Miscellaneous · Q3: Find $x, y, z$ with $A'A=I$

Basics you need

Solution

1. Write down $A'$ and check the orders

   The columns of A are

   $$\begin{aligned} \text{col 1} &: (0,\ x,\ x) \\ \text{col 2} &: (2y,\ y,\ -y) \\ \text{col 3} &: (z,\ -z,\ z) \end{aligned}$$

   and these become the rows of the transpose:

   $$A' = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}.$$

   $A'$ is $3\times3$ and A is $3\times3$⁠: columns of $A'$ $=3=$ rows of A, so $A'A$ is defined and is $3\times3$⁠, the same order as $I = I_3$⁠. The equation $A'A = I$ therefore makes sense.

   We read the transpose off column by column rather than trying to "flip" the picture in one go, because that avoids sign slips with the entries $-z$ and $-y$⁠.
   Definition 3 (Section 3.5): if $A = [a_{ij}]$⁠, then $A' = [a_{ji}]$⁠. Lesson: transpose

2. Row 1 of $A'A$

   Row 1 of $A'$ is $(0, x, x)$⁠. Multiply it by each column of A:

   $$\begin{aligned} p_{11} &= 0\cdot0 + x\cdot x + x\cdot x = 2x^2 \\ p_{12} &= 0\cdot2y + x\cdot y + x\cdot(-y) = 0 \\ p_{13} &= 0\cdot z + x\cdot(-z) + x\cdot z = 0 \end{aligned}$$
   In $p_{12}$ the terms $xy$ and $-xy$ cancel; in $p_{13}$ the terms $-xz$ and $xz$ cancel. These cancellations happen for every value of $x, y, z$⁠, not just special ones, because of how the signs in A are placed.
   Row $i$ of the first factor times column $j$ of the second gives the $(i,j)$ entry. Basics: pair up, multiply, add

3. Row 2 of $A'A$

   Row 2 of $A'$ is $(2y, y, -y)$⁠:

   $$\begin{aligned} p_{21} &= 2y\cdot0 + y\cdot x + (-y)\cdot x = 0 \\ p_{22} &= 2y\cdot2y + y\cdot y + (-y)(-y) \\ &= 4y^2 + y^2 + y^2 = 6y^2 \\ p_{23} &= 2y\cdot z + y\cdot(-z) + (-y)\cdot z \\ &= 2yz - yz - yz = 0 \end{aligned}$$
   In $p_{22}$ all three terms are squares, so they cannot cancel: $(-y)(-y) = +y^2$⁠. In $p_{23}$ the first term $2yz$ is cancelled by the two terms $-yz$⁠. Also $p_{21} = p_{12}$⁠, as it must be, because $A'A$ is always symmetric: $(A'A)' = A'(A')' = A'A$⁠.
   $(-y)(-y) = y^2$⁠: the product of two negatives is positive. Basics: pair up, multiply, add

4. Row 3 of $A'A$⁠, and the whole product

   Row 3 of $A'$ is $(z, -z, z)$⁠:

   $$\begin{aligned} p_{31} &= z\cdot0 + (-z)\cdot x + z\cdot x = 0 \\ p_{32} &= z\cdot2y + (-z)\cdot y + z\cdot(-y) \\ &= 2yz - yz - yz = 0 \\ p_{33} &= z\cdot z + (-z)(-z) + z\cdot z = 3z^2 \end{aligned}$$

   Assembling the nine entries:

   $$A'A = \begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix}.$$
   The product is a diagonal matrix for every choice of $x, y, z$⁠. The diagonal entries are the "lengths squared" of the columns (each column times itself), and the zeros say that different columns are "perpendicular" (their products cancel). This is exactly why the condition $A'A = I$ is easy to meet here.
   A diagonal matrix has all its non-diagonal entries zero (Section 3.3 (iv)). Lesson: types of matrices

5. Equate with I, entry by entry

   $$\begin{bmatrix} 2x^2 & 0 & 0 \\ 0 & 6y^2 & 0 \\ 0 & 0 & 3z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

   Both matrices are $3\times3$⁠, so they are equal exactly when all nine pairs of entries agree. The six off-diagonal pairs read $0 = 0$⁠, which is true whatever $x, y, z$ are, so they impose no condition. The three diagonal pairs give

   $$2x^2 = 1, \qquad 6y^2 = 1, \qquad 3z^2 = 1.$$
   It is worth saying out loud that the off-diagonal equations are automatically satisfied. If one of them had come out as, say, $xy = 0$⁠, it would have been an extra condition and could have ruled out some of the solutions below.
   Definition 2 (Section 3.3.1): equal matrices have equal orders and equal corresponding entries. Lesson: equality of matrices

6. Solve each equation, keeping both signs

   $$\begin{aligned} 2x^2 = 1 &\iff x^2 = \tfrac12 \iff x = \pm\tfrac{1}{\sqrt2} \\ 6y^2 = 1 &\iff y^2 = \tfrac16 \iff y = \pm\tfrac{1}{\sqrt6} \\ 3z^2 = 1 &\iff z^2 = \tfrac13 \iff z = \pm\tfrac{1}{\sqrt3} \end{aligned}$$

   The three equations are independent of one another (each involves one unknown), so the signs can be chosen freely: there are $2\times2\times2 = 8$ solutions $(x, y, z)$ in all. In rationalised form, $x = \pm\frac{\sqrt2}{2}$⁠, $y = \pm\frac{\sqrt6}{6}$⁠, $z = \pm\frac{\sqrt3}{3}$⁠.

   A positive number has two real square roots. Every sign choice works because only $x^2$⁠, $y^2$⁠, $z^2$ appear in $A'A$⁠. Changing the sign of $x$ changes the sign of the first column of A, and that does not change any of the column products in Steps 2 to 4.
   If $t^2 = c$ with $c > 0$⁠, then $t = \sqrt c$ or $t = -\sqrt c$⁠. Basics: solving equations

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