Prove for every matrix $A\in\mathbb{C}^{n\times n}$ in which the sum of every line is $\lambda$, $\lambda$ is also one of the matrix's eigenvalue. Prove for every matrix $A\in\mathbb{C}^{n\times n}$ in which the sum of every line is $\lambda$, $\lambda$ is also one of the matrix's eigenvalue.
Example:
$A=\begin{pmatrix}
1 & 2 & 3
\\ -1 & 4 & 3
\\ 8 & -1 & -1 \end{pmatrix}\Rightarrow \lambda=6$
Hint: Find the appropriate eigenvector.
Here's what I did so far:
For example matrix A the eigenvectors are $(\frac{1}{5}, \frac{-7}{5},1), 
(1,1,1), (\frac{-3}{7},\frac{-3}{7},1)$. I assume the "appropriate" eigenvector that they refer to is $(1,1,1)$. According to the definition of eigenvectors and eigenvalues:
$A\vec{v}= \begin{pmatrix}
1 & 2 & 3
\\ -1 & 4 & 3
\\ 8 & -1 & -1 \end{pmatrix}
\begin{pmatrix}1\\1\\1\end{pmatrix}
=6\begin{pmatrix}1\\1\\1\end{pmatrix}
=\lambda\vec{v}$
This shows that the sum of each of the lines of this specific matrix should always equal one of the eigenvalues of the matrix
So in order to prove that this law applies for any given matrix $A\in\mathbb{C}^{n\times n}$ in which the sum of every line is $\lambda$, I have to prove that one of the eigenvectors will always be $(1,1,1)$:  
In order for the sum of two different lines in a matrix to be the same, each element in both the lines must be multiplied by one. $\blacksquare$
Is my proof correct?
 A: Look at the vector
$e = (1, 1, \ldots, 1)^T; \tag{1}$
i.e., every entry of $e$ is $1$.  Then if $a_i$ is the $i$-th row of $A$, so that
$a_i = (a_{i1}, a_{i2}, \ldots, a_{in}), \tag{2}$
then
$a_i e = \sum_j a_{ij} = \lambda; \tag{3}$
this holds for every row of $A$.
Thus
$Ae = \lambda e, \tag{5}$
and $e$ is an eigenvector of $A$ with eigenvalue
$\lambda = \sum_j a_{ij}, \tag{6}$
independently of $i$.
Note: The OP CluelessButCurious' proof is basically sound, except that $(1, 1, 1)^T$ can only be an eigenvector in the case $n = 3$.
A: Your observation is very good, but there are some issues with your proof attempt. For one thing, it only works in the case $n=3.$
More generally, you're quite right that the $n\times 1$ matrix of $1$s is an eigenvector of such a matrix $A,$ associated to the eigenvalue $\lambda.$ Let $\vec e$ be the element of $\Bbb C^n$ whose entries are all equal to $1.$ What you must show is that each entry of $A\vec e$ is equal to $\lambda.$ To do so, use the definition of matrix multiplication, the definition of $\vec e,$ and the hypothesis about $A.$
A: Your proof is great until the last line, where it's not clear what you're saying.
The fact that $A(1,1,1)^T = (\lambda,\lambda,\lambda)$ follows from the given condition, because each of those lambdas is calculated by adding up the elements in a row of the matrix. That's what multiplication by $(1,1,1)^T$ is.
As noted in another answer, you should also generalize this proof to apply in dimensions other than $3$. Instead of calling it $(1,1,1)^T$, just call it $(1,\ldots,1)^T$.
