Hi all. Is there any explicit matrix expression for a general element of the special orthogonal group $SO(3)$? I have been searching texts and net both, but could not find it. Kindly provide any references.

Here is the standard quaternion answer: Given $(a,b,c,d)$ such that $a^2+b^2+c^2+d^2=1$, the matrix $$\begin{pmatrix} a^2+b^2c^2d^2&2bc2ad &2bd+2ac \\ 2bc+2ad &a^2b^2+c^2d^2&2cd2ab \\ 2bd2ac &2cd+2ab &a^2b^2c^2+d^2\\ \end{pmatrix}$$ is a rotation and every rotation matrix is of this form. Note that $(a,b,c,d)$ and $(a, b,c,d)$ give the same rotation. 


There is a good way to derive the sort of thing you're looking for: use the double cover $SU(2) \to SO(3)$. $SU(2)$ is diffeomorphic to the 3sphere $S^3 \subseteq \mathbb{C}^2$ $$ SU(2) = \left\{ \begin{pmatrix} a & \overline{b} \\ b & \overline{a} \end{pmatrix} : a^2 + b^2 = 1 \right\} $$ Now $SU(2)$ acts on its Lie algebra $\mathfrak{su}_2$ (which is 3dimensional) by conjugation. This action preserves the inner product $$ \langle X, Y \rangle =  \frac12 \mathrm{tr}(XY) = \frac12 \mathrm{tr}(X^*Y)$$ (which is a scalar multiple of the Killing form of $\mathfrak{su}_2$, FYI) and hence this gives a homomorphism $SU(2) \to SO(\mathfrak{su}_2) \simeq SO(3)$. (A priori this gives a map to $O(3)$, but $SU(2)$ is connected so the image lands in $SO(3)$. Now consider the orthonormal basis for $\mathfrak{su}_2$ given by $$ e_1 = \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix}, e_2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, e_3= \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}, $$ let $$ x = \begin{pmatrix} a & \overline{b} \\ b & \overline{a}\end{pmatrix},$$ and write down the adjoint action of $x$ on $e_1,e_2,e_3$. For instance you get $$ \begin{align} xe_1x^{1} & = \begin{pmatrix} i(a^2  b^2) & 2ia\overline{b} \\ 2i\overline{a} b & i(a^2  b^2) \end{pmatrix} \\ & = (a^2  b^2)e_1 + i(a\overline{b}  \overline{a}b)e_2 + (a \overline{b} + \overline{a}b)e_3. \end{align}$$ This gives you the first column of the matrix representation conjugation by $x$. I'll leave the others to you. But this way you can see where the formulas come from. This gives exactly David Speyer's answer (possibly modulo some reordering of the basis). His four real numbers $a,b,c,d$ would correspond to my complex numbers $a,b$ via $$a_{mine} = a + i b, \quad b_{mine} = c + id.$$ 


To give something explicit in sine and cosine, $$ \left( \begin{array}{ccc} \cos\theta \cos\psi & \cos\phi \sin\psi + \sin\phi \sin\theta \cos\psi & \sin\phi \sin\psi + \cos\phi \sin\theta \cos\psi \\\ \cos\theta \sin\psi & \cos\phi \cos\psi + \sin\phi \sin\theta \sin\psi & \sin\phi \cos\psi + \cos\phi \sin\theta \sin\psi \\\ \sin\theta & \sin\phi \cos\theta & \cos\phi \cos\theta \end{array} \right) $$ Note that three parameters are required. In odd dimension, there is a real eigenvalue. For $SO_n$ this eigenvalue is $+1.$ So there is a fixed vector in some direction. It takes two parameters to specify this point on the unit sphere. The Lie group element is then a rotation around this point. So it takes a third parameter specifying the amount of rotation about that axis. 

