# Every 3D rotation must have an axis?

For a 1D object 1D space we can translate $x\mapsto x+a$ but cannot define a rotation as $x\mapsto ax$ would not leave the distance invariant between two points and hence the onyl rigid tranformation is the translation. In 2D and 3D we can choose an axis and define a rotation using the rotation matrix. However, it seems that algebraically a rigid motion cannot be defined (as the sin and cos occur in pairs) unless we have one coordinate which remains unchanged $z=z'$ in order for $x^2+y^2+z^2 = x'^2+y'^2+z'^2$. I cannot prove this.

Prove that there exists or does not exist a rotation such that $x\neq x'$ , $y\neq y'$ and $z\neq z'$. Or, I think equivalently, prove or disprove that any rotation in 2D or 3D must have an axis.

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What's an "axis" in 2d? Just a point? Usually a rotation is defined as a distance-preserving map that fixes a point (and fixes orientation), so that would seem to immediately give an answer for 2d. –  MartianInvader Jul 19 '11 at 19:28
@MartianInvader a rotation in 2D has an axis in 3D space. A rotation in 3D has an axis in 3D space. Is it possible to have a rotation which does not have an axis in 3D space, but maybe in 4D? –  kuch nahi Jul 19 '11 at 19:33
@KuchNahi I think I see what you mean. In odd dimensions, rotations must always have a fixed direction, whereas in even dimensions this is not true (with the 2-dimensional case being an obvious example). –  Vhailor Jul 19 '11 at 19:36
The question is rather vague, and the title borders the absurd: a rotation 'is' about a point. –  leonbloy Jul 19 '11 at 19:40
@leonbloy could you suggest a better title. –  kuch nahi Jul 19 '11 at 19:50

It is true that any three-dimensional rotation must have a fixed line -- this is sometimes called Euler's Principal Axis Theorem.

[Note that my answer presumes that we are talking about linear maps. It is a classical theorem -- which I believe has been discussed on this site before -- that an isometry of $\mathbb{R}^n$ is linear iff it fixes the origin. If you have an isometry with any fixed points, then -- by translating your coordinate system -- you may assume that the isometry fixes the origin.]

I will show that this holds for any matrix $A \in \operatorname{SO}_n(\mathbb{R})$ when $n$ is odd: that is, $A$ is an orthogonal matrix of determinant $1$.

Step 1: In particular, $A$ viewed as a matrix over $\mathbb{C}$ is unitary, i.e., it preserves the standard Hermitian inner product $\langle, \rangle$ on $\mathbb{C}^n$. In particular we have for all $v \in \mathbb{C}^n$, $|Av|^2 = \langle Av, Av \rangle = \langle v, v \rangle = |v|^2$, so $|Av| = |v|$. In particular if $v$ is a nonzero eigenvector for $A$ -- so that $A v = \lambda$ -- we get $|v| = |\lambda v| = |\lambda| |v|$ and thus $|\lambda| = 1$. That is, every eigenvalue of $\lambda$ has complex absolute value $1$.

Step 2: The eigenvalues of $A$ are the roots of a polynomial with real coefficients -- the characteristic polynomial -- hence the complex roots come in conjugate pairs $\lambda, \overline{\lambda}$ and thus $\lambda \cdot \overline{\lambda} = |\lambda|^2 = 1$. Therefore the product of all of the complex eigenvalues is equal to $1$. Since the determinant of $A$ is $1$, the product of all the eigenvalues is equal to $1$, and since $n$ is odd and the number of complex eigenvalues is even, the number of real eigenvalues is odd and in particular positive. Moreover each real eigenvalue is either $\pm 1$ and the product of an odd number of them is equal to $1$, so $1$ occurs as an eigenvalue of $A$. That is, the $1$-eigenspace of $A$ is nonzero, so there is (at least) a one-dimensional subspace that $A$ leaves pointwise fixed.

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This is surprisingly clear answer for a question I could not express clearly. Thank you. –  kuch nahi Jul 19 '11 at 20:01
This effectively shows that any rotation in odd dimensions have an axis - and if we think the rotation as applied to the points of a (hyper)sphere surface, it shows that any rotation in 3 (5,7...) dimensions (or any composition of rotations, that is also a rotation) has at least one fixed point. Which it's obviously not true in 2D. –  leonbloy Jul 19 '11 at 20:08
@Pete: Yes, we totally agree, I was not objecting anything. Just pointing out the equivalence (perhaps too obvious, but as the OP was not very clear...) of saying that a rotation "has an axis" (thinking in the full 3D space, for a 3D rotation) or that it has "a fixed point" (thinking in the sphere surface). –  leonbloy Jul 19 '11 at 20:23

In 3 dimensions, the composition of rotations is always a rotation about some axis. This is due to a theorem of Euler.

One of your statements is not true, at least not in the way you formulated it. You can easily construct a rotation in 3D that doesn't fix any of the three coordinate axes x,y and z, simply compose a rotation about one of the axes with another one about another axis and you will get another rotation (by Euler's Theorem that I referenced above) which is not about a coordinate axis most of the time.

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A simple explicit example is the map $(x,y,z) \mapsto (z,x,y)$, which is a rotation by $2\pi/3$ about the axis $(1,1,1)$ and does not fix any of the three coordinate axes. –  Rahul Jul 19 '11 at 19:41
@Rahul But it does fix one axis (just not the original ones). The question was if it must always fix an axis. –  kuch nahi Jul 19 '11 at 20:16