This is a question from an algebra homework and I am just looking for some tips.
The question is:
We have to prove that $c$ is an eigenvalue of $M$.
I started out by writing out the form of the matrix and see if anything was popping up without much success.
I found a similar question except that they were working with a $2\times 2$ matrix and that the sum of entries in each column $= c$ as well as the sum of entries in each row. Also, in this question they are looking for eigenvectors. In this case the question is pretty easy as we can notice that each column or row is composed of the same entries (in different order or not).
link (p4): http://www.math.upenn.edu/~deturck/m260/hw3sols.pdf
Hence I looked at the sum in the columns to find some kind of pattern or hint that could help me solve this question without any success. It is obvious that the entries in each column might not be the same.
Anyhow I am stuck there.
Again I am not asking for the answer, just a hint.
Thank you for your time and good luck.
Showing that $c$ is an eigenvalue of $M$ is equivalent to showing that $M-cI$ is not invertible. But showing $M-cI$ is not invertible is equivalent to showing that $M-cI$ has a nontrivial nullspace. Let $v=e_1+\dotsb+e_n$. What is $\left(M-cI\right)v$?
Hint: Consider what happens if you pre-multiply $(1,\dots,1)$ to the matrix.
Hint: if you multiply $M$ on the left with a specific row vector, you can use the statement about the column sums effectively.
So v is an eigenvector with eigenvalue=c