Composition of Linear Transformations Resulting in $0$ How do I build linear transformations $T$ and $S$ such that 
$T\circ S=S\circ T=0$? 
I don't how to build these besides the trivial ones: $$T:\Bbb R^n\rightarrow \Bbb R^n,\quad T(v)=0,$$ $$S:\Bbb R^n\rightarrow \Bbb R^n, \quad S(v)=0,$$ -- then it's clear that  $T\circ S=S\circ T=0$
 A: Here is a simple way.  Let $\{v_1,  \cdots ,v_n\}$ be a basis for $\mathbb{R}^n$.  Suppose $1 < k < n$. If $x\in\mathbb{R}^n$ then we can write 
$$x = \sum_{k=1}^n c_k v_k,$$
and this representation is unique for $c_1, \cdots , c_n\in\mathbb{R}$.  The maps $x \mapsto \sum_{j=1}^k c_k v_k$ and $x \mapsto \sum_{j=k + 1}^n c_k v_k$
satisfy this property.
A: Define a linear transformation from $T$ from $\mathbb{R}^n \to \mathbb{R}^n$ as follows:
$$T(e_1)=0, T(e_2) = e_1, T(e_3)=e_1, \dots, T(e_{n-1})=e_1, T(e_n)=e_1$$ (where $e_i$ are the standard unit basis vectors).
Then $T \circ T = T^2 = T \circ T$ equals $0$, even though $T$ is not equal to 0. The key trick is to construct a linear transformation which is nilpotent.
For an example where $S\not=T$:
$T(e_1) = 0$ and $ T(e_i)=e_n $ for all $ i\not=1$.  
$S(e_i)=e_1$ for all $i\not=n$ and $S(e_n)=0$.
Then $S \circ T = 0 $ and $T \circ S = 0$, but the idea is essentially the same as the above.
A: Note that for example $S \circ T = 0$ if and only if $\operatorname{im}T \subseteq \ker S$. So you just need a pair of maps whose images are contained in the other's kernel. 
You can get a family of examples by choosing projections $S$ and $T$ that project onto orthogonal subspaces. For a specific example, let $S: \Bbb R^3 \to \Bbb R^3$ be the projection onto the $xy$-plane, and $T$ the projection onto the $z$-axis:
$$(x, y, z) \overset{S}{\mapsto}(x, y, 0) \qquad \text{and} \qquad (x, y, z) \overset{T}{\mapsto}(0, 0, z).$$
Then $S \circ T = T \circ S = 0$.
