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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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    @J.M., it's confusingly written, but I think the scope of the assertion that $U$ is unitary is only the paragraph _before_ the exercise. (It looks like definition 2.1.5 ends without fanfare in the middle of a paragraph with nothing but a full stop to mark it. The rest of the paragraph is discussion).2011-10-07

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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.)