If $L(x_1)$ = $L(x_2)$, does $x_1$ = $x_2$? Reviewing for an exam, and the book says False. But I can't seem think of a way to prove this wrong...

If $L(x_1)$ = $L(x_2)$, does $x_1$ = $x_2$?

where $L$ is a linear transformation from $\Bbb R^n$ to $\Bbb R^n$, and $x_1$ and $x_2$ are vectors.
 A: A trivial example would be the zero transformation
$$L: \mathbb{R}^n \to \mathbb{R}^n, L(x) = 0 \; \forall x \in \mathbb{R}^n.$$
Then $e_1 \neq e_2$ but $L(e_1) = 0 = L(e_2)$.
In general, not only in the context of vector spaces, $L(x_1) = L(x_2) \implies x_1 = x_2$ if and only if $L$ is injective. If, in particular, $L$ is a linear operator on vector spaces, then $L(x_1) = L(x_2) \implies x_1 = x_2$ if and only if $\text{ker}(L) = \{0\}$.
A: If $L$ is singular then $L(x_1)$ and $L(x_2)$ can be equal for different vectors $x_1$ and $x_2$. E.g.:
$$\mathcal{L}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$$
will map all vectors to $\begin{bmatrix}0\\0\end{bmatrix}$.
A: If $L$ is an injective linear function, then
$$
L(u)=0 \implies u=0.
$$ Observe that
$$
\begin{align}
L(x1) = L(x2)
\implies L(x1) - L(x2)&=0\\
L(x1-x2)&=0
\end{align}
$$ But you don't tell us that $L$ is injective...
A: More geometrical argument: 
Consider $L: \mathbb R^2 \mapsto \mathbb R^2$ that is perpendicular projection onto x-axis. Then for $x_1 = (1,1) , \ x_2= (1,2)$ we get $L(x_1) = L(x_2) = (1,0)$.
A: Consider x1 = -x2, and and L transforms the elements in a vector into absolute value. Then L(x1) = L(x2) even though x1 is not equal to x2. 
