Prove that $\sum_{j=1}^n a_{ij} = \lambda$ is an eigenvalue of $A$.

Question

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 $.

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

I don't see how to proceed to prove the sum of $a_{1j}$ is an eigenvalue.

Answer 1

Hint: see what happens when $A$ is multiplied on the right with a column vector of $1$.

Answer 2

you can use induction. for $n=1$ is abvious, for $n=2 $ you have :

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

so for finding eigenvalue you have to make the determinant of $\pmatrix{a-x & b\\ c & d-x}$ 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