Results 1 to 3 of 3

Thread: Find the eigenvalues of a matrix

  1. #1
    Apr 2009

    Find the eigenvalues of a matrix

    Suppose a matrix A(n) = ( n x n ) whos (i,j) entry is 1 if n < = i + j <= n+1 and zero other wise....

    find the eigenvalues of A(n)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    May 2008
    Quote Originally Posted by Khonics89 View Post

    Suppose a matrix A(n) = ( n x n ) whos (i,j) entry is 1 if n < = i + j <= n+1 and zero otherwise....

    find the eigenvalues of A(n)
    what i got is that f_n(\lambda)=\det(\lambda I_n - A(n)) satisfies this recurrence relation for all n \geq 2: \ \ f_n(\lambda)=(\lambda^2 - 1) f_{n-2}(\lambda) - \lambda f_{n-3}(\lambda), where: f_{-1}(\lambda)=f_0(\lambda)=1, \ f_1(\lambda)=\lambda-1.

    but i have no idea if this basically can help to find the roots of f_n(\lambda)=0.
    Last edited by mr fantastic; May 25th 2009 at 01:52 AM. Reason: Updated the quote with new username of OP
    Follow Math Help Forum on Facebook and Google+

  3. #3
    May 2009
    I don't know if this will be any more help than NonCommAlg's result but here's my two cents worth!

    As this is effectively an anti-diagonal matrix with an extra anti-diagonal added on I would multiply the eigenvalue equation by another A_n so that

    A_n A_{n} \, x = A_n \lambda x = \lambda^2 x .

    Define \Lambda = \lambda^2 and B_n = (A_{n})^2 so then we have the new eigenvalue equation

    B_n x = \Lambda x .

    The convenience of this is that B_n turns out to be a symmetric tri-diagonal matrix with elements b_{i,j} given by

    b_{i,j}= \left \lbrace \begin{array}{c}<br />
1 \quad :\quad i = j = n \\<br />
1 \quad : \quad i = j-1 \\<br />
1 \quad : \quad i = j+1 \\<br />
2 \quad : \quad i = j \neq n<br />
\end{array} \right.

    with f_n(\Lambda) = \left| B_n - \Lambda I_n \right| given by the recurrence relation

    <br />
f_n (\Lambda) = (2-\Lambda) f_{n-1} (\Lambda) - f_{n-2} (\Lambda)

    where f_1 (\Lambda) = 1- \Lambda and f_0 (\Lambda) = 1 .
    If you can find \Lambda then the eigenvalues of A_n will just be the square root of the different values of \Lambda. Of course you'll end up with twice the number required but it shouldn't be too difficult to sift out the ones you don't need.

    I only mention this alternative method as symmetric tri-diagonal matrices are well researched and have lots about them on the web aswell as many very efficient numerical methods for determining the eigenvalues if that's what you need to do.

    Also, the eigenvalues of B_n must be real (symmetric matrix) and their product is 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: November 27th 2010, 03:07 PM
  2. Replies: 2
    Last Post: October 17th 2010, 07:13 AM
  3. Replies: 2
    Last Post: August 11th 2010, 04:26 AM
  4. Replies: 4
    Last Post: April 25th 2010, 03:49 AM
  5. How to find the eigenvalues of a 3x3 matrix
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 17th 2009, 12:46 PM

Search Tags

/mathhelpforum @mathhelpforum