= Solution
If $\chi_T(\lambda)=0$, then $\det(\lambda I-A)=0$, so $\lambda I-A$ is singular. By the <rank-nullity theorem> it has a nonzero kernel vector $v$. Thus
$$
(\lambda I-A)v=0,\qquad\boxed{Tv=\lambda v\quad(v\neq0).}
$$
This is exactly the definition of an <eigenvalue>. Here $\lambda$ lies in the scalar field: for the stated real <vector space> it is real. A nonreal root instead gives an <eigenvalue> of the complexified operator, by the same argument over $\mathbb C$.
Back to article page
