M is a nxn matrix where all entries are 1's.

My conjecture is that n - 1 eigenvalues are 0 and the nth eigenvalues is one.

If this is correct, how do I go about proving it?

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- October 14th 2011, 07:42 PMdwsmithEigenvalues
M is a nxn matrix where all entries are 1's.

My conjecture is that n - 1 eigenvalues are 0 and the nth eigenvalues is one.

If this is correct, how do I go about proving it? - October 14th 2011, 08:38 PMDevenoRe: Eigenvalues
did you test your conjecture for small values of n? for a 2x2 matrix:

which appears to have eigenvalues 0 and 2. - October 14th 2011, 08:50 PMdwsmithRe: Eigenvalues
- October 14th 2011, 09:41 PMDevenoRe: Eigenvalues
try a 3x3 matrix next, maybe you'll get a theory?

- October 14th 2011, 10:28 PMFernandoRevillaRe: Eigenvalues
Using the steps:

we obtain

So, the eigenvalues are (n-1-uple) and (1-uple) if . If we have only the eigenvalue (1-uple). - October 14th 2011, 11:46 PMDevenoRe: Eigenvalues
i would have been inclined use induction and expand along the top row, but that works great!

- October 15th 2011, 01:42 AMFernandoRevillaRe: Eigenvalues
In general with the steps

we triangularize the determinant

- October 15th 2011, 03:38 PMdwsmithRe: Eigenvalues
- October 15th 2011, 03:41 PMDevenoRe: Eigenvalues
he started with 1-λ, and added (n-1) 1's. 1-λ + (n-1) = n-λ.

- October 15th 2011, 03:51 PMdwsmithRe: Eigenvalues
- October 15th 2011, 04:00 PMDevenoRe: Eigenvalues
first he subtracted the first row, from every other row. this changed all the 1's in the first column (except for the top entry) to -λ.

it also changed all the 1's in every other column to 0, except along the diagonal, where 1-λ becomes -λ.

then he sucessively added (it doesn't matter if you do it one at a time, or all at once) columns 2 through n, to the first column.

neither the row operation, nor the column operation changes the determinant, since no row or column was multiplied. - October 15th 2011, 04:03 PMdwsmithRe: Eigenvalues
- October 15th 2011, 04:08 PMdwsmithRe: Eigenvalues
Now the eigenvectors.

For , I obtained

However, I am finding it hard to solve for lambda = n

Is it just

- October 15th 2011, 04:40 PMDevenoRe: Eigenvalues
the eigenvalue n has a multiplicity of 1, so all we need to do is find ONE eigenvector. try (1,1,1,.....,1).

(how did i find this? note that (A-nI)(x1,x2,....,xn) = (0,0,....,0) implies that

x1+x2+....+xn = nxj for ALL j) - October 15th 2011, 04:52 PMdwsmithRe: Eigenvalues