Find linear transformations $U,T:\textbf{F}^{2}\to\textbf{F}^{2}$ such that $UT = T_{0}$ (the zero transformation), but $TU\neq T_{0}$.
My solution
Let us consider $T(x,y) = (x,0)$ and $U(x,y) = (y,y)$. Thus, for every $(x,y)\in\textbf{F}^{2}$, we have that
\begin{align*}
UT(x,y) = U(T(x,y)) = U(x,0) = (0,0)
\end{align*}
On the other hand, for all $(x,y)\in\textbf{F}^{2}$, we have that
\begin{align*}
TU(x,y) = T(U(x,y)) = T(y,y) = (y,0)
\end{align*}
and the desired properties are satisfied.
I am curious if there is a more general result which produces such pairs of linear transformations in $\textbf{F}^{n}$.
Anyone could tell me so? Any contribution is appreciated.
Best Answer
Let $A,B$ be matrices associated to $U,T$. You could look for two matrices such that $AB=0$, $BA\ne AB$, e.g. you could look for two diagonalizable, but not simultaneously diagonalizable, matrices (see Simultaneous diagonalization of commuting linear transformations).
Example: $U(x,y)=(-2x+y,-2x+y)$, $T=(x+2y,2x+4y)$, i.e. $$A=\begin{bmatrix} -2 & 1 \\ -2 & 1 \end{bmatrix},\quad B=\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$$ and $AB=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, but $BA=\begin{bmatrix} -6 & 3 \\ -12 & 6 \end{bmatrix}$. So $UT(x,y)=AB(x,y)=(0,0)$, but $TU(x,y)=BA(x,y)=(-6x+3y, -12x+6y)\ne (0,0)$ for all $x\ne y/2$.