# Eigenvalues

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• Oct 15th 2011, 05:06 PM
Deveno
Re: 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 PM
dwsmith
Re: Eigenvalues
Quote:

Originally Posted by Deveno
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).

What?
• Oct 15th 2011, 05:21 PM
Deveno
Re: 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 PM
dwsmith
Re: Eigenvalues
Quote:

Originally Posted by Deveno
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.

Is this not we obtain when we plug in n?

$\displaystyle \begin{bmatrix}0&1&\cdots & 1\\0&-n&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &-n \end{bmatrix}\Rightarrow\begin{bmatrix}0&0&\cdots & 0\\0&1&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &1 \end{bmatrix}$
• Oct 15th 2011, 05:46 PM
Deveno
Re: Eigenvalues
Quote:

Originally Posted by dwsmith
Is this not we obtain when we plug in n?

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

is:
$\displaystyle \begin{bmatrix}0&1&\cdots & 1\\0&-n&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &-n \end{bmatrix}$

the original matrix we are finding eigenvalues/eigenvectors for? no, it is not. that matrix is:

$\displaystyle \begin{bmatrix}1&1&\cdots & 1\\1&1&\cdots &1\\ \vdots & 1& \ddots \\ 1&\cdots & &1 \end{bmatrix}$
• Oct 15th 2011, 05:56 PM
dwsmith
Re: Eigenvalues
Quote:

Originally Posted by Deveno
is:
$\displaystyle \begin{bmatrix}0&1&\cdots & 1\\0&-n&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &-n \end{bmatrix}$

the original matrix we are finding eigenvalues/eigenvectors for? no, it is not. that matrix is:

$\displaystyle \begin{bmatrix}1&1&\cdots & 1\\1&1&\cdots &1\\ \vdots & 1& \ddots \\ 1&\cdots & &1 \end{bmatrix}$

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

From this matrix, I can obtain the previous one I posted by elementary row operations. This matrix will revert to the matrix in post 5 and from there it can be further simplified to the one I just posted but a 2 in the a_{11} position not a zero.
• Oct 15th 2011, 06:07 PM
Deveno
Re: 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 PM
dwsmith
Re: Eigenvalues
Quote:

Originally Posted by Deveno
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.

From simplifying I obtain, $\displaystyle (2-n)x_1=-x_3-\cdots -x_n$ and $\displaystyle x_1=x_2$ with the rest free variables. What can I do now if this is correct?
• Oct 15th 2011, 08:51 PM
Deveno
Re: Eigenvalues
is it not the case that:

$\displaystyle \begin{bmatrix}1&1&\cdots&1\\1&1&\cdots&1\\ \vdots&\vdots&\ddots&\vdots\\1&1&\cdots&1 \end{bmatrix}\begin{bmatrix}1\\1\\ \vdots\\1 \end{bmatrix} = \begin{bmatrix}n\\n\\ \vdots\\n \end{bmatrix} = n\begin{bmatrix}1\\1\\ \vdots\\1 \end{bmatrix}$

doesn't that make (1,1,1,...,1) an eigenvector corresponding to the eigenvalue n?
• Oct 18th 2011, 06:36 AM
dwsmith
Re: Eigenvalues
Dealing with the same matrix but the diagonals now all zeros, the characteristic equation would be $\displaystyle (n-1-\lambda)(-\lambda-1)^{n-1}$
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