Find linear transformations $U,T:\textbf{F}^{2}\to\textbf{F}^{2}$ such that $UT = T_{0}$ (the zero transformation), but $TU\neq T_{0}$.

Question

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.

Answer 1

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 $x\ne y/2$.

Answer 2

Here is a way to get two such maps if we are given a basis of a vector space $V$.

Let $v_1, …, v_k, v$ be a basis of $V$. Then observe that $v_1, …, v_k, v + v_1$ is a basis for $V$. Now define $U$ on the basis $v_1, …, v_k, v + v_1$ as follows: $v_i \mapsto 0$ for $1 \leq i \leq k$, and $v + v_1 \mapsto v + v_1$. Define $T$ on the basis $v_1, …, v_k, v$ as follows: $v_i \mapsto v_i$ for $1 \leq i \leq k$, and $v \mapsto 0$.

Then $\ker U = \text{span}(v_1, …, v_k) = \text{im} \, T$, so in particular $\text{im} \, T \subseteq \ker U$, hence $UT = 0$.

On the other hand, $\{ v + v_1 \}$ is a basis for $\text{im} \, U$ and $\{ v \}$ is a basis for $\ker T$, however $v + v_1$ is not in the span of $v$, so there is a vector in $\text{im} \, U$ that isn't in $\ker T$, hence $\text{im} \, U \not \subseteq \ker T$. This means $TU \neq 0$.