Chapter 6.1, Problem 15E

Let $A=\left(a_{ij}\right)$ be an $n\times n$ matrix with eigenvalues ${\lambda }_1,...,{\lambda }_n.$ Show that ${\lambda }_j=a_{jj}+{\displaystyle \underset{i\ne j}{\sum }\left(a_{ii}-{\lambda }_i\right)}$ for $j=1,...,n$