we used row-operations and column operations to evaluate det(A - λI).

the matrix we wound up with is NOT A-λI, it just has the same determinant.

it is clear that with your original matrix A(e1) = (1,1,1,.....,1) ≠ n(e1) = (n,0,0,.....,0).

Printable View

- Oct 15th 2011, 05:06 PMDevenoRe: Eigenvalues
we used row-operations and column operations to evaluate det(A - λI).

the matrix we wound up with is NOT A-λI, it just has the same determinant.

it is clear that with your original matrix A(e1) = (1,1,1,.....,1) ≠ n(e1) = (n,0,0,.....,0). - Oct 15th 2011, 05:09 PMdwsmithRe: Eigenvalues
- Oct 15th 2011, 05:21 PMDevenoRe: Eigenvalues
you're trying to find the eigenvectors of A = (a_ij) ,where every entry is 1, right?

if v is an eigenvector, then Av = λv, so (A-λI)v = 0.

when λ = n, this is: (A-nI)v = 0, a homogeneous linear system of equations.

e1 is not a solution to that system. - Oct 15th 2011, 05:24 PMdwsmithRe: Eigenvalues
- Oct 15th 2011, 05:46 PMDevenoRe: Eigenvalues
- Oct 15th 2011, 05:56 PMdwsmithRe: Eigenvalues
- Oct 15th 2011, 06:07 PMDevenoRe: Eigenvalues
the matrix in post 5 was not obtained by elementary row operations, but by elementary row AND column operations.

column operations DO NOT preserve the solution space. - Oct 15th 2011, 06:31 PMdwsmithRe: Eigenvalues
- Oct 15th 2011, 08:51 PMDevenoRe: Eigenvalues
is it not the case that:

doesn't that make (1,1,1,...,1) an eigenvector corresponding to the eigenvalue n? - Oct 18th 2011, 06:36 AMdwsmithRe: Eigenvalues
Dealing with the same matrix but the diagonals now all zeros, the characteristic equation would be