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)
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 so that
.
Define and so then we have the new eigenvalue equation
.
The convenience of this is that B_n turns out to be a symmetric tri-diagonal matrix with elements given by
with given by the recurrence relation
where and .
If you can find then the eigenvalues of will just be the square root of the different values of . 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 must be real (symmetric matrix) and their product is 1.