Example of a linear operator on some vector space with more than one right inverse. In my preparations for an upcoming exam, I'm working through the past exams available from my department.  One of the questions asks for an example of a linear operator on some vector space with more than one right inverse.
In another part of the same problem, I proved that right-inverse uniqueness implies invertibility, but I feel like I'm missing how that result will help me find or construct an example (assuming that it's applicable.)
I've considered various square matrices (because I'm looking for an operator) over various fields, but nothing has worked yet and I don't feel like I'm going about it the right way.
 A: $d/dx$ is an operator on the vector space of differentiable functions with lots of right-inverses: $\int_a^x\quad dx$ - one for each $a$. 
(Not sure what the right/left convention is here, but they are one-sided inverses anyway.)
A: Let $V$ be the space of infinite sequences of real numbers, and let $T$ take the sequence $x_1,x_2,x_3,\dots$ to the sequence $x_1+x_2,x_3,x_4,\dots$.  Then $T$ has at least two right inverses: for example, $U$, which takes $y_1,y_2,y_3,y_4\dots$ to $0,y_1,y_2,y_3\dots$ and $V$, which takes $y_1,y_2,y_3,y_4\dots$ to $y_1,0,y_2,y_3\dots$.  
If you're allowed operators $T:V\to W$ where $V\neq W$, then you can use a matrix such as $T=\begin{pmatrix}1 & 0\end{pmatrix}:\mathbb R^2\to\mathbb R:\begin{pmatrix}a\\b\end{pmatrix}\mapsto a$.  Then $S(x)=\begin{pmatrix}1\\x\end{pmatrix}:c\mapsto\begin{pmatrix}c\\xc\end{pmatrix}$ is linear and an inverse for $T$, for any $x$.  
If $V$ is a finite-dimensional vector space, then no linear map $T:V\to V$ will have more than one right inverse.  Suppose that $S_1,S_2$ are two right inverses for $T$.  Then $TS_1x=TS_2x=x$, so $T(S_1-S_2)x=0$ for any $x\in V$.  So, for each $x\in V$, $(S_1-S_2)x\in\ker T$.  By the rank-nullity theorem, $\dim\ker T = \dim\textrm{Im} T-\dim V$.  But we must have $\dim\textrm{Im} T=\dim V$, because, for any $x\in V$, $x=Ix=TS_1x=T(S_1x)\in\textrm{Im} T$.  So $\dim\ker T=0$, so the kernel of $T$ is the singleton set $\{0\}$.  So $(S_1-S_2)x=0$ for every $x\in V$; i.e., $S_1-S_2=0$, so $S_1=S_2$.  
