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I am reading about unitary matrices in Horn and Johnson's Matrix Analysis. On page 68, the exercise asks, letting $T(\theta)=\begin{pmatrix}\cos\theta & \sin\theta \\-\sin\theta & \cos\theta \\ \end{pmatrix}$ where $\theta$ is a real parameter:

If $U\in M_{2}(\mathbb{R})$ is a unitary matrix, show that $U$ is real orthogonal iff $U=T(\theta)$ or $U=\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}T(\theta)$

Since $U$ is unitary, it we know that it is an isometry, i.e. if $x\in \mathbb{R}$, $Ux=y \implies \|Ux\| = \|y\|$. But what does does this have to do with sine and cosine functions?

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The sine and cosine functions in coordinate form $(\cos\theta,\sin\theta)$ trace out the unit circle. Now if you have a unit vector in $\mathbb{R}^2$, what are the two unit vectors perpendicular to it? – anon Oct 7 '11 at 0:46
Can you solve the "if" direction? That will show you how the trig functions work in this context. – Henning Makholm Oct 7 '11 at 0:50
I think the sine and cosine come into the matrix simply because the norm of the matrix columns is 1 and $\cos^2 \theta+\sin^2\theta=1$. – Shiyu Oct 7 '11 at 1:07
@Henning: I think that OP simply isn't transcribing the problem in full. – anon Oct 7 '11 at 1:22
up vote 3 down vote accepted

Brief explanation:

If $U$ is a unitary we have that $Ue_1$ and $Ue_2$ are unit vectors, where $e_1=(1,0)^T,e_2=(0,1)^T$. Let's call these two column vectors $a=Ue_1$ and $b=Ue_2$, so that $U=[\;a\;\;\; b\;]$ is the unitary matrix. Now the orthogonal property tells us $U^TU=I$, so $a$ and $b$ must be perpendicular in order for the off-diagonal entries of $U^TU$ to be zero (write it out and you should see). Since $(\cos\theta,\sin\theta)$ parametrizes all unit vectors, we can say $a=(\cos\theta,\sin\theta)^T$, which determines $b$ up to sign as you calculated in the comments. (The placement of negative signs in the form given to you doesn't quite agree with this explanation, but you should be able to fix that with the substitution $\phi=-\theta$ when need be, noting that cosine is even and sine is odd.)

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