Consider the following figure (the length of the sides in the square is 1):
The unit square is transformed into the rectangle by using scaling and rotation.
What is the transformation matrix for the transformation that moves the unit square to the rectangle depicted above? (Hint: $A=A_\theta \,*A_s$)
I'm stuck here. How am I supposed to approach this?
On
We have that $$A_{\theta} = \left(\begin{array}{c} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\right)$$ realize a rotation of $\theta$ in $\{e_1,e_2\}$ and $$A_s = \left(\begin{array}{c} a & 0 \\ 0 & b \end{array}\right)$$ maps $e_1$ into $a\cdot e_1$ and $e_2$ into $b\cdot e_2$.
Then, $$A=A_{\theta} \cdot A_s = \left(\begin{array}{c} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\right)\left(\begin{array}{c} a & 0 \\ 0 & b \end{array}\right) = \left(\begin{array}{c} a\cos\theta & -b\sin\theta \\ a\sin\theta & b\cos\theta \end{array}\right).$$
Take the standard basis $\{e_1,e_2\}$ for $\Bbb{R}^2$. First, you have to rotate it by an angle of $\theta=\frac{3\pi}{4}\text{rad}$ (why?). So, you're mapping $e_1$ into $v_1$ and $e_2$ into $v_2$.
Then, you have to map $v_1$ into $w_1=\sqrt{2} v_1$ and $v_2$ into $w_2=2\sqrt{2}v_2$ (why?). And then, you're done.
Ps.: the order that you do this is not important
Edit: There is another way to solve this, a way that "hides" those steps (but I really appreciate if you understand the first one, it's more constructive).
Think about where you're going to send each vector of the basis $\{e_1,e_2\}$. We're looking for a transformation $A$ that maps $(1,0)\mapsto(-2,2)$ and $(0,1)\mapsto(-1,-1)$, that is, $$A=\left(\begin{array}{c} -2 & -1 \\ 2 & -1 \end{array}\right)$$ and we're done.