Which matrices perform dilation?

Question

For each angle $\theta \in \mathbb{R}$, we get two corresponding matrices:

$$\mathrm{Rot}^\theta = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos\theta\end{bmatrix}, \qquad \mathrm{Refl}^\theta = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos2\theta\end{bmatrix}.$$

The associated linear transformations:

1. Rotate by an angle $\theta$ about the origin

2. Reflect in the line through the origin at an angle $\theta$ from the $x$-axis

Now another way of thinking about $\mathrm{Refl}^\theta$ is that it decomposes each vector into two components, a parallel and perpendicular component, then scales the perpendicular component by $-1$, and adds the resulting vectors. This suggests that there should be a generalization of $\mathrm{Refl}^\theta$ that we'll denote $\mathrm{Dil}^{\theta}_a$ in which we dilate by a factor of $a$ perpendicular to the relevant line. It ought to be the the case that $\mathrm{Refl}^{\theta} = \mathrm{Dil}^\theta_{-1}.$

Question. What matrix decribes $\mathrm{Dil}^\theta_a$?

Answer 1

In case $x$-axis is the reflecting line the matrix you are seeking is $A=\pmatrix {1 & 0\cr 0 & a\cr}$. For a reflecting line that makes angle $\theta$ with $x$-axis the matrix would be got by change-of-basis formula: $ \mathrm{Rot}^\theta * A *\mathrm{Rot}^{-\theta}$. (here $*$ denotes matrix multiplication}

Answer 2

$\def\dil{\mathop{\rm Dil}\nolimits^\theta_a}$A standard and useful way to do this kind of thing is to write vectors in terms of a convenient basis. For your "$\dil$" problem a convenient basis will consist of $$v_1=\pmatrix{\cos\theta\cr \sin\theta\cr}\quad\hbox{and}\quad v_2=\pmatrix{-\sin\theta\cr \cos\theta\cr}\ ,$$ so that $v_1$ is parallel to your line and $v_2$ is perpendicular to it. Then $$\dil v_1=v_1\quad\hbox{and}\quad \dil v_2=av_2\ .$$ The matrix of $\dil$ with respect to the basis $\{v_1,v_2\}$ is $$\pmatrix{1&0\cr 0&a\cr}\ ,$$ and the matrix with respect to standard bases is $$\pmatrix{\cos\theta&-\sin\theta\cr \sin\theta&\cos\theta\cr} \pmatrix{1&0\cr 0&a\cr}\pmatrix{\cos\theta&-\sin\theta\cr \sin\theta&\cos\theta\cr}^{-1} =\pmatrix{\cos^2\theta+a\sin^2\theta&(1-a)\cos\theta\sin\theta\cr (1-a)\cos\theta\sin\theta&a\cos^2\theta+\sin^2\theta\cr}\ . $$

Answer 3

David gives an explicit formula.

Furthermore, using the identities $$\cos^2\theta - \sin^2 \theta = \cos 2\theta, \qquad \sin \theta \cos \theta = \frac{1}{2}\sin 2\theta$$ we can rewrite David's formula to resemble the matrix $\mathrm{Refl}^\theta$ more closely. An especially elegant derivation follows P Vanchinathan's approach, and we find that:

Proposition. $$\mathrm{Rot}^\theta \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \mathrm{Rot}^{-\theta} = \frac{a+b}{2}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \frac{a-b}{2}\mathrm{Refl}^\theta$$

Proof.

In particular, observe that for any two real numbers $a$ and $b$, we have: $$(a,b) = \frac{a+b}{2}(1,1) + \frac{a-b}{2}(1,-1).$$ This allows us to argue as follows:

$$\begin{align*} \mathrm{Rot}^\theta \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \mathrm{Rot}^{-\theta} &= \mathrm{Rot}^\theta \left(\frac{a+b}{2}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}+\frac{a-b}{2}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\right) \mathrm{Rot}^{-\theta} \\ &= \frac{a+b}{2}\mathrm{Rot}^\theta \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\mathrm{Rot}^{-\theta} + \frac{a-b}{2}\mathrm{Rot}^\theta \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \mathrm{Rot}^{-\theta} \\ &= \frac{a+b}{2}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \frac{a-b}{2}\mathrm{Refl}^\theta \end{align*} $$

Corollary. To dilate a point $P$ by a factor of $b$ from a line, first dilate $P$ by a factor of $(1+b)/2$ from the origin, then take the reflection of $P$ in the relevant line and dilate that by a factor of $(1-b)/2$, and then add the two.