# Find the eigenvalues of a matrix

• May 9th 2009, 06:51 PM
Khonics89
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)
• May 10th 2009, 12:06 PM
NonCommAlg
Quote:

Originally Posted by Khonics89

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 $\displaystyle f_n(\lambda)=\det(\lambda I_n - A(n))$ satisfies this recurrence relation for all $\displaystyle n \geq 2: \ \ f_n(\lambda)=(\lambda^2 - 1) f_{n-2}(\lambda) - \lambda f_{n-3}(\lambda),$ where: $\displaystyle f_{-1}(\lambda)=f_0(\lambda)=1, \ f_1(\lambda)=\lambda-1.$

but i have no idea if this basically can help to find the roots of $\displaystyle f_n(\lambda)=0.$ (Wondering)
• May 14th 2009, 12:01 PM
the_doc
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 $\displaystyle A_n$ so that

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

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

$\displaystyle B_n x = \Lambda x$ .

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

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

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

$\displaystyle f_n (\Lambda) = (2-\Lambda) f_{n-1} (\Lambda) - f_{n-2} (\Lambda)$

where $\displaystyle f_1 (\Lambda) = 1- \Lambda$ and $\displaystyle f_0 (\Lambda) = 1$ .
If you can find $\displaystyle \Lambda$ then the eigenvalues of $\displaystyle A_n$ will just be the square root of the different values of $\displaystyle \Lambda$. 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 $\displaystyle B_n$ must be real (symmetric matrix) and their product is 1.