Exercise 3.D.16

Proof. Suppose $T$ is a scalar multiple of the identity. There exists $c\in 𝔽$ such that $T=\mathit{cI}$. Let $S\in L(V)$. Then, for any $v\in V$, we have

Hence $\mathit{ST}=\mathit{TS}$.

Conversely, suppose that $\mathit{ST}=\mathit{TS}$ for any $S\in L(V)$. Let $v$ be a non-zero vector in $V$ and $S$ an operator on $V$ such that $v$ is a basis of $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->S$ (it is easy to show a map like this exists). We have

Hence $\mathit{Tv}\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->S$. But $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->S=\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v)$, therefore $\mathit{Tv}=\mathit{cv}$ for some $c\in 𝔽$. We but need to show that this scalar $c$ is the same for any $v$.

Let $u\in V$ such that $u=\mathit{av}$, where $a\in 𝔽$. Because $T$ is linear, we have that

Hence $c$ is the same for any multiple of $v$. Let $w\in V$ such that $w$ is not a scalar multiple of $v$, i.e, $v$ and $w$ are linearly independent. There is a scalar $b$ such that $\mathit{Tw}=\mathit{bw}$ and a scalar $d$ such that

But, because $T$ is linear, also

Therefore

Because $v$ and $w$ are linearly independent, it follows that $c=d=b$.

Hence the scalar is the same for all vectors and $T$ is a scalar multiple of the identity. □