# Doubt in the proof of Euler's rotation theorem

The question arises the way Goldstein proves Euler theorem (3rd Ed pg 156 ) which says:

" In three-dimensional space, any displacement of a rigid body such that a point on the rigid body remains fixed, is equivalent to a single rotation about some axis that runs through the fixed point".

The Matrix proof essentially takes an arbitrary $$3 \times 3$$ orthogonal matrix with real entries and shows that there is at least one vector $$n\neq 0$$ with $$A n=n$$ that is an eigenvector with +1 as its eigenvalue .

The author states that this proves the Eulers theorem, which I am not sure why this is true. It seems that all we have shown is that are some vectors that are invariant under $$A$$ along some line. That doesn't necessarily mean $$A$$ is a rotation along that line.

• Hint! => the rotation axis is invariant to the rotation. Commented Nov 7, 2021 at 19:24
• Yes agreed, rotation axis is invariant to rotation. But if we have a transformation $A$ that keeps a line of vectors invariant, does it necessarily mean $A$ is a rotation? That is the essence of my question. Can you please add something more. Thank you :) Commented Nov 8, 2021 at 3:47
• @Kashmiri That's clearly not true for a generic transformation. For example, $A$ could be a projection onto that line. It is true for orthogonal linear transformations with determinant 1, since all orthogonal linear transformations with determinant 1 are rotations. Commented Nov 8, 2021 at 3:53

Let $$A$$ be "a displacement of a rigid body such that a point on the rigid body remains fixed", and let $$B$$ be "a single rotation about some axis." Let's assume we are in three dimensions.

To prove $$A$$ and $$B$$ are equivalent, we have to show that $$A$$ implies $$B$$ and $$B$$ implies $$A$$.

The matrix proof you are describing shows that $$B$$ implies $$A$$. In other words, a general rotation (described by an orthogonal matrix) has an axis on which points are invariant under the rotation.

What's left is to show that $$A$$ implies $$B$$. At a physicist level of rigor, I think it's fairly obvious that the only way to move a rigid object with one point fixed, is to rotate the object around an axis passing through that point, so at a physicist level of rigor I would consider that proven. Perhaps Goldstein has a more rigorous justification. One way to justify it would be to formalize the idea that a rigid body has 6 degrees of freedom (3 to give the position of one point, and 3 to define its orientation in space). Fixing a point removes three of the degrees of freedom. You can then use 2 of the remaining degrees of freedom to specify an axis, and 1 to specify an angle of rotation about that axis.

• I couldn't understand How did 'The matrix proof show that B implies A" could you please elaborate if possible. Thank you Commented Nov 8, 2021 at 4:38
• @Kashmiri The proof you cite "takes an arbitrary 3×3 orthogonal matrix with real entries and shows that there is at least one vector 𝑛≠0 with 𝐴𝑛=𝑛 that is an eigenvector with +1 as its eigenvalue." If I replace "3x3 orthogonal matrix with real entries" with "rotation matrix" (since these are the same thing), and replace "At least one vector $n \neq 0$ with $An = n$" with "a vector that is invariant under the rotation", then your proof shows that "a rotation matrix leaves one vector invariant under the rotation." Statement $B$, 'rotation about some axis', corresponds to acting (...) Commented Nov 8, 2021 at 5:15
• (...) with the rotation matrix. Statement $A$, 'a point on the rigid body is fixed', corresponds to the fixed point on the rigid body lying on a vector that remains invariant. Then your theorem is pretty much in the form $B$ implies $A$ -- acting with a rotation matrix is leaving some axis fixed (and the fixed point of the rigid body lies on this axis). Does that make sense? Or can you pinpoint a place where the explanation is confusing? Commented Nov 8, 2021 at 5:17
• Thank you it's getting clearer. Could you say how you said "an orthogonal matrix with real entries" is same as "rotation matrix" (since these are the same thing)? Is it because you already know that the determinant of the matrix is +1 and it's orthogonal so it's a rotation matrix? Commented Nov 8, 2021 at 10:25
• @Kashmiri I should have said "an orthogonal matrix with real entries and determinant one" is the same as a "rotation matrix." Commented Nov 8, 2021 at 13:10