Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra / 1 / b / ii / 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 $v$. ThusThis 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 $\mathbb{C}$.