# Eigenvalue of 3D rotation matrix

## Main Question or Discussion Point

It was pretty cool to stumble upon Euler's formula as the eigenvalues of the rotation matrix.

det(Rot - kI) = (cos t - k)2 + sin2t
=k2-2(cos t)k + cos2t + sin2t
=k2-2(cos t)k + 1

k = {2cos t +/- $$\sqrt{4cos^2(t) - 4}$$}/2
k = cos t +/- $$\sqrt{cos^2(t) - 1}$$
k = cos t +/- $$\sqrt{cos^2(t) - cos^2t - sin^2(t)}$$
k = cos t +/- $$\sqrt{-sin^2(t)}$$
k = cos t +/- i sin t = e(+/-)it

I was wondering what the eigenvalues are for the rotation matrix in 3D, and if there's a 3D equivalent to Euler's formula.

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CompuChip
Homework Helper
Actually, I suppose it is not that strange when you look at a rotation as complex multiplication.
If you write a two-dimensional vector v = (x, y) as z = x + iy, then rotation over an angle t can be written either as R(t) v, where R is the 2d rotation matrix. But you can also write it as eit z.

A 3D rotation matrix, in the appropriate basis, looks like
$$R_3(t) = \begin{pmatrix} R_2(t) & 0 \\ 0 & 1 \end{pmatrix}$$
where R2(t) is the 2-dimensional rotation matrix and the z-axis is the rotation axis (i.e. rotation in the (x,y) plane). So, by expansion along the last row or column,
det(R3 - k I) = (1 - k) det(R2 - k I)
which you can work out in terms of your previous result. (And then add that an arbitrary change of basis does not alter the eigenvalues).

In addition to the two imaginary eigenvalues, the most intuitive one (for me) is the real valued one -- it's 1, with eigenvector along the axis of rotation. You can see that one just by noting that rotating won't change a vector along the axis of rotation.

edit:
if there's a 3D equivalent to Euler's formula.
Quaternions could be considered an extension of Euler's formula ...

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