Boulderbeast

@James
I think you're on the right track!

@msw

Okay.
That makes sense, but as you say, my concern now becomes "how do I prove that orthogonal matrices map between orthonormal bases?"
Also, I'd like to know why it is that
[ COS(Rz), -SIN(Rz) ; COS(Rz), SIN(Rz)]
rotates vectors around the z-axis and not some other axis.

notation:

i=(1,0,0)
j=(0,1,0)
k=(0,0,1)
I=3x3 identity matrix
M(trans)=the transpose of the matrix M
*=matrix multiplication
i'=M*i=the image of i
j'=M*j=the image of j
k'=M*k=the image of k
(dot)=the dot product
0=0
M1c=the vector formed by the first column of M
M2c=the vector formed by the 2nd column of M
M3c=the vector formed by the 3rd column of M

We must first realize that the nth column of any matrix (not just orthogonal ones) is just the image of the nth basis vector under matrix multiplication. To see this, just take any random 3x3 matrix and try M*i, M*j, and M*k. You'll just get the columns of the original matrix. Using my above notation:
M1c=M*i=i'
M2c=M*j=j'
M3c=M*k=k'

Okay, with that understood, this becomes easy. If your definition of orthogonal matrix is that M*M(trans)=I, then consider:

Getting the elements of I is the same as (dot)ing a column of M with a row of M(trans), but rows of M(trans) are columns of M. So the ones along the diagonal of I are telling you that:
M1c(dot)M1c=1
M2c(dot)M2c=1
M3c(dot)M3c=1

which we can reinterpret as:
i'(dot)i'=1
j'(dot)j'=1
k'(dot)k'=1
which means these images are all of length 1

Similarly, all the 0's in I are just telling us, multiple times over, that:
i'(dot)j'=0
i'(dot)k'=0
j'(dot)k'=0
which is what it means (algebraically) for these various images to be orthogonal to each other.

And so we can see that M*M(trans)=I is nothing more than a nice compact algebraic way of saying that M transforms the standard orthonormal basis into another one.

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I am Srime

Knowing that M1c=i' and so forth, it is also nearly trivial to recognize when a matrix happens to be rotating around a specific coordinate axis. I mean, take the x and y axes and rotate them by theta about the z-axis, and you can convince yourself easily that the image of i is (cos theta, sin theta) and the image of j is (-sin theta, cos theta). Thus, this is how such a two dimensional matrix is built.

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I am Srime

@ World
Any other guesses?

@ Math man
Cool!
Thanks, math man.
That proof makes sense to me, although, as you say, I'll have to sit down and convince myself about that z-axis rotation.
Er ... nevermind, I think I just got it

I think your name is "Math (is) So Wack". And then there's a number because that's totally wack.

Glad to have been able to help. Just don't start calling me "Math Man" or "Math is So Wack 188"

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I am Srime

Ok dude, but just so you know, I was just handing you one of the coolest nick names of all time.