linear transformation $T$ such that $TS = ST$

Here is a proof of a different nature, proving the contrapositive. It also avoids matrices and indexing. Probably because of this, it is also valid for infinite dimensional $V$. Assuming $\dim V>1$, Let $T$ be a transformation that is not $cI_V$ for any $c$. We will find a transformation $S$ such that $ST\neq TS$.

1. If $T$ has a nontrivial kernel, then let $\vec{u}$ be a nonzero vector in the kernel, and $\vec{v}$ be a vector not in the kernel. Let $S$ be any transformation that sends $\vec{u}$ to $\vec{v}$. Then $ST(\vec{u})=\vec{0}$, while $TS(\vec{u})\neq\vec{0}$.

2. Now since we can assume that $\ker T=\left\{\vec{0}\right\}$ and $T$ is not $cI_V$ for any $c$, we can find two nonzero vectors $\vec{u}$ and $\vec{v}$ such that either

   • $T(\vec{u}) = c\vec{u}$ and $T(\vec{v}) = d\vec{v}$ with $c\neq d$. Let $S$ be any transformation that permutes the vectors $\vec{u}$ and $\vec{v}$. Then $ST(\vec{u})=S(c\vec{u})=c\vec{v}$ . While $TS\vec{u}=T(\vec{v})=d\vec{v}$. Since $d\neq c$, $ST\neq TS$.
   • $T(\vec{u})\neq c\vec{u}$ for any $c$, making $\vec{v}=T(\vec{u})$ a nonzero vector (since $T$'s kernel is trivial) that is linearly independent from $\vec{u}$. Let $S$ be any transformation that sends $\vec{v}$ to $\vec{u}$ and annihilates $\vec{u}$. Then, $ST(\vec{u})= S(\vec{v})=\vec{u}$. Meanwhile $TS(\vec{u})=T(\vec{0})=\vec{0}$. Again, we have $ST\neq TS$.

Notice that if two linear operators $A,B$ commute, they preserve one another's eigenspaces. It remains to show that a transformation which preserves every subspace (so in particular, every vector is its eigenvector) is a multiple of identity.

Note that for this argument, you don't need to assume that $V$ is finite-dimensional.

Choose a basis $\,\{v_1,...,v_n\}\,$ of $\,V\,$ , and define $$S_i:V\to V\,\,,\,\,S_i(v_j)=\left\{\begin{array}{} v_i\,\,&\,\text{if}\,\,\,i=j\\0\,\,&\,\text{if}\,\,\,i\neq j\end{array}\right.$$

Now: $$TS_i(v_j)=\left\{\begin{array}Tv_i,&\,\text{if}\,\,\,i=j\\T(0)=0\,&\,\text{if}\,\,\,i\neq j\end{array}\right.$$ Suppose $$T(v_j)=\sum_{k=1}^na_{jk}v_k\Longrightarrow S_iT(v_j)=S_i\left(\sum_{k=1}^na_{jk}v_k\right)=a_{ji}v_i$$

Well, compare the different cases and deduce $\,c=a_{ji}\,$