How to show an image gives the unit square? Let $A$ be the parallelogram distended by two vectors $\begin{pmatrix}a \\\ b\end{pmatrix}$ and $\begin{pmatrix}c \\\ d\end{pmatrix}$ which are not parallel, and let $M = \begin{pmatrix}a & c \\\ b & d\end{pmatrix}$. Show that the image $$\begin{pmatrix}x \\\ y \end{pmatrix} \:= \:\textbf{T} \begin{pmatrix}u \\\ v \end{pmatrix} \:=\: M\begin{pmatrix}u \\\ v\end{pmatrix}$$ images the unit square $K$ distended by $\textbf{e}_1$ and $\textbf{e}_2$ on $A.$ The only work i can show is: $M\begin{pmatrix}u \\\ v\end{pmatrix} \:=\:\begin{pmatrix}a & c \\\ b & d\end{pmatrix}\begin{pmatrix}u \\\ v\end{pmatrix} \: = \:\begin{pmatrix}au \: +\:cv\\\bu \: + \: dv \end{pmatrix}$. I would like to know exactly what they are asking, so a clarification would be nice in adition to some hints on how this is supposed to be done. My apologize in advance if i'm not making myself clear, i have a very limiting vocabulary in linalg:) Thanks in advance!

$$M\textbf{e}_1=M\begin{pmatrix}1 \\\ 0\end{pmatrix} \:=\:\begin{pmatrix}a & c \\\ b & d\end{pmatrix}\begin{pmatrix}1 \\\ 0\end{pmatrix} \: = \:\begin{pmatrix}a \\\b \end{pmatrix}$$ and: $$M\textbf{e}_2=M\begin{pmatrix}0 \\\ 1\end{pmatrix} \:=\:\begin{pmatrix}a & c \\\ b & d\end{pmatrix}\begin{pmatrix}0 \\\ 1\end{pmatrix} \: = \:\begin{pmatrix}c\\\d \end{pmatrix}$$ Thus: $$M( \lambda\textbf{e}_1+\mu\textbf{e}_2)=\lambda\begin{pmatrix}a \\\b \end{pmatrix}+\mu\begin{pmatrix}c\\\d \end{pmatrix}$$ Do you see now how it goes from here?