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

Question

Let $V$ be a finite-dimensional vector space over $F$. Let $T:V \rightarrow V$ be a linear transformation such that $ST=TS$ for all linear transformations $S:V \rightarrow V$. Show that $T = cI_v$ for some $c \in F$.

Answer 1

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 $ST\neq TS$.

Answer 2

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.

Answer 3

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: $$ TS_i(v_j)= \begin{cases} Tv_i &\text{if } i=j \\ T(0)=0 &\text{if } i\neq j \end{cases} $$

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 $\,c=a_{ji}\,$

Answer 4

This may be very late, but I think I can contribute yet another method (which may, in essence, be similar to @2'5 9'2 's answer):

Definition: A subspace $U$ of vector space $V$ is said to be invariant under $T$ if $T(U) \subseteq U$.

Lemma: If $V \in fdvs_n(\mathbb{F})$, and $T:V \to V$ is an operator such that every subspace $U$ of $V$ is invariant under $T$, then $T=\lambda I$ for some scalar $\lambda$.

Proof:

Let $\{v_1,v_2 \cdots v_n\}$ be a basis of $V$. Definition of $T$ on these vectors defines $T$ on all vectors in $V$ (Let us, for notation sake, denote span($S$):=$\langle S \rangle$). Since $\langle v_j \rangle$ is a subspace, it is invariant under $T$, which means $T(c_jv_j)=\gamma_jc_jv_j$ for some scalar $\gamma_j$, for any scalar $c_j$ (Why?). $T(c_jv_j+c_iv_i)=\gamma_ic_iv_i+\gamma_jc_jv_j$. But again, $\langle c_jv_j+c_iv_i \rangle$ itself forms a 1-dimensioned subspace, invariant under $T$. This means $T(c_jv_j+c_iv_i)=\gamma(c_iv_i+c_jv_j)$. But note that this implies, in the previous case, $\gamma_i=\gamma_j=\gamma$. Therefore, we can conclude that every vector maps to $\gamma$ times itself, therefore, $T=\gamma I$.

The contraposition of the lemma states that, if for every $\lambda \in \mathbb{F}$, $T \neq \lambda I$, then there exists a subspace of $V$ that is not invariant under $T$.

Now, onto the theorem itself:

Claim: $T=\lambda I$ for some scalar $\lambda$ if and only if $ST=TS$ for every $S \in L(V)$.

Proof:

$\impliedby$) This is pretty straightforward, $S(\lambda I)=\lambda I(S)$

$\implies$) Assume that $T \neq \lambda I $ for any $\lambda$. This means that there exists a subspace $U$ of $V$ that is not invariant under $T$. Specifically, this means that there exists a vector $u \in U$ such that $T(u)$ falls outside the span of $u$. This means that $u$ and $Tu$ are linearly independent in $V$. Define a map $S$ as follows: $S(u)=u $ and $S(Tu)=0$. $T(S)(u)=T(u)$ and $T(S(Tu))=0$. But $S(T)(u)=0$, which immediately means $TS \neq ST$. We are done.