Exercise 5.B.18

Proof. Note that $f$ can only output integer values. Thus, if $f$ is not constant, there will be a jump discontinuity at some point. We will prove $f$ is not constant.

If $T$ is invertible, then the existence of an eigenvalue of $T$ (guaranteed by 5.21) implies that $T-\mathit{\lambda I}$ is not surjective for some $\lambda \in 𝔽$. Hence $f(0)=\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T>\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->(T-\mathit{\lambda I})=f(\lambda )$.

If $T$ is not invertible, choose $\lambda$ such that it is not an eigenvalue of $T$. Then, for any non-zero $v\in V$, $(T-\mathit{\lambda I})v\ne 0$, showing that $T-\mathit{\lambda I}$ is injective and, therefore, surjective. Hence $f(0)=\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T<\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->(T-\mathit{\lambda I})=f(\lambda )$. □