Proving that linear transformations map parallelograms onto parallelograms 
Suppose $T: \mathbb{R}^2 \to \mathbb{R}^2$ is linear and such that $T(\textbf{x})=A\textbf{x}$ is a $2 \times 2$ matrix. Show that if $\det A \neq 0$, then $T$ takes parallelograms onto parallelograms.

I'm kinda iffy on linear algebra, so I'd really like to know if I got this right. 
We know that the general parallelogram in $\mathbb{R}^2$ can be described as $\textbf{q} = \textbf{p} + \lambda \textbf{v} + \mu \textbf{w}$, with $\textbf{v},\textbf{w}$ linearly independent and $\lambda, \mu \in (0,1)$. We want to show that
$$\begin{pmatrix} 
a_{11} & a_{12} \\
a_{21} & a_{22} 
\end{pmatrix} \begin{pmatrix} 
p_1 + \lambda v_1 + \mu w_1 \\
p_2 + \lambda v_2 + \mu w_2 
\end{pmatrix} = \textbf{p'} + \lambda \textbf{v'} + \mu \textbf{w'}$$
where $\textbf{v'}=(a_{11}v_1 + a_{12}v_2,a_{21}v_1 + a_{22}v_2), \textbf{w'}=(a_{11}w_1 + a_{12}w_2, a_{21}w_1 + a_{22}w_2)$ are linearly independent. 
Suppose, then,
$$\alpha_0 \textbf{v'} + \alpha_1 \textbf{w'}= \textbf{0}$$
this implies
$$\begin{pmatrix} 
a_{11} & a_{12} \\
a_{21} & a_{22} 
\end{pmatrix} \big( \alpha_0 \textbf{v} + \alpha_1 \textbf{w} \big) = \textbf{0}$$
Since $A$ is invertible, the equation $A\textbf{x}=\textbf{0}$ has only trivial solutions. But $\alpha_0 \textbf{v} + \alpha_1 \textbf{w} \neq \textbf{0}$ because $\textbf{v},\textbf{w}$ are by definition linearly independent. Hence, $T(\textbf{q})$ is also a parallelogram.
 A: The idea of your proof is correct, though I think it's preferable when concluding to say "since $v$ and $w$ are linearly independent, we must have $\alpha_0=0$ and $\alpha_1=0$, which shows that $v'$ and $w'$ are linearly independent".
You succeeded in proving that the image of a parallelogram is a parallelogram, but you forgot to show that this mapping is surjective, and that's what is meant in the question which says "then $T$ takes parallelograms onto parallelograms". I'm sure you can do it ;)
Finally, notice that your proof is about proving "image of parallelogram is parallelogram" and you use injectivity of $T$, while for the second point that you should prove, i.e $T$ takes parallelograms onto parallelograms, you will need to use surjectivity of $T$ (the matrix corresponding to $T$ is $A$ and $\det A\neq 0$). You can combine both and write a more compact proof where you use the bijectivity of $T$.
A: An alternative geometric approach:
Show that $T$ maps any pair of parallel lines to another pair of parallel lines. Since a parallelogram is defined by two pairs of parallel lines, and any two pairs of parallel lines define a parallelogram, then $T$ will map a parallelogram to another parallelogram. And since $T$ is invertible, the mapping will be one-to-one.
A: A proof without calculations: $T$ transforms a couple of parallel lines into parallel lines, because if that were not true than there would be two different points transformed into the same point, and that's impossible because $T$ is invertible ($\det T\ne 0$).
