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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- Oct 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? - Oct 14th 2011, 08:38 PMDevenoRe: Eigenvalues
did you test your conjecture for small values of n? for a 2x2 matrix:

$\displaystyle \begin{vmatrix}1-x&1\\1&1-x\end{vmatrix} = (1-x^2) - 1 = 1 - 2x + x^2 - 1 = x^2 - 2x = x(x-2)$

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

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

$\displaystyle (i)\;R_i\to R_i-R_1\;(i>1)\qquad (ii)\;C_1\to C_1+C_2+\ldots C_n$

we obtain

$\displaystyle \displaystyle\begin{aligned}\begin{vmatrix} 1-\lambda & 1 & \ldots & 1\\ 1 &1-\lambda & \ldots & 1 \\ \vdots&&&\vdots \\ 1 & 1 &\ldots & 1-\lambda\end{vmatrix}&=\begin{vmatrix} 1-\lambda & \;\;1 & \ldots & \;\;1\\ \lambda &-\lambda & \ldots & \;\;0 \\ \vdots&&&\vdots \\ \lambda & \;\;0 &\ldots & -\lambda\end{vmatrix}\\&=\begin{vmatrix} n-\lambda & 1 & \ldots & 1\\ 0 &-\lambda & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & -\lambda\end{vmatrix}\\&=(n-\lambda)(-\lambda)^{n-1}\end{aligned}$

So, the eigenvalues are $\displaystyle \lambda=0$ (n-1-uple) and $\displaystyle \lambda=n$ (1-uple) if $\displaystyle n\geq 2$ . If $\displaystyle n=1$ we have only the eigenvalue $\displaystyle \lambda=1$ (1-uple). - Oct 14th 2011, 11:46 PMDevenoRe: Eigenvalues
i would have been inclined use induction and expand along the top row, but that works great!

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

$\displaystyle (i)\;R_i\to R_i-R_1\;(i>1)\qquad (ii)\;C_1\to C_1+C_2+\ldots C_n$

we triangularize the determinant

$\displaystyle \begin{vmatrix} a & b & \ldots & b\\ b & a & \ldots & b \\ \vdots&&&\vdots \\ b & b &\ldots & a\end{vmatrix}=\ldots=[a+(n-1)b](a-b)^{n-1}$ - Oct 15th 2011, 03:38 PMdwsmithRe: Eigenvalues
- Oct 15th 2011, 03:41 PMDevenoRe: Eigenvalues
he started with 1-λ, and added (n-1) 1's. 1-λ + (n-1) = n-λ.

- Oct 15th 2011, 03:51 PMdwsmithRe: Eigenvalues
- Oct 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. - Oct 15th 2011, 04:03 PMdwsmithRe: Eigenvalues
- Oct 15th 2011, 04:08 PMdwsmithRe: Eigenvalues
Now the eigenvectors.

For $\displaystyle \lambda =0$, I obtained

$\displaystyle \left\{\begin{bmatrix}-1\\n\\0\\ \vdots\\0\end{bmatrix},\begin{bmatrix}-1\\0\\n\\ \vdots\\0\end{bmatrix},\cdots,\begin{bmatrix}-1\\0\\0\\ \vdots\\n\end{bmatrix}\right\}$

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

Is it just

$\displaystyle \begin{bmatrix}1\\0\\0\\ \vdots\\0\end{bmatrix}$ - Oct 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) - Oct 15th 2011, 04:52 PMdwsmithRe: Eigenvalues
Shouldn't we get this:

$\displaystyle \begin{vmatrix}0&1&\cdots & 1\\0&-n&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &-n \end{vmatrix}$

$\displaystyle x_2=-x_3-x_4-\cdots -x_n$

$\displaystyle x_i=0 \ \ i=2,...,n$

$\displaystyle x_1$ is a free variable.

So shouldn't it just be the n dimensional $\displaystyle e_1$ vector?