular should follow from the fact that $V = UW$ for some invertible upper-triangular matrix $U$. And when I say "fact", I mean "this looks like it's true, but again I don't have the time right now". | |
| Aug 20, 2018 at 17:19 | comment | added | darij grinberg | @Wolfgang: We can define two $n \times n$-matrices $W$ and $U$ by $W = \left( \dbinom{i-1}{j-1}\right) _{1\leq i\leq n,\ 1\leq j\leq n}$ and $U = \left( 2^{n+1-2j+i}\dbinom{n+1-j}{j-i}\right) _{1\leq i\leq n,\ 1\leq j \leq n}$, and then we have $U = W^{-1} B W$. At least I'm quite sure about that, though I still haven't found the time to prove it. Note that the matrix $W$ and the claim that $W^{-1} B W$ is upper-triangular are due to Suvrit in the thread linked by Johann Cigler, but he didn't find the above-diagonal entries of $W^{-1} B W$ explicitly. | |
| Aug 20, 2018 at 17:09 | comment | added | Wolfgang | @MTyson My bad, I had taken the wrong matrix for $B$. Now I have found experimentally that the diagonals of $V^{-1}BV$ are proportional to those of a matrix $A$ with $a_{i,j}=2^{n+1-j}\binom{j-1}{i-1}$, i.e. (more elegant with shifting indices) $A=(2^{n-j}\binom ji)_{i,j=0}^{n-1}$. But the ratios have big prime factors. :( | |
| Aug 20, 2018 at 14:19 | comment | added | MTyson | @Wolfgang Yes, I believe those conventions should work. | |
| Aug 20, 2018 at 8:04 | comment | added | Wolfgang | @MTyson Can you be more precise? E.g. $V=\begin{pmatrix}1&1&1&1&1\\1&2&4&8&16\\ .&.&. \end{pmatrix}$, and then you mean $V^{-1}BV$ should be upper-triangular, or how? (I can't reproduce it) | |
| Aug 20, 2018 at 1:56 | comment | added | MTyson | It seems also that the Vandermonde matrix $V(1,2,\dots,n)$ upper-triangularizes $B$, but I do not see the pattern in the resulting matrix's entries. | |
| Aug 19, 2018 at 20:35 | comment | added | DVD | Also, please, see math.stackexchange.com/questions/2884380/… | |
| Aug 19, 2018 at 20:34 | comment | added | MTyson | The right eigenvectors seem to be of this form. Fit a degree $n-1$ polynomial $p$ that takes the value $(-1)^{i-1}/{n-1\choose i-1}$ at $i$. Then the $i$th coordinate of the eigenvector corresponding to $2^{k+1}$ is $p^{(k)}(i)$. | |
| Aug 19, 2018 at 19:27 | answer | added | Wolfgang | timeline score: 4 | |
| Aug 19, 2018 at 14:11 | history | edited | darij grinberg | CC BY-SA 4.0 |
correction & simpler notation (since i no longer need to induct on n)
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| Aug 19, 2018 at 13:50 | answer | added | T. Amdeberhan | timeline score: 11 | |
| Aug 19, 2018 at 12:52 | comment | added | darij grinberg | @JohannCigler: Thank you! This is noticeably simpler than my proof. | |
| Aug 19, 2018 at 12:51 | comment | added | Johann Cigler | There is some connection with mathoverflow.net/questions/258284/… | |
| Aug 19, 2018 at 10:48 | history | edited | darij grinberg | CC BY-SA 4.0 |
added 277 characters in body
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| Aug 19, 2018 at 10:43 | history | asked | darij grinberg | CC BY-SA 4.0 |