Composite Linear Transformations

Give an explicit example of a pair of linear transformations $T : V \to W$ and $S : W \to U$ between vector spaces $V$, $W$, and $U$, so that neither $T$ nor $S$ is the zero linear transformation, but the composition $ST$ is the zero linear transformation.

Hint: What’s the relationship between the range of $T$ and the kernel of $S$?

I am struggling with this problem. Using the rank nullity theorem, I found that the range of $T$ as well as the the kernel of $S$ should have the same dimension as the domain of $T$. However, I'm confused as how to proceed from there. Any tips are appreciated!

-
Think of projecting onto a coordinate in $\Bbb R^2$. –  David Mitra Dec 7 '12 at 19:50

Your conclusion from the rank-nullity theorem is inaccurate. It's quite alright for the range of $T$ to have smaller dimension than its domain (so long as it doesn't have dimension $0$), and it's quite alright for the kernel of $S$ to have larger dimension than the domain of $T$ (so long as it isn't all of the domain of $S$). We really need the range of $T$ to be a subspace of the kernel of $S$, though. If we don't have that, then there's some $y$ in the range of $T$ such that $Sy\neq 0$, but $y$ being in the range of $T$ means $y=Tx$ for some $x$ in the domain of $T$, whence $STx=Sy\neq 0$, and so our desired result fails.

In summary: Saying that $ST=0$ while $S,T\neq 0$ (where $T:V\to W$ and $S:W\to V$) is equivalent to the following conditions holding:

(i) $\text{rank } T>0$ (equivalently, $\ker T\neq V$)

(ii) $\text{rank } S>0$ (equivalently, $\ker S\neq W$)

(iii) $\text{ran } T\subseteq\ker S$.

Coming up with an example shouldn't be too hard. Can you figure out two $2\times 2$ matrices $A,B$ such that $AB$ is the $2\times 2$ zero matrix, but neither of $A,B$ is a zero matrix?

-
so what would be the relationship between the range of T and the kernel of S? –  tamefoxes Dec 7 '12 at 20:10
I found the matrices that the product of them is the zero matrix, but neither of them are the zero matrix –  tamefoxes Dec 7 '12 at 20:11
The desired relationship is condition (iii). –  Cameron Buie Dec 7 '12 at 20:12
If you've found such matrices $A$ and $B$, then let $U=V=W=\Bbb R^2$, and define $T(x)=Bx$ and $S(x)=Ax$ for all $x\in\Bbb R^2$. Then $ST(x)=ABx=0x=0$, but neither $T$ nor $S$ is the zero linear transformation since each takes one of $[1,0]^t$ or $[0,1]^t$ to a non-$0$ vector. –  Cameron Buie Dec 7 '12 at 20:13
thanks for the clarification cam! –  tamefoxes Dec 7 '12 at 20:21

Hint: Define $T$ such that its image is contained in the kernel of $S$.

Take $V=\langle v\rangle$ one-dimensional generated by $v$.

Take $W=\langle w_1,w_2\rangle$ two-dimensional generated by $w_1,w_2$.

Take $U=\langle u_1,u_2\rangle$ two-dimensional generated by $u_1,u_2$.

Define $T(v)=w_1$ and $S(w_1)=0, S(w_2)=u_1$ and extend them by linearity.

-

Hint: What you need to do is find any map $T$ with a non-trivial kernel (so that the image isn't the whole space) and then let $S$ be any map that sends the whole of the image of $T$ to zero.

Partial answer: One example of this that works quite nicely is letting $T$ and $S$ be projections down onto two different lines.

Full answer:Say for example we have a bases $\{v_i\}$ of $V$, $\{w_i\}$ of $W$ and $\{u_i\}$ of $U$. Let $T(\sum x_iv_i) = x_1w_1$ and $S(\sum y_iw_i) = y_2u_2$. Both maps are linear and $ST$ is the zero map, but neither $S$ nor $T$ are the zero maps.

-