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In my lecture my professor spoke of this function $R$ that takes a vector $\vec u=\left(\begin{matrix}a\\b\end{matrix}\right)$ and rotates by $\frac{\pi}{6}$ radians counter clockwise. Then he talked about a matrix $M=\left(\begin{matrix}\frac{\sqrt{3}}{2} & -\frac12\\\frac12 & \frac{\sqrt{3}}{2}\end{matrix}\right)$. He showed finding the lengths on a triangle with a hypotenuse length $1$, and I have a feeling that the $x$ and $y $ coordinates is in the first column, but I'm not sure where this second column comes from. If I need to provide any more information let me know. I really don't know what else to say without making this look like a jumbled mess.

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See – Guest 86 Jan 24 '13 at 18:32
You're right...the first column is the $x$ and $y$ coordinates of the point of the circle $(1,0)$ moves when we rotate by $\frac{\pi}{6}$...the second column is the coordinates of where the point $(0,1)$ rotates to, which are the cosine (top) and sine (bottom) of the angle $\frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3}$. – David Wheeler Jan 24 '13 at 18:40
up vote 2 down vote accepted

A terse answer goes like this:

Rigid rotations of the plane look like this:

$$ A(\theta)=\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix} $$

Matrices with orthonormal columns, like $A(\theta)$ has, produce all the rotations of the plane around the origin. This matrix produces a rotation of $\theta$ radians when placed to the right of a 1-by-2 row vector. In your case, $\theta=\pi/6$.

You can explore this by examining where the point $(1,0)$ goes under this rotation, and then argue that everything else rigidly follows. (After rotation, this could be viewed as the length-1 hypoteneuse of a right triangle, which is probably how the proof you saw went.)

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so really any rotation is going to have the form of $A(\theta)$ that you posted? That's what I think I got out of that when I read up on wikipedia. – TheHopefulActuary Jan 24 '13 at 18:47
Yes, all the rotations that don't reflect the plane, anyhow. – rschwieb Jan 24 '13 at 18:50

You know that the effect of the matrix $M$ on a vector $\vec{u}$ is to rotate it counterclockwise by $\frac{\pi}{6}$ radians. To determine the matrix $M$, let $\vec{e_1} = (1,0)$ and $\vec{e_2} = (0,1)$ be the standard basis vectors. Then the first column of $M$ is $M \vec{e_1}$ and the second column of $M$ is $M \vec{e_2}$.

To figure out $M \vec{e_1}$, you simply have to rotate $\vec{e_1}$ and write down the resulting vector in coordinates. For that, you wil need to recall some basic triangle goemetry. Then you need to perform a similar calculation to determine $M \vec{e_2}$.

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why is it that we have $\vec e_1$ and $\vec e_2$? I'm not understanding that part. – TheHopefulActuary Jan 24 '13 at 18:49
That's just notation for the usual unit vectors in the "$+x$" and "$+y$" directions. The matrix of any linear transformation can be figured out by seeing what happens to each of these vectors. In fact, that is what the columns of the matrix tell you. Or, in your problem, you know how to figure out where each of these vectors gets sent to, so that allows you to fill in the columns of your matrix. – Michael Joyce Jan 24 '13 at 18:52

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