Jtdylktuy

$begin{pmatrix}cos alpha & -sin alpha \ sin alpha & cos alphaend{pmatrix}$ defines a rotation of any angle on a vector in $mathbb{R}^2$, but how does it work in $mathbb{R}^3$ ? (I'd imagine this idea isn't very useful in dimensions other than 2 and 3)

There is another very efficient answer : by using matrix exponentials.

This is best understood by analogy with the fact that, identifying $mathbb{R}^2$ with $mathbb{C}$, your transformation :

$$begin{pmatrix}x'\y'end{pmatrix}=begin{pmatrix}cos theta & -sin theta \ sin theta & cos thetaend{pmatrix}begin{pmatrix}x\yend{pmatrix}$$

becomes, with complex notations :

$$x'+iy'=cos(theta)x-sin(theta)y + i(sin(theta)x+cos(theta)y)$$

which can be simplified into

$$x'+iy'=(cos(theta)+isin(theta))(x+iy)$$

Said otherwise : $$z'=e^{itheta}z. tag{1}$$

It happens that expression (1) has a nice generalization in $mathbb{R^3}$ under an analogous form $X'=exp(...)X$ as follows:

The matrix of the 3D rotation with angle $theta$ around (any) unit vector $N=[a,b,c]$ is given by :

$$exp(theta [N]_{times}),tag{2}$$

with the skew-symmetric matrix :

$$ [N]_{times} := begin{pmatrix}0&-c&b\c&0&-a\-b&a&0end{pmatrix} tag{3}.$$

Most languages have a built-in implementation of the exponential of a matrix. Let us take an example with Matlab : rotation with angle $2pi/3$ around the axis defined by the nonunitary vector $U=(1,1,1)$ is obtained in this way :

U=[1,1,1];

U=U/norm(U);

a=U(1);b=U(2);c=U(3);

V=[0,-c,b;

   c,0,-a;

   -b,a,0];

R=expm((2*pi/3)*V); % (expm for exp-matricial)

% giving the result :

% 0 0 1

% 1 0 0

% 0 1 0

as can be awaited (canonical axes $e_1 to e_2 to e_3 (to e_1)$).

Let us give some precision about :

• The exponential of a matrix $M$, which is defined by the same series expansion as the ordinary exponential :

$$exp(M)=I+M+frac12M^2+cdots+frac{1}{n!}M^n+cdots tag{4}$$

• The matrix $[N]_{times}$ defined in (3) which will no longer look mysterious when it is associated with the cross product under the following form :

$$Z=[N]_{times}Y = N times Y$$

as can be verified with the following explicit computation :
$$begin{pmatrix}u\v\wend{pmatrix}=begin{pmatrix} 0&-c& b\ c&0&-a\-b& a&0end{pmatrix}begin{pmatrix}x\y\zend{pmatrix}=begin{pmatrix}bz-cy\cx-az\ay-bxend{pmatrix}$$

A connection with Rodrigues formula (given by @Artic tern) is straightforward by decomposing even and odd terms in (3), then taking into account the fact that $([N]_{times})^3=-[N]_{times}$: see for example : https://en.wikipedia.org/wiki/Axis%E2%80%93angle_representation . See also an answer of mine :{https://math.stackexchange.com/q/1917093}.

Final remark : this expression of a 3D rotation is made for computers, of course. But nowadays very few people compute 3D rotations by hand...

Let ${mathbf{u},mathbf{v},mathbf{w}}$ be an oriented orthonormal basis and let $R$ be the rotation around $mathbf{w}$ by $theta$.

Since the cross product $mathbf{w}timesmathbf{x}$ has the effect of projecting $mathbf{x}$ onto the $mathbf{uv}$-plane and rotating by a right angle, the combination $(costheta)mathbf{x}+(sintheta)mathbf{w}timesmathbf{x}$ is almost $R(mathbf{x})$, but with the $mathbf{w}$-component of $mathbf{x}$ scaled by $costheta$; to compensate, add in $mathbf{w}$ scaled by $(1-costheta)$ times the $mathbf{w}$-component of $mathbf{x}$.

This yields the Rodrigues' rotation formula

$$ R(mathbf{x})=(costheta)mathbf{x}+(sintheta)(mathbf{w}timesmathbf{x})+(1-costheta)(mathbf{w}cdotmathbf{x})mathbf{w}. $$

Feel free to write out an explicit formula from this.

Because the $2$-sphere $S^2$ parametrizes the possible "axes" $mathbf{w}$ and $theta$ fills in the rest of the rotation $R$, the group of all $3times3$ rotation matrices is $(2+1)=3$-dimensional.

An orthogonal matrix is one which satisfies any of the following:

• 
  $|Amathbf{x}|=|mathbf{x}|$ for all vectors $mathbf{x}$ (i.e. preserves lengths)


• 
  $(Amathbf{x})cdot(Amathbf{y})=mathbf{x}cdotmathbf{y}$ for all vectors $mathbf{x},mathbf{y}$ (preserves lengths and angles)


• 
  $A^TA$ is the identity matrix $I$
  


• The columns of $A$ are orthonormal vectors


• The rows of $A$ are orthonormal vectors

The orthogonal matrices with determinant $-1$ reverse the orientation of space, such as reflections or the inversion-about-the-origin matrix $-I$. The orthogonal matrices with determinant $+1$ preserve the orientation of space and are called special orthogonal - these are precisely the $3times 3$ rotation matrices.

So to form any rotation matrix $R$, you can pick the first column vector to be any unit vector $mathbf{u}$ (essentially a choice of where you want $R$ to send $mathbf{e}_1$), in the second column you can pick any unit vector $mathbf{v}$ orthogonal to the first vector $mathbf{v}$ (a choice of where $R$ sends $mathbf{e}_2$), and finally $mathbf{w}=mathbf{u}timesmathbf{v}$ is the unique unit vector which makes ${mathbf{u},mathbf{v},mathbf{w}}$ a correctly-oriented orthonormal basis (and $mathbf{w}$ is a choice of where to send $mathbf{e}_3$).

Since $mathbf{u}$ is chosen from $S^2$ and $mathbf{v}$ is chosen from the circle's worth of perpendicular unit vectors, the $3times 3$ rotation group we can again conclude is $(2+1)=3$-dimensional.

(Note the vectors $mathbf{u},mathbf{v},mathbf{w}$ represent different things in these two descriptions.)