# Proving a certain orthogonal matrix is a rotation matrix

1. Jul 18, 2015

### ELB27

1. The problem statement, all variables and given/known data
Let $U$ be a $2\times 2$ orthogonal matrix with $\det U = 1$. Prove that $U$ is a rotation matrix.

2. Relevant equations

3. The attempt at a solution
Well, my strategy was to simply write the matrix as
$$U = \begin{pmatrix} a & b\\ c & d \end{pmatrix}$$
and using the given properties to solve for $a,b,c$ and $d$. I have 4 equations:
1. Determinant = 1
2. $U^TU = I$ where $I$ is identity
where the last property gives me 3 equations (one of the entries is redundant). Thus, I have 4 equations in 4 unknowns. When I solve them, I get 1 free variable, and my matrix turns out to be of the form:
$$U = \begin{pmatrix} x & \mp \sqrt{1-x^2}\\ \pm \sqrt{1-x^2} & x \end{pmatrix}$$
and since all entries must be real (orthogonal matrix) we have the constraint on $x$:
$$-1≤x≤1$$
Clearly, these properties are satisfied if we let $x=\cos\theta$ or $x=\sin\theta$, thus obtaining a rotation matrix in its standard notation. However, I do not see how to prove that these trigonometric functions are the only possible solutions. Also, how does one define formally and rigorously a rotation matrix? Only as a matrix of cosines and sines (with the appropriate values of determinant etc.)?

Any suggestions/comments will be greatly appreciated!

2. Jul 18, 2015

### ShayanJ

There is no need for that. You've done all things needed. You have shown that for all allowed values of x, there exists a $\theta$ that either $x=\sin\theta$ or $x=\cos\theta$. It means for all allowed values of x, this matrix corresponds to a rotation. It may correspond to many other things, but that doesn't matter. All we care about now, is that it corresponds to a rotation. So you're done with this question.

3. Jul 18, 2015

### ELB27

Ah, I think I get it. Basically, the question can be rephrased as "Prove the $U$ can be represented as a rotation matrix."? Thus my proof will end as:
$x=\cos\theta \ ∀x$ for some angle $\theta$ and thus, $U$ is always a rotation matrix about some angle $\theta$.