Exercise 5.A.27

Proof. We will prove the contrapositive.

Suppose $T$ is not a scalar multiple of the identity. Note that the linear map $0$ is also a scalar multiple of the identity. There are non-zero and linearly independent vectors $w,v\in V$, such that $\mathit{Tv}=w$. Let $n=\dim <!--\; FUNCTION\; APPLICATION\; -->V$. Extend $w,v$ to a basis $w,v,v_1,\dots ,v_{n-2}$ of $V$. The dimension of $\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v,v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_{n-2})$ equals $\dim <!--\; FUNCTION\; APPLICATION\; -->V-1$, but this subspace is clearly not invariant under $T$, completing the proof. □