Inverse by left multiplication but not right? Suppose $T: V \rightarrow W$ and  $U: W \rightarrow V$ are linear transformations. It is known that $ U = T^{-1}$ if $UT = I_V$ and $TU = I_W$.
Is it possible to then also have a transform $Z: W \rightarrow V$ such that $ZT = I_V$ but $TZ \neq I_W$ (and likewise, $TZ = I_W$ but $ZT \neq I_V$)?
 A: It is indeed possible.  For instance, take $W = \Bbb R^2, V = \Bbb R$,
$$
Z(x,y) = x
$$
If we define $T(x) = (x,0)$, then we indeed have $ZT = I_{V}$ but $TZ \neq I_W$.  The reverse happens if we exchange the roles of $Z$ and $T$.
For a fixed $V,W$ with $\dim(V) = \dim(W) < \infty$, no such pair exists.
In the infinite dimensional setting, the left and right shift operators are an example of such a pair for which $V = W$.
A: $\begin{bmatrix}1&0\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix}=1=I_1$  but $\begin{bmatrix}1\\0\end{bmatrix} \begin{bmatrix}1&0\end{bmatrix}= \begin{bmatrix}1&0\\0&0\end{bmatrix}\neq I_2$
It’s also worth pointing out that (for transformations over fields, anyway) that theorem only holds when $V$ and $W$ have equal dimension.
A: If W, V are not the same dimension...e.g. suppose $V$ is $\mathbb R^3$ and $W$ is $\mathbb R^2$
Then there can exist $T, U$ such that $TU = I_W$  but since the rank of $T$ has rank less than $3$ it is not possible for $UT$ to be rank $3$ and be an the identity matrix of $\mathbb R^3$ 
