Exercise 5.4.21

If we have some vector $v\in V$ such that $v$ and $T(v)$ are not parallel, we know the $T$-cyclic subspace generated by $v$ has dimension $2$. This means $V$ is a $T$-cyclic subspace of itself generated by $v$. Otherwise for every $v\in V$, we may find some scalar ${\lambda }_v$ such that $T(v)={\lambda }_{v}v$. Now, if ${\lambda }_v$ is a same value $c$ for every nonzero vector $v$, then we have $T=\mathit{cI}$. If not, we may find ${\lambda }_v\ne {\lambda }_u$ for some nonzero vector $v$ and $u$. This means $v$ and $u$ lies in different eigenspace and so the set $\{u,v\}$ is independent. Thus they forms a basis. Now let $w=v+u$. We have both

and

By the uniqueness of representation of linear combination, we must have

a contradiction. So in this case we must have $T=\mathit{cI}$.