Problem on orthogonal matrices

An orthogonal matrix is one satisfying $A A^t = I$. Suppose $$A = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ a & b & c \end{pmatrix}.$$

If $A$ is orthogonal, show that $(a, b, c)$ is perpendicular to $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)$ and $(0,\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$
If $A$ is orthogonal, show that $(a,b,c)$ is of unit length.
Find two values of $(a, b, c)$ so that $A$ is orthogonal.

Solution
Part (a)
Recall that
Dot product of perpendicular vectors is zero

Two vectors are perpendicular if their dot product is zero.
Hence, we are trying to show that $(a,b,c)\cdot(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)=0$ and $(a,b,c)\cdot(0,\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})=0$.
Matrix transpose

Now, let's explore what it means for $A$ to be orthogonal. Recall that $A^t$ is the transpose of $A$.
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
Matrix multiplication

The $(i,j)$ entry of the matrix product $\bfA \mathbf{B}$ is the dot product of the $i$th row of $\bfA$ with the $j$th column of $\mathbf{B}$.
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:
Dot product and vector length

The length of the vector $(a,b,c)$ is $\left| (a,b,c) \right| = \sqrt{ (a,b,c) \cdot (a,b,c) }$.
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 $A$.
(see details)

We can show that the only two unit vectors perpendicular to $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)$ and $(0,\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ are $$\Bigl(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \Bigr) \ \text{ and } \ \Bigl(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \Bigr).$$
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}

Problem on orthogonal matrices

Solution

Part (a)

Part (b)

Part (c)

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