For some experimental and practical reason, I have created a new coordinate system in the form


where $T_{ij}$ isn't a square matrix. $x_i$ is standard Cartesian coordinates, and $x^\prime_j$ is a point in the new system. I have to mention that the new system's axes are not linearly independent. So the last relation can be written as

$$\left(\matrix{x_0^\prime\\x_1^\prime\\x_2^\prime\\x_3^\prime}\right)=\left( \matrix{T_{11} & T_{12} & T_{13} \\ T_{21} & T_{22} & T_{23} \\ T_{31} & T_{32} & T_{33} \\ T_{41} & T_{42} & T_{43} } \right)\cdot \left(\matrix{x\\y\\z}\right)$$

The matrix $T_{ij}$ is well defined.

What I need is a rotation operator that will transform a point in the primed system, as the standard rotation operator does. So say I have the standard rotation matrix in Cartesian coordinates around the z-axis:

$$R_{ij}= \left( \matrix{\cos{\theta}&-\sin{\theta}&0\\ \sin{\theta}&\cos{\theta}&0\\0&0&1} \right)$$

So to rotate a point in Cartesian coordinates, we use the standard operator formula:


where $P_j$ is the point before rotation, and $P^\prime_i$ is the point after rotation.

How can I write this rotation formula for a point in the new coordinates system that uses 4 points? How will the rotation matrix look like? I expect a rotation matrix that is $4\times4$, but I don't know how to derive it. Please help in that.

  • $\begingroup$ My feeling it that your problem is underconstrained, because in the ' system you have a degree of freedom more. $\endgroup$
    – Bernhard
    Apr 18, 2014 at 17:06
  • $\begingroup$ By the way, your notation is confusing with the primes, that are use to indicate rotation and 4d-system. $\endgroup$
    – Bernhard
    Apr 18, 2014 at 17:07
  • $\begingroup$ @Bernhard I'm not certain, but I suspect that there isn't a new degree of freedom on the primed system because the axes aren't linearly independent. Given three of the coordinates, the last would be uniquely defined. $\endgroup$ Apr 18, 2014 at 17:18
  • $\begingroup$ @Bernhard Never mind, I'm wrong. $\endgroup$ Apr 18, 2014 at 17:23
  • 1
    $\begingroup$ @Draksis I've written the proof in my answer (not sure if it is proof from a mathematicians point of view, but for physicist it is hopefully enough ;)) $\endgroup$
    – Bernhard
    Apr 18, 2014 at 17:25

1 Answer 1


To be consistent with notation, I use the $x'$ for the transformation to the new system and $\tilde{x}$ for the rotation. Thus, as you defined

$$x'_i=T_{ij}x_j,$$ $$\tilde{x}_i=R_{ij}x_j.$$

We know that


You are looking for the transformation matrix $Q_{ij}$, such that




Naively, one could now write from (1) and (2)

$$T R = Q T ,$$ $$Q=T R T^{-1}.$$

However, $T$ is not a square matrix, and does not invertable. In other words, such a matrix $Q$ can not be determined uniquely.

Or, looking at it as $$T_{ij}R_{jk} = Q_{pq}T_{qr}$$

You know that these are twelve equations, because both procut matrices are $4\times3$. $Q_{ij}$ is a $4\times4$ matrix, with, thus, $16$ unknowns. In other words, there are infinitely many possibilities. Unless, of course, you add constraints.


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