Exercise 3.2 · Q18: Show $I+A=(I-A)$ times a rotation matrix

Basics you need

Solution

1. When the statement makes sense, and the orders

   $A$ contains $\tan\frac{\alpha}{2}$⁠, which exists only when $\cos\frac{\alpha}{2} \ne 0$⁠, that is, when $\frac{\alpha}{2}$ is not an odd multiple of $\frac{\pi}{2}$⁠. Equivalently,

   $$\alpha \ne (2n + 1)\pi, \quad n \in \mathbb{Z}.$$

   We assume this throughout (the question tacitly does, since it uses $A$⁠). Write

   $$t = \tan\frac{\alpha}{2}, \qquad A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix}.$$

   Orders: $I$⁠, $A$ and $R$ are all $2\times2$⁠. So $I + A$ and $I - A$ are $2\times2$⁠, and $(I - A)R$ is $(2\times2)(2\times2) = 2\times2$⁠. Both sides have the same order, so they can be compared.

   Naming $t = \tan\frac{\alpha}{2}$ keeps every line short and turns the trigonometry into algebra in one letter. The domain restriction matters later too: we will divide by $\cos^2\frac{\alpha}{2}$⁠, which is allowed precisely because $\cos\frac{\alpha}{2} \ne 0$⁠.
   $\tan\theta = \frac{\sin\theta}{\cos\theta}$ is undefined where $\cos\theta = 0$⁠, at $\theta = \pm\frac{\pi}{2}, \pm\frac{3\pi}{2}, \dots$ Basics: compound angles

2. Write $I + A$ and $I - A$

   Entry by entry, with $I$ having rows $(1, 0)$⁠, $(0, 1)$⁠:

   $$I + A = \begin{bmatrix} 1 + 0 & 0 - t \\ 0 + t & 1 + 0 \end{bmatrix} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$$ $$I - A = \begin{bmatrix} 1 - 0 & 0 - (-t) \\ 0 - t & 1 - 0 \end{bmatrix} = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$$
   $I + A$ is the target: at the end of Step 4 the product must come out as this matrix. $I - A$ is the left factor of the product we must compute. Note the sign in the $(1,2)$ entry of $I - A$⁠: subtracting $-t$ gives $+t$⁠.
   $P \pm Q = [p_{ij} \pm q_{ij}]$⁠. Lesson: addition of matrices

3. Write $\cos\alpha$ and $\sin\alpha$ in terms of $t$

   Let $\theta = \frac{\alpha}{2}$⁠, so $\alpha = 2\theta$ and $t = \tan\theta$⁠. Using the double-angle formulas and dividing by $1 = \cos^2\theta + \sin^2\theta$⁠:

   $$\cos\alpha = \frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta + \sin^2\theta}$$ $$\sin\alpha = \frac{2\sin\theta\cos\theta}{\cos^2\theta + \sin^2\theta}$$

   Now divide the top and bottom of each fraction by $\cos^2\theta$⁠, which is not zero by Step 1, and use $\frac{\sin\theta}{\cos\theta} = t$⁠:

   $$\cos\alpha = \frac{1 - t^2}{1 + t^2}, \quad \sin\alpha = \frac{2t}{1 + t^2}.$$

   The denominator $1 + t^2$ is at least 1, so it is never zero.

   Dividing by $\cos^2\theta + \sin^2\theta$ changes nothing (it equals 1), but it makes every term have degree 2 in $\sin\theta$ and $\cos\theta$⁠, so dividing top and bottom by $\cos^2\theta$ turns everything into powers of $\tan\theta$⁠. This is how a single variable $t$ captures both $\cos\alpha$ and $\sin\alpha$⁠.
   $\cos 2\theta = \cos^2\theta - \sin^2\theta$⁠, $\sin 2\theta = 2\sin\theta\cos\theta$⁠. Basics: compound angles

4. Multiply $(I - A)R$⁠, every entry

   $$(I - A)R = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$$

   Rows of $I - A$⁠: $(1, t)$ and $(-t, 1)$⁠. Columns of $R$⁠: $(\cos\alpha, \sin\alpha)$ and $(-\sin\alpha, \cos\alpha)$⁠. Each entry, with the formulas of Step 3 substituted and the common denominator $1 + t^2$⁠:

   Entry (1,1): row $(1, t)$ times column $(\cos\alpha, \sin\alpha)$⁠.

   $$\begin{aligned} e_{11} &= \cos\alpha + t\sin\alpha\\ &= \frac{1 - t^2 + t\cdot 2t}{1 + t^2}\\ &= \frac{1 + t^2}{1 + t^2} = 1 \end{aligned}$$

   Entry (1,2): row $(1, t)$ times column $(-\sin\alpha, \cos\alpha)$⁠.

   $$\begin{aligned} e_{12} &= -\sin\alpha + t\cos\alpha\\ &= \frac{-2t + t(1 - t^2)}{1 + t^2}\\ &= \frac{-t - t^3}{1 + t^2}\\ &= \frac{-t(1 + t^2)}{1 + t^2} = -t \end{aligned}$$

   Entry (2,1): row $(-t, 1)$ times column $(\cos\alpha, \sin\alpha)$⁠.

   $$\begin{aligned} e_{21} &= -t\cos\alpha + \sin\alpha\\ &= \frac{-t(1 - t^2) + 2t}{1 + t^2}\\ &= \frac{t + t^3}{1 + t^2}\\ &= \frac{t(1 + t^2)}{1 + t^2} = t \end{aligned}$$

   Entry (2,2): row $(-t, 1)$ times column $(-\sin\alpha, \cos\alpha)$⁠.

   $$\begin{aligned} e_{22} &= t\sin\alpha + \cos\alpha\\ &= \frac{2t^2 + 1 - t^2}{1 + t^2}\\ &= \frac{1 + t^2}{1 + t^2} = 1 \end{aligned}$$

   Assembling:

   $$(I - A)R = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$$
   Every entry ends with $1 + t^2$ on top and bottom, and the cancellation is legal because $1 + t^2 \ne 0$ (Step 3). In $e_{22}$⁠, the first product is $(-t)(-\sin\alpha) = t\sin\alpha$⁠: two minus signs make a plus.
   Factor before cancelling: $-t - t^3 = -t(1 + t^2)$⁠. Basics: pair up, multiply, add

5. Compare with $I + A$

   From Step 2, $I + A$ has rows $(1, -t)$ and $(t, 1)$⁠. From Step 4, $(I - A)R$ has rows $(1, -t)$ and $(t, 1)$⁠. All four corresponding entries are equal, so

   $$I + A = (I - A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$$

   for every $\alpha$ with $\alpha \ne (2n + 1)\pi$⁠, which is every $\alpha$ for which $A$ is defined.

   The argument used an arbitrary allowed $\alpha$ and no special values, so it proves the identity for all such $\alpha$ at once.
   Equal order and equal entries means equal matrices (Definition 2). Lesson: equality of matrices

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