How is my textbook finding this rotation? I have this transformation $\mathbf x\mapsto A\mathbf x $ which is the composition of a rotation and a scaling. I need to give the angle $\varphi$ of the rotation and give the scale factor $r$. Here is the matrix:$\left[\begin{matrix}0&2\\ -2&0\end{matrix}\right]$. So after finding that my eigenvalue is $\lambda = \pm 2i$ I find that $r$ is given as $\sqrt{a^2+b^2}$ and $a$ and $b$ come from the complex number $a+bi$. So now that I have found that $r=2$, I'm stuck on how to find the angle of rotation. I know that normally I would plot $(a,b)$ in the complex plane, but here $(0,2)$ doesn't exactly make a triangle. My text says that the answer is $-\frac{\pi}{2}$, but I don't see where it's getting that from.
 A: One way to see that the angle is $-\frac{\pi}{2}$ is to note that after you factor out the $2$ from that matrix, you end up with: $$\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right]$$Since rotation matrices are of the form: $$\left[\begin{array}{rr} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\right]$$what must $\theta$ be?
A: Recall that the matrix associated with a rotation of angle $\varphi$ is
$$\begin{pmatrix}
\cos \phi & -\sin \phi\\
\sin \phi & \cos\phi
\end{pmatrix}$$
while after dividing by the scale factor your matrix is
$$\begin{pmatrix}
0 & 1\\
-1 & 0
\end{pmatrix}$$
so you are looking for $\varphi$ such that $\cos(\varphi)=0$ and $\sin(\varphi)=-1$. From basic trigonometry you should know that the only solution to this (up to a multiple of $2\pi$) is $-\pi/2$.
A: The best way to figure out how a transformation matrix affects a vector is to perform different transformations on the unit circle. The unit circle has $x=\cos\theta$ and $y=\sin\theta$. Thus, the vector $v=\begin{bmatrix} x\\y \end{bmatrix}$ can be re-written as $v=\begin{bmatrix} \cos\theta\\\sin\theta \end{bmatrix}$. Now you can transform this vector using the equation $Mv=w$. Begin with the identity matrix, and then begin changing a single value at a time and seeing how it transforms the unit circle into an elliptical shape. Before long, you can build a theory about linear transformations. For a given transformation matrix $M=\begin{bmatrix} a&b\\c&d\end{bmatrix}$, the variables $a$ and $d$ scale the $\hat{i}$ and $\hat{j}$ unit vectors, respectively. On the other hand, $c$ skews $\hat{i}$ into the $+y$ direction, and $b$ skews $\hat{j}$ into the $+x$ direction.  For the transformation matrix $M=\begin{bmatrix} 0&2\\-2&0 \end{bmatrix}$, it's clear that now the unit vector $\hat{x}$ points into the direction $(0,-2)$, which is straight down, and the unit vector $\hat{y}$ points in the direction $(2,0)$, which is  to the right. It appears as though the transformation rotated the axes clockwise by $\pi/2$ and scaled them by $2$.I believe that this is a more intuitive way of thinking of linear transformations. See if this doesn't make is easier to understand what is going on.
