# #A_n# is a matrix in size #n xx n#. Diagonal entry of A are #0# and the other entry are #-1#. How to determine eigenvalue of #A_n# ?

##### 2 Answers

From your question, we have the matrix

#A_n = [(0,-1, cdots, cdots , -1),(-1,0, -1, cdots, vdots),(vdots,-1,ddots, " ", vdots), (vdots, " "," ", ddots , -1), (-1,cdots,cdots, -1 , 0)]#

Using the determinant definition of:

#|lambdaI_n - A_n| = 0# ,

we have:

#|lambdaI_n - A_n| = |(lambda,1, cdots, cdots , 1),(1,lambda, 1, cdots, vdots),(vdots,1,ddots, " ", vdots), (vdots, " "," ", ddots , 1), (1,cdots,cdots, 1, lambda)| = 0#

For simplicity, let us consider a

#A_3 = [(0,-1,-1),(-1,0,-1),(-1,-1,0)]#

with the determinant...

#|(lambda,1,1),(1,lambda,1),(1,1,lambda)| = 0#

If you diagonalize this matrix, you should get

If you evaluate this as a determinant, you should easily get

The

If you notice, the eigenvalues all add up to zero, while the trace of the matrix is also zero.

*Thus, the trace of the matrix is the sum of the eigenvalues.*

This is true because diagonalizing the matrix preserves the eigenvalues, and the eigenvalues of a diagonal matrix can be obtained straight from treating each diagonal entry as its own matrix block set equal to

It does not, however, mean that all the eigenvalues are zero. It just means that whatever they are, they add up to zero. As Cesareo has mentioned, a general formula for this is that

#(lambda - 1)^(n-1)(lambda + n - 1) = 0#

is the characteristic equation for

Alternative approach

For possible eigenvalues

We are now looking for the eigenvalues for:

A' is singular of rank 1 and has

Because

The corresponding eigenvalues for