Prove that $\sum_{j=1}^n a_{ij} = \lambda$ is an eigenvalue of $A$. Let $\lambda \in \mathbb{R} $ and $A = (a_{ij})_{1 \leq i \leq j \leq n}$ is a $n 
 \times n$ matrix. 
  
  Suppose $a_{i1} + a_{i2} + ... + a_{in} = \lambda $.
  
  For all $i \in \{1,2,...,n\}$, prove that $\lambda$ is an eigenvalue of $A$. 


I started by making the sum of the first row so I can use:   $a_{i1} + a_{i2} + ... + a_{in} = \lambda $.

$\begin{align} \det(A - \lambda I_n)= \begin{vmatrix} 
a_{11} - \lambda &amp; a_{12} &amp; \dots &amp; a_{1n} \\
a_{21} &amp; a_{22} - \lambda &amp; \dots &amp; a_{2n} \\
\vdots &amp;  &amp; \ddots &amp; &amp;  \\
a_{n1} &amp; \dots &amp; &amp; a_{nn} - \lambda
\end{vmatrix} = 
\begin{vmatrix} 
\sum_{j=1}^{n} a_{1j} - \lambda &amp; a_{12} &amp; \dots &amp; a_{1n} \\
a_{21} &amp; a_{22} - \lambda &amp; \dots &amp; a_{2n} \\
\vdots &amp;  &amp; \ddots &amp; &amp;  \\
a_{n1} &amp; \dots &amp; &amp; a_{nn} - \lambda
\end{vmatrix} =
 \begin{vmatrix} 
0 &amp; a_{12} &amp; \dots &amp; a_{1n} \\
a_{21} &amp; a_{22} - \lambda &amp; \dots &amp; a_{2n} \\
\vdots &amp;  &amp; \ddots &amp; &amp;  \\
a_{n1} &amp; \dots &amp; &amp; a_{nn} - \lambda
\end{vmatrix}
\end{align}$

I don't see how to proceed to prove the sum of $a_{1j}$ is an eigenvalue.
Hint: see what happens when $A$ is multiplied on the right with a column vector of $1$.
you can use induction.
for $n=1$ is abvious, for $n=2 $ you have :

$ \pmatrix{b &amp; a\\ c &amp; d} $ with $ a+b=c+d=K$

so for finding eigenvalue you have to make the determinant of $\pmatrix{a-x &amp; b\\ c &amp; d-x}$ equal zero. i.e

$$(a-x)(d-x)-cb=0$$ $$ad-ax-xd+x^{2}-cb=0$$, we have b=K-a and c=K-d so

$$ad-ax-xd+x^{2}-(K-a)(K-d)$$ so 
$$ad-ax-xd+Kd+aK-ad-K^{2}+x^{2}=0$$, so

$$ K=x $$ is eigenvalue .

now you should assume its true for a $n \times n$ matrix , and prove for $ n+1\times n+1$ . it is very easy