Thread: 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.

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) =



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



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



can you find the determinant of this last matrix? this should give you the only possible eigenvalues rather easily.

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:



Edited: Sorry, I didn't see Deveno's post.

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.

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?

y can only be 0 or N. what does that mean x can equal, given that y = Nx?