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:

$\displaystyle AA^T = I = A^TA$

For your matrix, the transpose is:

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

So multiply it by the original matrix
$\displaystyle \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) $$\displaystyle = \left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array} \right) \displaystyle \left( \begin{array}{ccc} \bigg(\frac{\sqrt{3}}{2}\bigg)^2+\bigg(\frac{1}{2} \bigg)^2 & \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} \\ \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} & m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2 \end{array} \right)$$\displaystyle = \left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array} \right)$

$\displaystyle \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)$$\displaystyle = \left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array} \right)$

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