Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm trying to efficiently calculate the eigenvalues for this matrix. It looks like this:

http://www.math.pitt.edu/~sussmanm/2071Spring08/lab06/img2.png

It's diagonalized by the DST matrix in a similar way that a circulant matrix is diagonalized by the DFT. In the case of the circulant matrix, you can take the DFT of the first column to get the eigenvalues. I'm trying to do a similar thing for the second difference matrix using the DST function in matlab. The best I can get are half the eignvalues by taking the dst of the middle column. Anyone have any ideas how I can do this (Without explicitly forming the DST matrices and second difference matrix)? Thanks.

share|improve this question
    
$2cos(πj/(n+1)−2$ for $j=1,…,n$. But your question is out of place on this site, see the faq. –  Chris Godsil Mar 1 '13 at 0:23
    
Chris, can you elaborate how you got this solution please? Thanks. Also, this is minor but I think it should be: 2-2cos(πj/(n+1)) for j=1,…,n. –  user31848 Mar 1 '13 at 0:29
    
@unkown(google): see this older MO question for a general approach, although this particular case may be simpler: mathoverflow.net/questions/91161/… –  Yemon Choi Mar 1 '13 at 1:23
1  
@unknown: The matrix you have is $A-2I$, where $A$ is the adjacency matrix of the path. The eigenvalues of the path are easily found using google. For more, the key is to note characteristic polynomial of $A$ is a form of Chebyshev polynomial, and these eigenvalues are calculated in many places. –  Chris Godsil Mar 1 '13 at 1:40
    
@Chris. If you post your response as a question I can accept it, although you said this question wasn't suited for this site so it's up to you. –  user31848 Mar 1 '13 at 1:57
show 1 more comment

1 Answer

You can see the solution here: http://www.math.purdue.edu/~eremenko/dvi/beads.pdf This is from a linear algebra course that I teach.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.