Author |
Topic: Eigenvalues (Read 556 times) |
|
ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489
|
Given a 2n x 2n matrix whose entries are all odd integers, show that its eigenvalues cannot be odd integers.
|
|
IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
Sameer
Uberpuzzler
Pie = pi * e
Gender:
Posts: 1261
|
|
Re: Eigenvalues
« Reply #1 on: Dec 5th, 2006, 9:58pm » |
Quote Modify
|
Consider matrix A = a1,1 a1,2 ... a1,2n a2,1 a2,2 ... a2,2n ... a2n,1 a2n,2 ... a2n,2n Then eigenvector can be defines as L such that AX=LX So (a1,1+a2,1+...a2n,1) = L1 (a1,2+a2,2+...a2n,2) = L2 ... (a1,2n+a2,2n+....+a2n,2n) = L2n Since all the elements are odd summation of 2n odd numbers is also even. Thus all L1,L2...L2n is even.
|
« Last Edit: Dec 5th, 2006, 9:59pm by Sameer » |
IP Logged |
"Obvious" is the most dangerous word in mathematics. --Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself. Sir Arthur Eddington, quoted in Bridges to Infinity
|
|
|
towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Eigenvalues
« Reply #2 on: Dec 6th, 2006, 1:21am » |
Quote Modify
|
If I had to hazard a guess I'd say A2 has all even integers, so it's eigenvalues are all even, and they're the squares of the eigenvalues of A, which can thus not be odd. On the other hand, that's assumming the eigenvectors aren't all even. But then, if they were you could divide them by the highest common factor of two. So errr, I'm left to hope there isn't a (reduced) eigenvector with an even number of odd numbers..
|
« Last Edit: Dec 6th, 2006, 1:22am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Eigenvalues
« Reply #3 on: Dec 6th, 2006, 6:11am » |
Quote Modify
|
on Dec 6th, 2006, 1:21am, towr wrote:So errr, I'm left to hope there isn't a (reduced) eigenvector with an even number of odd numbers.. |
| The number of odd numbers can be even all it wants, it just can't be 0. My solution: Suppose A has an odd integer eigenvalue L. Then Ax=Lx for some rational vector x; multiplying by an integer, and dividing by a power of 2 if necessary, we may assume x is integral, with at least one odd coordinate. Now, Ax=Lx means that, mod 2, xi = Lxi = [sum] aijxj = [sum] xj for all i. Since the RHS is independent of i, all the xi must be odd. But then [sum] xj is even, a contradiction. (On the other hand, an odd array of odd size can have odd eigenvalues, but they can't all be odd integers, since the determinant is even, except in the 1x1 case.)
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Eigenvalues
« Reply #4 on: Dec 6th, 2006, 6:30am » |
Quote Modify
|
on Dec 6th, 2006, 6:11am, Eigenray wrote:The number of odd numbers can be even all it wants, it just can't be 0. |
| Sometimes I look back at something I wrote, and wonder what I was thinking..
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
flamingdragon
Uberpuzzler
Gender:
Posts: 671
|
|
Re: Eigenvalues
« Reply #5 on: Dec 6th, 2006, 3:35pm » |
Quote Modify
|
LOL
|
|
IP Logged |
"The fool doth think he is wise, yet the wise man knoweth himself to be a fool"
"He who commands the past, commands the future. He who commands the future, commands the past."
|
|
|
|