Timeline for If graph is tree what can be said about its adjacency matrix ?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 12, 2012 at 16:21 | vote | accept | Alexander Chervov | ||
| Jun 7, 2012 at 2:59 | comment | added | tweetie-bird | Actually, you can show that the elementary symmetric functions $e_i$ for $i$ odd are all 0 by expressing them in terms of the power sum symmetric functions and seeing that each summand is 0. Then shows that the coefficients for odd powers of $x$ in your original $P(x)$ are 0, implying $P(x)$ is an even function. This is interesting to think about what happens when you take the symmetric functions and mod out by the elements of odd degree in a basis. Thanks! | |
| Jun 7, 2012 at 2:31 | comment | added | Gjergji Zaimi | I had a typo, I meant $\prod (1-x\lambda_i)$. | |
| Jun 7, 2012 at 0:24 | comment | added | tweetie-bird | Thanks! I'm a little slow in understanding your comment though I'm afraid. I don't yet see how to deduce that $log P(x)$ is even. I'll keep thinking about it. | |
| Jun 7, 2012 at 0:08 | comment | added | Gjergji Zaimi | @tweetie-bird: Look at the characteristic polynomial $P(x)=\prod(x-\lambda_i)$. Since your odd-power sums vanish then $\log P(x)$ is even. This implies $P$ is even so the roots come in pairs $\pm \lambda$. | |
| Jun 6, 2012 at 23:52 | history | edited | tweetie-bird | CC BY-SA 3.0 |
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| Jun 6, 2012 at 23:40 | comment | added | tweetie-bird | Now I'm also curious about the general statement for real numbers $\lambda_1, \dots ,\lambda_n$ of whether knowing $\sum_{j=1}^n \lambda_j^i =0$ for all positive, odd integers $i$ implies that the nonzero reals come in pairs $\mu_j, \nu_j$ of equal magnitude and opposite sign. It looks like this might follow from a variant on the Vandermonde determinant being nonzero where the usual Vandermonde matrix entry $a_i^j$ is replaced by $\mu_i^j + \mu_i^{j-1}\nu_i + \mu_i^{j-2}\nu_i^2 + \cdots + \nu_i^j$. | |
| Jun 6, 2012 at 20:27 | comment | added | tweetie-bird | I see -- you have given a proof of my conjecture in the specific case of adjacency matrices of bipartite graphs (rather than working with the relations among real numbers), noting that the standard basis vectors are indexed by the nodes of the graph and that applying the adjacency matrix sends a node to the sum of standard basis vectors indexing its neighbors. That's a nice proof! Thanks! | |
| Jun 6, 2012 at 20:12 | comment | added | Douglas Zare | You can modify any eigenvector with eigenvalue $\lambda$ by negating the values on one of the parts to get an eigenvector with eigenvalue $-\lambda$. | |
| Jun 6, 2012 at 20:01 | history | edited | tweetie-bird | CC BY-SA 3.0 |
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| Jun 6, 2012 at 19:42 | history | answered | tweetie-bird | CC BY-SA 3.0 |