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I am a math undergraduate student taking a course called "Geometry and symmetry" and I have something I don't understand with the definition the lecture gave in class.

Definition: $T\,:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is called a linear rotation operator if there exist an orthonormal basis $B$ s.t $$\begin{pmatrix}cos(\theta) & -sin(\theta)\\ sin(\theta) & cos(\theta)\\ & & 1\\ & & & 1\\ & & & & 1\\ & & & & & & .\\ & & & & & & & .\\ & & & & & & & & .\\ & & & & & & & & 1\\ & & & & & & & & & 1 \end{pmatrix}$$.

Definition: $T\,:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is called a linear reflection operator if there exist an orthonormal basis $B$ s.t $$\begin{pmatrix}1\\ & 1\\ & & .\\ & & & .\\ & & & & .\\ & & & & & 1\\ & & & & & & -1 \end{pmatrix}$$.

The definitions for matrix $P$ are that a matrix is called linear reflection matrix if there is an orthogonal matrix $P$ s.t $P^{-1}AP$ is in one of the above forms, accordingly.

Later on, there is a theorem that says that is $T\,:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is orthogonal then there exist an orthonormal basis $B$ s.t $$[T]_{B}=\begin{pmatrix}R_{\theta_{1}}\\ & R_{\theta_{2}}\\ & & .\\ & & & .\\ & & & & .\\ & & & & & R_{\theta_{k}}\\ & & & & & & 1\\ & & & & & & & 1\\ & & & & & & & & .\\ & & & & & & & & & 1\\ & & & & & & & & & & -1\\ & & & & & & & & & & & .\\ & & & & & & & & & & & & .\\ & & & & & & & & & & & & & -1 \end{pmatrix};R_{\theta}:=\begin{pmatrix}cos(\theta) & -sin(\theta)\\ sin(\theta) & cos(\theta) \end{pmatrix}$$

My question is: if $T$ is orthogonal then if $A$ is a matrix s.t $Tv=Av$ then $A$ is orthogonal hence $|A|=1$ or $|A|=-1$. in the first case I learned that by $A$ is a linear rotation matrix, in the second case I learned that $A$ is linear reflection operator, so I deduce that there is an orthonormal basis in which $A$ have the form of the two matrix above - i.e rotating around some $2$ dimensional space fixing the other $n-2$ subspace of $\mathbb{R}^{n}$ or rotating around some vector fixing the other $n-1$ subspace of $\mathbb{R}^{n}$.

It seems that the matrix in the theorem is very different than the first two matrix in the question since it has $k$ (that can be $>1$) rotations, and it can reflects around more then $1$ vector...

Can someone please explain whats going on here? I expected the theorem to be the $[T]_{B}$ is one of the first two matrices in the question, instead I have in front of me a strange looking matrix that I can not classify as a rotation or a reflection matrix, so where in my reasoning I am wrong (I determined that $[T]_{B}$ is the first matrix in the question if $A$ above have det $1$ and the second if $A$ have det -1) ?

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You can use \ddots to get three dots in a diagonal, like so: $\ddots$ You should also use \cos and \sin to write $\cos\theta$ instead of $cos\theta$. –  Rahul Jul 13 '12 at 23:25
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I think your confusion stems from the mistaken belief that every orthogonal matrix with positive determinant is what you call a "linear rotation operator". That happens to be true in three dimensions, but not more. However, every orthogonal matrix with positive determinant is a composition of such linear rotation operators, also known as simple rotations. So you can think of the latter as the set of "atomic" rotations that all other rotations are composed of.

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