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?
3 Answers
$\begingroup$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 $\theta$.
$\endgroup$ $\begingroup$Here are some preliminary facts to recall, which we'll find useful when solving this problem:
- Every vector $\vec{v}$ has magnitude and direction.
- When you apply a matrix $\mathbf{M}$ and get $\mathbf{M}\vec{v}$, the resulting vector may have a new magnitude and/or a new direction.
- The vectors that have a new magnitude but keep the same direction are called the eigenvectors of $\mathbf{M}$.
- If the magnitude of the vector changes by a factor of $\lambda$, the value $\lambda$ is called the eigenvalue corresponding to that eigenvector.
- The formula for the dot product of two vectors is $\vec{v}\cdot \vec{w} = |v|\,|w|\,\cos{\theta}$ where $\theta$ is the angle between them. Turn this formula around and you get a definition for the angle between two vectors:$$\cos{\theta} = \frac{|v|\,|w|}{\vec{v}\cdot \vec{w}}$$
In the rest of this answer, let's consider your particular matrix $\mathbf{A}$, which consists of a rotation and a scaling effect. A key idea is comparing a vector $\vec{v}$ before and after the transformation $\mathbf{A}$.
If the effect of $\mathbf{A}$ is a rotation and a scaling, then you can take any vector $\vec{v}$ and know that the after-vector $\mathbf{A}\vec{v}$ will be a rotated and scaled version of it.
Notice that if there is a scaling factor, you can find it by taking the ratio of any vector's lengths before and after: $\frac{||\mathbf{A}v||}{||\vec{v}||}$.
And if there's a rotation factor, you could find it if you could take any vector and compute the angle between its positions before and after the transformation—find a vector where you can compute the angle between $\vec{v}$ and $\mathbf{A}\vec{v}$.
Let's start by computing the scaling factor. We need to pick a vector $\vec{v}$ and compute the ratio of its lengths before and after. That is, $\frac{||\mathbf{A}v||}{||\vec{v}||}$.
Based on our knowledge of eigenvalues, we know that it will be especially easy to compute this ratio if we pick $\vec{v}$ to be an eigenvector of the transformation $\mathbf{A}$. This is because if $\vec{v}$ is an eigenvector with eigenvalue $\lambda$, then the ratio is:
$$\frac{||\mathbf{A}v||}{||\vec{v}||} = \frac{||\lambda v||}{||\vec{v}||} = \frac{|\lambda|\cdot ||\vec{v}||}{||\vec{v}||} = |\lambda|$$
Here, we've used the fact that the length of any scaled vector $||\lambda \vec{v}||$ is equal to the magnitude $|\lambda|$ times the original length of the vector $||\vec{v}||$. When $\lambda =a+bi$ is a complex number, its magnitude is defined as $\sqrt{a^2 + b^2}$.
In short, we've shown that:
If $\mathbf{A}$ scales the lengths of all vectors by the same amount, and $\vec{v}$ is an eigenvector of $\mathbf{A}$ with complex eigenvalue $\lambda=a+bi$, the magnitude of the scaling effect must be $r\equiv \sqrt{a^2+b^2}$.
Now let's compute the angle of rotation. We need to pick a vector $\vec{v}$ and compute the angle between its positions before and after. We can use the formula $$\cos{\theta} = \frac{|v|\cdot |\mathbf{A}\vec{v}|}{\vec{v}\cdot \mathbf{A}\vec{v}}$$ to do it.
A convenient choice of vector is $\vec{v}=\vec{e}_1 = [1,0]$, because it has a 1 in one position and zeroes everywhere else, so it's easy to do matrix multiplication with. You can straightforwardly compute:$\mathbf{A}\vec{e}_1 = [4\sqrt{3}, 4].$
The length of $\vec{e}_1$ is 1, and the length of $\mathbf{A}\vec{e}_1$ is 2. (One formula for the length of a vector $[a_1, \ldots, a_n]$ is $\sqrt{a_1^2 + \ldots + a_n^2}$).
Their dot product is $\vec{e}_1 \cdot \mathbf{A}\vec{e}_1 = 4\sqrt{3}$. (One formula for the dot product of two vectors $[a_1, \ldots, a_n]$ and $[b_1, \ldots, b_n]$ is $a_1b_1+\ldots+a_nb_n$.)
So applying our formula, we find:
$$\cos{\theta} = \frac{1\cdot 2}{4 \sqrt{3}} = \frac{1}{2\cdot \sqrt{3}} = \frac{\sqrt{3}}{6}$$
Note that $|\vec{v}| \cdot |\mathbf{A}\vec{v}|$ from the dot product formula is the product of the lengths before and after. The scaling factor $r$ is the ratio of the lengths $|\mathbf{A}\vec{v}| / |\vec{v}|$. In general they won't be equal— except, of course, when you pick a vector with length one: $|\vec{v}|=1$.
$\endgroup$ 4 $\begingroup$Rotation (of angle) $\pi/6$ matrix should be unitary, so factor out.
$$A=\left[\begin{matrix}4\sqrt{3}&-4\\4&4\sqrt{3}\\\end{matrix}\right] = 8\left[\begin{matrix}\sqrt{3}/2&-1/2\\1/2&\sqrt{3}/2\\\end{matrix}\right]$$
$\endgroup$ 1
