Exercise 3.3 · Q6: Verify $A'A=I$ for two rotation-type matrices

Basics you need

Part (i)

1. (i) Write $A'$ and check orders

   Rows of $A$ become columns: row 1 $\cos\alpha, \sin\alpha$ becomes column 1, row 2 $-\sin\alpha, \cos\alpha$ becomes column 2.

   $$A' = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}.$$

   $A'$ and $A$ are both $2\times2$⁠: columns of $A'$ (2) = rows of $A$ (2), so $A'A$ is defined and is $2\times2$⁠, the same order as $I$⁠.

   Transpose, Definition 3; product orders, Section 3.4.5. Lesson: transpose

2. (i) Compute the four entries of $A'A$

   Row $i$ of $A'$ times column $k$ of $A$⁠. Column 1 of $A$ is $\cos\alpha, -\sin\alpha$⁠; column 2 is $\sin\alpha, \cos\alpha$⁠.

   $$\begin{aligned} c_{11} &= \cos\alpha\cos\alpha + (-\sin\alpha)(-\sin\alpha) \\ &= \cos^2\alpha + \sin^2\alpha = 1, \\[1mm] c_{12} &= \cos\alpha\sin\alpha + (-\sin\alpha)\cos\alpha \\ &= 0, \\[1mm] c_{21} &= \sin\alpha\cos\alpha + \cos\alpha(-\sin\alpha) \\ &= 0, \\[1mm] c_{22} &= \sin\alpha\sin\alpha + \cos\alpha\cos\alpha \\ &= \sin^2\alpha + \cos^2\alpha = 1. \end{aligned}$$
   In $c_{12}$ and $c_{21}$ the two products are the same number with opposite signs, because multiplication of numbers is commutative ($\sin\alpha\cos\alpha = \cos\alpha\sin\alpha$⁠). So they cancel for every $\alpha$⁠, with no trigonometric identity needed. The diagonal entries need the Pythagorean identity (Basic 3).
   Row-times-column rule, Section 3.4.5 (Basic 1). Basics: pair up, multiply, add

3. (i) Assemble and compare with $I$

   $$A'A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I.$$

   Every entry matches the identity matrix of order 2, so $A'A = I$ for every real $\alpha$⁠.

   The identity matrix (Basic 4), Section 3.3 (Lesson: types of matrices).

Part (ii)

1. (ii) Write $A'$ and check orders

   Row 1 of $A$⁠, $\sin\alpha, \cos\alpha$⁠, becomes column 1; row 2, $-\cos\alpha, \sin\alpha$⁠, becomes column 2:

   $$A' = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}.$$

   As before, $A'A$ is defined and is $2\times2$⁠.

2. (ii) Compute the four entries of $A'A$

   Column 1 of $A$ is $\sin\alpha, -\cos\alpha$⁠; column 2 is $\cos\alpha, \sin\alpha$⁠.

   $$\begin{aligned} c_{11} &= \sin\alpha\sin\alpha + (-\cos\alpha)(-\cos\alpha) \\ &= \sin^2\alpha + \cos^2\alpha = 1, \\[1mm] c_{12} &= \sin\alpha\cos\alpha + (-\cos\alpha)\sin\alpha \\ &= 0, \\[1mm] c_{21} &= \cos\alpha\sin\alpha + \sin\alpha(-\cos\alpha) \\ &= 0, \\[1mm] c_{22} &= \cos\alpha\cos\alpha + \sin\alpha\sin\alpha \\ &= \cos^2\alpha + \sin^2\alpha = 1. \end{aligned}$$
   This matrix is the matrix of Part (i) with $\sin$ and $\cos$ interchanged. The same two facts do all the work: the diagonal entries are $\sin^2\alpha + \cos^2\alpha$⁠, and the off-diagonal products cancel in pairs.

3. (ii) Assemble and compare with $I$

   $$A'A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I.$$

   So $A'A = I$ for every real $\alpha$ in Part (ii) as well.

   A matrix with $A'A = I$ is one whose columns each have "length" 1 ($\cos^2\alpha+\sin^2\alpha = 1$⁠) and are "perpendicular" to each other (their row-times-column product is 0). In both parts, the two columns are $(\cos\alpha, -\sin\alpha)$ and $(\sin\alpha, \cos\alpha)$⁠, or the same with sin and cos swapped, which have exactly this property. (One can check in the same way that $AA' = I$ too, though the question does not ask for it.)
   Pythagorean identity (Basic 3). Basics: trigonometric tools

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