Exercise 4.1.10

Proof. (Just use the hint) The characteristic polynomial of $n\times n$ square matrix $A$ is $\det <!--\; FUNCTION\; APPLICATION\; -->(A-\mathit{\lambda I})$ and we consider the roots of it in complex space. According to the fundamental theorem of algebra, $\det <!--\; FUNCTION\; APPLICATION\; -->(A-\mathit{\lambda I})$ has $n$ roots counting multiplicity and can be factorized as $\det <!--\; FUNCTION\; APPLICATION\; -->(A-\mathit{\lambda I})=({\lambda }_1-\lambda )({\lambda }_2-\lambda )...(\lambda -{\lambda }_n)$. (Recall the formal definition of determinant, the highest order term of $\lambda$, ${\lambda }^n$, is generated by the diagonal product ${\mathrm{\Pi }}_{i=1}^n(a_{\mathit{ii}}-\lambda )$. Thus the sign of the factorization is correct.) Then let $\lambda =0$, we will get $\det <!--\; FUNCTION\; APPLICATION\; -->A={\lambda }_1{\lambda }_2...{\lambda }_n$. □