determinant, eigenvalue

• Dec 18th 2011, 06:20 AM
alexandrabel90
determinant, eigenvalue
How do you find the left eigenvalue for a matrix where P is an N xN matric where all its entries are 1/N?

i know that i need to use characteristic polynomial but how do i find the determinant for it? its too tedious by this method.
• Dec 18th 2011, 06:42 AM
Deveno
Re: determinant, eigenvalue
suppose that A is the matrix in question. we have A = (1/N)B, where B is a matrix with all 1's.

so det(A-xI) = det((1/N)B - xI) = det((1/N)(B - xNI) = det((1/N)(B - NxI)) = ((1/N)^N)(det(B - NxI)).

if we let y = Nx, we have (1/N)^N(det(B - yI)).

so let's look at that determinant, det(B-yI) =

$\displaystyle \begin{vmatrix}1-y&1&\dots&1\\1&1-y&\dots&1\\ \vdots&\vdots&\ddots&\vdots\\1&1&\dots&1-y\end{vmatrix}$

if we subtract the first row from every other row (which does not change the determinant), we get:

$\displaystyle \begin{vmatrix}1-y&1&\dots&1\\y&-y&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\y&0&\dots&-y\end{vmatrix}$

if we add every column to the first (which also does not change the determinant), we get:

$\displaystyle \begin{vmatrix}N-y&1&\dots&1\\0&-y&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\dots&-y\end{vmatrix}$

can you find the determinant of this last matrix? this should give you the only possible eigenvalues rather easily.
• Dec 18th 2011, 06:46 AM
FernandoRevilla
Re: determinant, eigenvalue
Quote:

Originally Posted by alexandrabel90
How do you find the left eigenvalue for a matrix where P is an N xN matric where all its entries are 1/N?

i know that i need to use characteristic polynomial but how do i find the determinant for it? its too tedious by this method.

You obtain a triangular determinant with the following steps:

$\displaystyle (i)\;\; R_2-R_1,\ldots,R_n-R_1\quad (ii)\;\;C_1+C_2+\ldots+C_n$

Edited: Sorry, I didn't see Deveno's post.
• Dec 18th 2011, 06:50 AM
ILikeSerena
Re: determinant, eigenvalue
Hi alexandrabel90! :)

Your matrix P has N identical columns.
This means that the column space is 1-dimensional.
In turn this means that the null space is (N-1)-dimensional.
Therefore 0 is an eigenvalue of P with multiplicity (N-1).

So you have only 1 other eigenvalue that is not zero.
The trace of P is the sum of the eigenvalues.
Can you guess what the value of this last eigenvalue must be?

Edit: Ah, it seems we have been stacking replies.
• Dec 18th 2011, 06:50 AM
alexandrabel90
Re: determinant, eigenvalue
therefore def(B-yI) = (N-y) (-y)^(N-1)
and thus det (A-xI) = (N-Nx) (-Nx)^ (N-1) (1/N)^(N)

if i let this equal to 0, how do i find the value that x can take given that N is unknown?
• Dec 18th 2011, 09:13 AM
Deveno
Re: determinant, eigenvalue
y can only be 0 or N. what does that mean x can equal, given that y = Nx?