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