# Orthogonal Matrix

• Feb 17th 2009, 03:40 AM
mr_motivator
Orthogonal Matrix
• Feb 17th 2009, 03:53 AM
Mush
Quote:

Originally Posted by mr_motivator

An Orthogonal matrix is one whos transpose is its inverse.

In other words:

$AA^T = I = A^TA$

For your matrix, the transpose is:

$\left( \begin{array}{ccc}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
m & -\frac{\sqrt{3}}{2}
\end{array} \right)$

So multiply it by the original matrix
$\left( \begin{array}{ccc}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
m & -\frac{\sqrt{3}}{2}
\end{array} \right)\left( \begin{array}{ccc}
\frac{\sqrt{3}}{2} & m \\
\frac{1}{2}
& -\frac{\sqrt{3}}{2}
\end{array} \right)
$
$= \left( \begin{array}{ccc}
1 & 0 \\
0
& 1
\end{array} \right)$

$\left( \begin{array}{ccc}
\bigg(\frac{\sqrt{3}}{2}\bigg)^2+\bigg(\frac{1}{2} \bigg)^2 & \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} \\
\frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} & m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2
\end{array} \right)
$
$= \left( \begin{array}{ccc}
1 & 0 \\
0
& 1
\end{array} \right)$

$\left( \begin{array}{ccc}
1 & \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} \\
\frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} & m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2
\end{array} \right)
$
$= \left( \begin{array}{ccc}
1 & 0 \\
0
& 1
\end{array} \right)$

Hence $\frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} =0$ and $m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2 = 1$