Prove eigenvalue is equal to sum of elements in every column

[Enemy of my enemy]

We have matrix A (n x n), I have to prove that one of the eigenvalues in every square matrix is equal to sum of elements in column of this matrix. I know it works, but I dont know how can I prove it.

 

[Romsek]

There must be some other restriction on the matrix you aren't mentioning because your statement as written isn't true.

 

[Enemy of my enemy]

Yes, you right... I forgot to write we have F - division ring. And a is element of it, so let A will be: sum of elements in column is equal to a... prove a is eigenvalue of matrix. I am sorry, I misunderstand it.

 

[Romsek]

\(\displaystyle \text{consider the vector $v = (1,1,1)$}\)

\(\displaystyle A^Tv = a(1,1,1) = a v\)

\(\displaystyle \text{So $a$ is an eigenvalue of $A^T$ with eigenvector $(1,1,1)$}\)
\(\displaystyle \text{We know that $A$ and $A^T$ share the same eigenvalues thus $a$ is also an eigenvalue of $A$}\)

 

[Enemy of my enemy]

Thank you sir! Now I see the problem I was doing.