Solving Orthogonal Matrix Problem

[Niles]

[SOLVED] Ortogonal matrix

Homework Statement

I have a 2x2 matrix A:

[alfa beta]
[beta -alfa],

where alfa and beta are real parameters. I have to find out for which values of alfa and beta A is an orthogonal matrix.

The Attempt at a Solution

A matrix is orthogonal if it satisfies Q*Q^T = I.

So I will multiply A with A^T and equal it to I, and I get the condition alfa^2 + beta^2 = 1. Are there any other conditions I need?

 

[rohanprabhu]

ok.. so u have a matrix:

<br /> \begin{bmatrix}<br /> \alpha &amp; \beta \\<br /> \beta &amp; -\alpha<br /> \end{bmatrix}<br />

on multiplying with it's transpose, you have:

<br /> \begin{bmatrix}<br /> \alpha^2 + \beta^2 &amp; 0\\<br /> 0 &amp; \beta^2 - \alpha^2<br /> \end{bmatrix} = <br /> \begin{bmatrix}<br /> 1 &amp; 0\\<br /> 0 &amp; 1<br /> \end{bmatrix}<br />

Look at the matrix now, equated with the identity matrix. You've taken one equation correctly. But, does the relation we have now provide you with another equation?

HINT: For two matrices to be equal, all their elements should be equal.

 

[Niles]

Hmm, when multiplying with it's inverse (transpose), I get:[a^2+b^2 0 ]
[ 0 a^2+b^2].

 

[HallsofIvy]

Yes, you were right the first time. Given only the information that A is a 2x2 orthogonal matrix you only have \alpha^2+ \beta^2= 1. That's because there are an infinite number of such matrices, not just one. Any \alpha and \beta satisfying \alpha^2+ \beta^2= 1 will give you an orthogonal matrix.

 

[Niles]

Cool, thanks.