Section 39.4 Examples
Here we will provide two examples of constructing product-preserving matrices.
Example 39.4.1. An orthogonal matrix.
The vectors
\begin{align*}
\uvec{v}_1 \amp = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} \text{,} \amp
\uvec{v}_2 \amp = \left[\begin{array}{r} -1 \\ 1 \\ 0 \\ 2 \end{array}\right] \text{,} \amp
\uvec{v}_3 \amp = \left[\begin{array}{r} -1 \\ 1 \\ 0 \\ -1 \end{array}\right] \text{,} \amp
\uvec{v}_4 \amp = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}
\end{align*}
form an orthogonal basis of \(\R^4\text{.}\)
We can normalize these vectors into an orthonormal basis:
\begin{align*}
\uvec{v}_1' \amp = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \\ 0 \end{bmatrix} \text{,} \amp
\uvec{v}_2' \amp = \begin{bmatrix} -\frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \\ 0 \\ \frac{2}{\sqrt{6}} \end{bmatrix} \text{,} \amp
\uvec{v}_3' \amp = \begin{bmatrix} -\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ 0 \\ -\frac{1}{\sqrt{3}} \end{bmatrix} \text{,} \amp
\uvec{v}_4' \amp = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \text{.}
\end{align*}
Placing these four vectors as columns in a \(4 \times 4\) matrix results in an orthogonal matrix:
\begin{equation*}
A = \begin{bmatrix}
\frac{1}{\sqrt{2}} \amp -\frac{1}{\sqrt{6}} \amp -\frac{1}{\sqrt{3}} \amp 0 \\
\frac{1}{\sqrt{2}} \amp \frac{1}{\sqrt{6}} \amp \frac{1}{\sqrt{3}} \amp 0 \\
0 \amp 0 \amp 0 \amp 1 \\
0 \amp \frac{2}{\sqrt{6}} \amp -\frac{1}{\sqrt{3}} \amp 0
\end{bmatrix}\text{.}
\end{equation*}
Example 39.4.2. A unitary matrix.
The vectors
\begin{align*}
\uvec{v}_1 \amp = \begin{bmatrix} 1 \\ 1 \\ \ci \\ \ci \end{bmatrix} \text{,} \amp
\uvec{v}_2 \amp = \left[\begin{array}{r} 1 \\ -3 \\ \ci \\ \ci \end{array}\right] \text{,} \amp
\uvec{v}_3 \amp = \begin{bmatrix} \ci \\ 0 \\ 1 \\ 0 \end{bmatrix} \text{,} \amp
\uvec{v}_4 \amp = \left[\begin{array}{r} \ci \\ 0 \\ -1 \\ 2 \end{array}\right]
\end{align*}
form an orthogonal basis of \(\C^4\text{.}\)
We can normalize these vectors into an orthonormal basis:
\begin{align*}
\uvec{v}_1' \amp = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ \frac{\ci}{2} \\ \frac{\ci}{2} \end{bmatrix} \text{,} \amp
\uvec{v}_2' \amp = \begin{bmatrix} \frac{1}{2\sqrt{3}} \\ -\frac{\sqrt{3}}{2} \\ \frac{\ci}{2\sqrt{3}} \\ \frac{\ci}{2\sqrt{3}} \end{bmatrix} \text{,} \amp
\uvec{v}_3' \amp = \begin{bmatrix} \frac{\ci}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \\ 0 \end{bmatrix} \text{,} \amp
\uvec{v}_4' \amp = \begin{bmatrix} \frac{\ci}{\sqrt{6}} \\ 0 \\ -\frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \end{bmatrix} \text{.}
\end{align*}
Placing these four vectors as columns in a \(4 \times 4\) matrix results in a unitary matrix:
\begin{equation*}
A = \begin{bmatrix}
\frac{1}{2} \amp \frac{1}{2\sqrt{3}} \amp \frac{\ci}{\sqrt{2}} \amp \frac{\ci}{\sqrt{6}} \\
\frac{1}{2} \amp -\frac{\sqrt{3}}{2} \amp 0 \amp 0 \\
\frac{\ci}{2} \amp \frac{\ci}{2\sqrt{3}} \amp \frac{1}{\sqrt{2}} \amp -\frac{1}{\sqrt{6}} \\
\frac{\ci}{2} \amp \frac{\ci}{2\sqrt{3}} \amp 0 \amp \frac{2}{\sqrt{6}}
\end{bmatrix}\text{.}
\end{equation*}