an eigen value problem of matrix Let A be a matrix of order n such that sum of elements of each row is zero. Prove that 0 is an eigenvalue of the matrix. I think logically like this.. As 0 is the sum of all elements of row 1/0 is the sum of all elements of the row of inverse of A. which contradicts the existence of the inverse. Now if 0 is the eigenvalue of A 1/0 is an eigenvalue of the inverse matrix.. which also contradict the existence of inverse.. hence 0 is eigenvalue of A. Maybe it is wrong method of proof..

Here are the things that are wrong with your proof:


*

*It is not true that if the sum of the first row of $A$ is zero, then the sum of the first row of $A^{-1}$ is zero (if $A^{-1}$) exists.

*You can't divide by zero.  Just because you decided to divide by zero, however, doesn't mean you've reached a contradiction.


Here's an easy method of proof:
We note that $\lambda$ is an eigenvlaue of $A$ if and only if $A - \lambda I$ is non-invertible.  Note that $A(1,\dots,1)^T = 0$, so $A = A - 0I$ is not invertible. So, $0$ is an eigenvalue of $A$.
Alternatively, see that the vector $v = (1,\dots,1)^T$ satisfies $Av = 0 \cdot v = 0$, so that $v$ is an eigenvector of $A$ associated with the eigenvlaue $0$.  So, $0$ is an eigenvalue of $A$.
