# Composition of Linear Maps satisfying $ST\neq TS$

Is the following proof Correct?

Theorem. Suppose $V$ is finite-dimensional with $\dim V \ge 2$. Prove that there exist $S,T\in\mathcal{L}(V,V)$ such that $ST \neq TS$.

Proof. Since $V$ is finite-dimensional and $\dim V = m\ge 2$ it follows that there exist a list of vectors $v_1,v_2,...,v_m$ basis for $V$, arguing from cases

Case-1$(m =2)$: Consider the linear transformations $S$ and $T$ defined as follows $$Tv_1=v_1,\ Tv_2=0$$ $$Sv_1=v_2,\ Sv_2 = v_1$$ consequently $$S(Tv_1)=Sv_1=v_2,\ S(Tv_2)=S(0) = 0$$ $$T(Sv_1)=Tv_2=v_1,\ T(Sv_2)=Tv_1 = v_1$$ evidently $ST\neq TS$.

Case-2$(m > 2)$: Consider now the linear transformations $S$ and $T$ defined by $$Tv_{j}=v_{m-j+1}$$ $$Sv_{j}=v_{j\bmod m+1}$$ consequently $$T(Sv_j) = v_{m-(j\bmod m+1)+1}$$ $$S(Tv_j) = v_{(m-j+1)\bmod m+1}$$

Assume now that there exists some $j\in\{1,2,...,m\}$ such that $$m-(j\bmod m+1)+1=(m-j+1)\bmod m+1$$ but then

$$\implies m-(j\bmod m+1)=(m-j+1)\bmod m$$ $$\implies m-(j\bmod m)-1=(m-j+1)\bmod m$$ $$\implies m-1=(m-j+1)\bmod m+(j\bmod m)$$ $$\implies m-1=(m-j+1+j)\bmod m$$ $$\implies m-1=(m+1)\bmod m$$ $$\implies m-1=1$$ $$\implies m=2$$

resulting in a contradiction we may now conclude that $$\forall j\in\{1,2,...,m\}(S(Tv_j)\neq T(Sv_j))$$ implying that $ST\neq TS$.

$\blacksquare$

• Minor mistake so far. $T(S(v_{1})) = T(v_{2}) = 0$. You have the last equality as $v_{1}$. Commented Aug 7, 2017 at 18:20
• I personally found nothing else wrong with the proof aside from the comment above. What I find weird, though, is that I interpreted $TS \neq ST$ to mean that for some $v$, $TS(v) \neq ST(v)$. You ended up proving something stronger than my intepretation: that $TS(v) \neq ST(v)$ for all $v$. Commented Aug 7, 2017 at 18:25
• @layman Indeed! Commented Aug 7, 2017 at 18:45
• Is this exercise from Linear Algebra Done Right by Sheldon Axler? If not what book is it from? Commented Aug 7, 2017 at 18:47
• @layman, assuming $v\ne0,$ :P Commented Aug 7, 2017 at 18:50

Your proof looks correct besides the small mistake that layman spotted out. As layman said, it's impressing that you went as far as finding $T,S\in\mathcal{L}(V)$ with $TSv\neq STv$ for every $v$ when $\dim V>2$. I'd just like to point out two small things:
1- In your proof for the case $\dim V=2$, you didn't have to bother yourself calculating the images of $v_2$ by $ST$ and $TS$. Since $STv_1\neq TSv_1$ this shows that $ST\neq TS$.
2- You could use generalize your construction in the case of $\dim V=2$. Define $S$ and $T$ for $v_1$ and $v_2$ just as you did, and then for $v_3,\dots,v_n$ (if the dimension is larger than $2$) define their values in whatever way (say $Sv_i=Tv_i=v_i$ for $i\ge 3$, or $Sv_i=Tv_i=0$ for $i\ge 3$, or whatever). You will still have $TSv_1=0$ and $STv_1=v_2$, showing that $TS\neq ST$.