# Any reference(a book) that defines the $n$-dimensional rotation matrix?

I want to refer to a mathematics book that explains the n-dimensional rotation matrix or rotation transformation.

Wikipedia concentrates most on 2D or 3D. There are things that one can say definition here and there, but I think it is not a good idea to use the definition there. Actually they don't seem to be definitions.

Strang's "Linear Algebra", Barret O'neill's "Elementary Differential Geometry" deal only with 2D or 3D cases. I think physicist are more interested in the general case, due to the theory of relativity. I found one explanation in "Geometrical Methods of Mathematical Physics" by Bernard Schutz. But I think it doesn't define the rotation matrix.

Artin's "Geometric Alegebra" defines the rotation group as an isometry $$\sigma:V\to V$$ such that $$\det\sigma=1$$. But the language there is so abstract that I can't catch any of them.

Can anyone give a reference that defines rotation transformation on $$\mathbb R^n$$ and state as a property that $$A$$ is a rotation matrix if and only if $$A\in SO(n)$$?

This is the end of the question and the below is what I wanted to do. I wanted to prove that if $$A\in SO(n)$$, then $$A$$ is a rotation about a line through the origin in $$\mathbb R^n$$. So I need to define the rotation transformation(or matrix) in $$n$$ dimensional Euclidean space.

The definition Artin gives is probably the most transparent and standard:

1. It is a linear isometry, meaning that it preserves distances and lengths as given by a bilinear form (You may want to restrict yourself to the dot product in $$\mathbb R^n$$).
2. It has determinant $$1$$, meaning that it doesn't change the orientation of the space by reflecting it ($$\mathbb R$$-linear isometries must have determinant $$\pm 1$$, so we are excluding half with this condition.)

I think you should spend time trying to understand this rather than discarding it as "too abstract." It's really quite concrete.

But I think it doesn't define the rotation matrix.

To split hairs for a second, the most important nature of a rotation is that it is a transformation, not just a matrix. A matrix is just a particular way to represent a transformation. That's why the definition above emphasizes the qualities that make it a rotation: it does not mess with distances, and it does not mess with orientation. (Out of infinitely many choices of bases it can have infinitely many different matrix representations.)

I wanted to prove that if $$A\in SO(n)$$, then $$A$$ is a rotation about a line through the origin in $$\mathbb R^n$$.

That isn't even true in $$\mathbb R^2$$, as I'm sure you can see.

It happens to be true for rotations in $$\mathbb R^n$$ for odd $$n$$ due to the fundamental theorem of algebra, because it says each rotation (like in Artin's definition) has an eigenvector. But even then I doubt it's what you wanted: some rotations in $$\mathbb R^n$$ preserve much more than just one line through the origin.

And for even $$n$$'s, you still might not preserve any line: take, for example

$$\begin{bmatrix}0&-1&0&0 \\ 1&0&0&0 \\ 0&0&0&-1\\ 0&0&1&0\end{bmatrix}$$ as an $$\mathbb R$$ linear transformation.

• Thanks. Actually, I postulated that the above statement holds for odd $n\ge3$. In my argument, I used the fact that $A\in SO(n)$ has 1 as an eigenvalue, which followed from $\det(-B)=-\det B$. – shyzealot08 May 26 at 15:13
• @shyzealot08 Yes, that much is true. The eigenspace for $1$ could have many more dimensions than $1$, though, if you've accounted for that. – rschwieb May 26 at 15:15
• (About the original question) If the definition of rotation matrix were an isometry whose determinant(regardless of the basis of $V$) is one, then it seems to be the same as the definition of being an element of $SO(n)$, provided $SO(n)$ is a collection of orthogonal matrices whose determinant is $1$. So the statement I wanted to show becomes ridiculous, since it is "if p, then p". So, now I'm not sure that I can state anything that is meaningful. – shyzealot08 May 26 at 15:20
• @shyzealot08 It's slightly different, in that you have fixed a basis to produce the matrices. – rschwieb May 26 at 15:21
• So, can I define the rotation transformation as follow? "It is a linear isometry. If I fix a basis for $V$ or ($\mathbb R^n$), then the corresponding matrix has determinant 1." – shyzealot08 May 26 at 15:30