Constructing two matrices that do not commute I need to construct square matrices $A$ and $B$ such that $AB=0$ but $BA \neq 0$. 
I know matrix multiplication is not commutative, but I don't know how to construct such matrices. Thanks in advance.
Edit: looking for some simple way

Pick $\mathrm u, \mathrm v, \mathrm w \in \mathbb R^n \setminus \{0_n\}$ such that $\neg (\mathrm u \perp \mathrm v)$ and $\mathrm v \perp \mathrm w$. Define
$$\mathrm A := \mathrm u \mathrm v^{\top} \qquad \qquad \qquad \mathrm B := \mathrm w \mathrm v^{\top}$$
whose traces are
$$\mbox{tr} (\mathrm A) = \mathrm v^{\top} \mathrm u \neq 0 \qquad \qquad \qquad \mbox{tr} (\mathrm B) = \mathrm v^{\top} \mathrm w = 0$$
Hence
$$\mathrm A \mathrm B = \mathrm u \underbrace{\mathrm v^{\top} \mathrm w}_{= \mbox{tr} (\mathrm B)} \mathrm v^{\top} = \mbox{tr} (\mathrm B) \cdot \mathrm u \mathrm v^{\top} = \mbox{tr} (\mathrm B) \cdot \mathrm A = \mathrm O_n$$
$$\mathrm B \mathrm A = \mathrm w \underbrace{\mathrm v^{\top}\mathrm u}_{= \mbox{tr} (\mathrm A)} \mathrm v^{\top} = \mbox{tr} (\mathrm A) \cdot \mathrm w \mathrm v^{\top} = \mbox{tr} (\mathrm A) \cdot \mathrm B \neq \mathrm O_n$$
