Solving Orthogonal Matrix Problem

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[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?
 
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ok.. so u have a matrix:

[tex] \begin{bmatrix}<br /> \alpha & \beta \\<br /> \beta & -\alpha<br /> \end{bmatrix}[/tex]

on multiplying with it's transpose, you have:

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

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.
 
Hmm, when multiplying with it's inverse (transpose), I get:[a^2+b^2 0 ]
[ 0 a^2+b^2].
 
Last edited:
Yes, you were right the first time. Given only the information that A is a 2x2 orthogonal matrix you only have [itex]\alpha^2+ \beta^2= 1[/itex]. That's because there are an infinite number of such matrices, not just one. Any [itex]\alpha[/itex] and [itex]\beta[/itex] satisfying [itex]\alpha^2+ \beta^2= 1[/itex] will give you an orthogonal matrix.