I was given the following question without the material appearing first in the book (I am learning independently). I would gladly look into the matter, but I am not sure where to look – because I don't know what kind of problem this is. I do know how to find eigenvectors, but I don't know what the rotation/scale piece is referring to.
List the eigenvalues of A. The transformation x$\rightarrow$ Ax is the composition of a rotation and a scaling. Give the angle $\phi$ of the rotation, where $-\pi\leq\phi\leq\pi$, and give the scale factor r.$$A=\left[\begin{matrix}4\sqrt{3}&-4\\4&4\sqrt{3}\\\end{matrix}\right]$$
*Edit:
I found the answer by using this question as a guide. I found the eigenvalues of $A$ and the scale factor using $r=\sqrt{a^2+b^2}$ with $a$ and $b$ coming from the fact that a complex number is $a+bi$. I then used the comment from @MSDG to find the angle of rotation. This comment told me to find $e_1$ and $Ae_1$ and find the angle between them, using the formula $v\bullet w=|v||w|\cos(\phi)$. I noticed that $|v||w|$ is the same thing as my r.
Is this always the case? Is there a similar trick to find the $v\bullet w$? Also, what is the reasoning behind the formula used for finding r?
Best Answer
Your matrix takes the form
$$ A = rR = r\begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta &\cos\theta \end{pmatrix}, $$ where $r > 0$ is your scaling, which can be assumed positive without loss of generality, and $R$ is the rotation.
The eigenvalues of this matrix are the roots of the characteristic polynomial: $$ \det(A-\lambda I) = (r\cos\theta - \lambda)^2 + r^2\sin^2\theta = r^2\cos^2\theta - 2\lambda r \cos\theta + \lambda^2 + r^2\sin^2\theta = \lambda^2 - 2\lambda r \cos\theta + r^2 = 0, $$ so $$ \lambda_{1,2} = r(\cos\theta \pm i\sin\theta). $$
Thus, $\lvert r \rvert = \lvert \lambda_{1,2} \rvert$, and $\theta = \arg \lambda_1$.
To answer your other questions: rotations are isometries w.r.t. the Eucledian norm, i.e. Eucledian distance preserving, which means that for any vector $x$ and rotation $R$, we have that $$ \lVert Rx \rVert = \lVert x \rVert, $$ where $\lVert\cdot \rVert$ is the 2-norm. Furthermore, norms are homogeneous, which means that $\lVert r x \rVert = \lvert r \rvert \lVert x \rVert$. You can easily prove these properties your self, and is in fact a nice exercise.
Therefore, for a unit vector $x$ we have that $$ \lVert Ax \rVert = \lvert r \rvert \lVert x \rVert = \lvert r \rvert, $$ which gives you another way to calculate $r$. On the other hand, from the definition of the dot product: $$ \cos\theta = \frac{x^\top A x}{\lVert x \rVert\lVert Ax \rVert} = \frac{x^\top A x}{\lvert r \rvert}, $$ which gives you another way to compute $\theta$.