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Let $P=(x_1,x_2,...,x_n)$ and you want to rotate around the $x_1$ axis by $\theta$, what are the new $P'= (x_1',x_2',...,x_n')$ coordinates?
In 2D and 3D it's easy, because we can write up the matrix for rotating and just take the matrix product.

Matthew
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user128576
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    Shouldn' you rotate around an $n-2$ dimensional hyperplane rather than an axis? – David Jul 29 '19 at 10:17
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    Given's rotation is what you're looking for – Ahmad Bazzi Jul 29 '19 at 10:50
  • yes, Givens rotation looks like what I need, David's matrix in his answer is correct if you substitute $n-1,n$ into $G(i,j,theta)$ (except for the minus sign is swapped, but it gives the same result I think). – user128576 Jul 29 '19 at 11:10
  • You can only rotate about an axis in an odd-dimensional space. – amd Jul 29 '19 at 23:23

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$\begin{bmatrix} 1 & 0 & 0 &... & 0 & 0 & 0 \\ 0 & 1 & 0 &... & 0 & 0 & 0 \\ 0 & 0 & 1 &... & 0 & 0 & 0 \\ &&& ... \\ 0 & 0 & 0 &... & 1 & 0 & 0 \\ 0 & 0 & 0 &... & 0 & \cos{\theta} & \sin{\theta} \\ 0 & 0 & 0 &... & 0 & -\sin{\theta} & \cos{\theta} \\ \end{bmatrix}$

is the associated matrix for the linear transformation that rotates the $x_{n-1}, x_n$ plane an angle $\theta$ counterclockwise. Rotation of other planes would have the same Jordan canonical form, although the matrix may appear different

David
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  • But this is rotation about the origin. OP is asking about an axis which is basically rotation around a line. – Your IDE Jul 29 '19 at 10:28
  • Rotations occur in planes! – David Jul 29 '19 at 10:28
  • If I reduce your matrix for 2d plane by considering the lower right 4 elements, then it'd be rotation about the origin. If it is XY plane, rotating about X-axis would lead to the point being shifted into XYZ plane, but the matrrix above wouldn't compute for the z-coordinate. – Your IDE Jul 29 '19 at 10:32
  • @YourIDE I think I haven't understood you. Why doesn't this compute one of the coordinates? You canno make rotations around an axis in 2D space – David Jul 29 '19 at 10:36