## Problem on orthogonal matrices

An orthogonal matrix is one satisfying $A A^t = I$. Suppose $$A = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ a & b & c \end{pmatrix}.$$
1. If $A$ is orthogonal, show that $(a, b, c)$ is perpendicular to $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)$ and $(0,\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$
2. If $A$ is orthogonal, show that $(a,b,c)$ is of unit length.
3. Find two values of $(a, b, c)$ so that $A$ is orthogonal.

• ## Solution

#### Part (a)

Recall that
Hence, we are trying to show that $(a,b,c)\cdot(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)=0$ and $(a,b,c)\cdot(0,\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})=0$.
We are told $A A^t =I$, which is to say $$\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}} \\ a & b & c \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & 0 & a \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}& b \\ 0 & \frac{1}{\sqrt{2}} & c \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
Recall that
Observing that the third row of $A$ is $(a,b,c)$ and the first column of $A^t$ is $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)$, the $(3,1)$ entry of the matrix multiplication tells us that $(a,b,c) \cdot (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)=0$:$$\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \color{blue}{a} & \color{blue}{b} & \color{blue}{c} \end{pmatrix} \begin{pmatrix} \color{blue}{\frac{1}{\sqrt{2}}} & 0 & a \\ \color{blue}{\frac{1}{\sqrt{2}}} & \frac{1}{\sqrt{2}} & b \\ \color{blue}{0} & \frac{1}{\sqrt{2}} & c \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \color{blue}{0} & 0 & 1 \end{pmatrix}$$
Similarly, the $(3,2)$ entry of the matrix multiplication tells us that $(a,b,c)\cdot(0,\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})=0$:$$\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \color{blue}{a} & \color{blue}{b} & \color{blue}{c} \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & \color{blue}{0} & a \\ \frac{1}{\sqrt{2}} & \color{blue}{\frac{1}{\sqrt{2}}} & b \\ 0 & \color{blue}{\frac{1}{\sqrt{2}}} & c \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & \color{blue}{0} & 1 \end{pmatrix}$$

#### Part (b)

Because the matrix multiplication has a direct interpretation as a dot product, we recall the relationship between dot product and length:
Observe, that the $(3,3)$ entry of the matrix multiplication above tells us that $(a,b,c) \cdot (a,b,c) = 1$:$$\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \color{blue}{a} & \color{blue}{b} & \color{blue}{c} \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & 0 & \color{blue}{a} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \color{blue}{b} \\ 0 & \frac{1}{\sqrt{2}} & \color{blue}{c} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \color{blue}{1} \end{pmatrix}$$
Hence, the length of $(a,b,c)$ is also $1$.

#### Part (c)

Combining parts (a) and (b), we need to find two separate vectors $(a,b,c)$ that have unit length and are perpendicular to the first two rows of $A$.
Hence, if $A$ is orthogonal, \begin{align} (a,b,c) &= \Bigl(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \Bigr), \text{ or } \\ (a,b,c) &= \Bigl(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \Bigr). \end{align}