3D rotation group It is known that the group $\text{SO}(3)$ of rotation-matrices (matrices $A$ with $\det(A)=1$)
are generated from three parameters. This can be expressed by the fact, that
any rotation matrix is a composition of axis rotations
$$
\begin{pmatrix}
\cos(\phi)&-\sin(\phi)&0\\
\sin(\phi)&\cos(\phi)&0\\
0&0&1\\
\end{pmatrix},
\begin{pmatrix}
\cos(\phi)&0&\sin(\phi)\\
0&1&0\\
-\sin(\phi)&0&\cos(\phi)\\
\end{pmatrix},
\begin{pmatrix}
1&0&0\\
0&\cos(\phi)&-\sin(\phi)\\
0&\sin(\phi)&\cos(\phi)\\
\end{pmatrix}
$$
The question is: Why is the second matrix (Usually called rotation around the $y$-axis ) in almost any textbook written like this?
Related to the other two matrices, I would say that the negative 
$$
\begin{pmatrix}
\cos(\phi)&0&-\sin(\phi)\\
0&1&0\\
\sin(\phi)&0&\cos(\phi)\\
\end{pmatrix},
$$
is conceptual more straight forward. Any help or guidance will be appreciated. 
 A: Note that all the matrices listed will rotate vectors by the angle $\phi$ around the $x,y$ and $z$ axis respectively. The alternating signs is a result of the right hand screw rule. Let 
\begin{equation*}
A=\bigl(\begin{smallmatrix}
\cos(\phi) & 0 & \sin(\phi) \\ 
0 & 1 & 0 \\ 
-\sin(\phi) & 0 & \cos(\theta)
\end{smallmatrix}\bigr).
\end{equation*}
Note that to be a rotation matrix, we must have $A^T=A^{-1}$ and $\det A=1$ which you can check holds by an elementary computation. The locations of all the elements in the $y-$axis rotation matrix are placed so that we have a rotation around the $y-$axis. For example, suppose we are in $\mathbb{R}^3$ and we want to rotate the vector $(0,0,1)$ (aligned with the $z-$axis) $90^o$s. Then multiplying $A$ evaluated at $\phi=90$ by this unit vector gives $(1,0,0)$ which geometrically is a $90^o$ anticlockwise direction around the $y-$axis.
A: I’m not going to be even remotely rigorous, but to get a better geometric mnemonic of the sign-flip’s purpose:
Consider that the ordering of the x,y,z axes is cyclical: cyclical ordering implies that only cyclical permutation has certain kinds of meaning. The expression that gives us the number of cyclical permutations is (n-1)!. In other words, there are precisely 2 self-cyclically-isomorphic orderings of ℝ^3 base vectors, and they happen to be the reverse of each-other. Transitioning between one cyclical permutation and the other is geometrically isomorphic to a chirality-switching sign-flip in an ℝ^3 set of basis vectors.
In other words, there are precisely 2 cyclical permutations of a matrix’s columns (or rows). Each of the two cyclical permutations contains all three linear permutations that have the same sign/chirality (visualize cycling from one to another by exchanging every base cyclically: no such transform requires a mirror-flip).
The (thoroughly non-rigorous, incorrect-if-generalized) mnemonic arises when you consider that the y-axis rotation matrix “skips over” the center row & column. Consider this non-adjacent-elements 3-choose-2 operation as a kind of “sign-flipped” version of the two other adjacent-elements 3-choose-2 operations, belonging to the other, sign-flipped cyclical permutation. Alternatively, consider the “cycling back around” 3-choose-2 operation to be sign-flipping.
The whole “precisely two cyclical permutations, precisely two chiralities” analogy breaks down in higher dimensions, but every (linear?) permutation still has an associated chirality sign (probably applicable in a correct generalization, and equivalent to the sign-obtaining function used in finding a determinant). Also, higher dimensions have entire separate classes of rotation, and I don’t know a lot about it.
