sufficient condition for matrices to be a certain rotation matrices One has for example given a matrix, let's say: $$ {1\over \sqrt 2 }  \begin{pmatrix}
        1 & 1 \\
        1 &-1 \\      
        \end{pmatrix}$$
and wants to know what happens with a point ( x | y ), if you multiply it with the matrix. 
It is necessary that the distance to 0 stays the same after the multiplication, your result you get after the multiplication also that it is a linear function.
Those 2 condition should tell one, if a matrix is a rotation matrix or not and those conditions seem to be easy to verify for a given matrix. 
But I don't know, how to prove, hence I am not sure, if my thoughts are correct. 
Can someone give me a good a first approach on how to prove that, if it is correct or give me and better (faster) way to check, if a matrix is actually a rotation matrix, which could also be applied in higher dimensions.   
 A: Hint: check if your matrix is actually of the form $$\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} , $$ for some $\theta $. Rotation matrices are orthogonal, that is, $AA^t = {\rm Id}$.
A: a rotation matrix keeps the length of a vector invariant(isometry) and keeps the same orientation. that is if you go from $x$ to $y$ clockwise so should $Ax$ and $Ay.$  reflection is also an isometry but changes the orientation. keeping lengths invariant implies the angles between vectors are also invariant. we will use the two facts that angles and lengths are invariant under isometry. we will also need the facts for two dimensional vector $a = (a_1, b_1)^T, b = (b_1, b_2)^T$ we have the inner(dot) product $a.b = a_1b_1 + a_2b_2 = |a||b| \cos \theta, |a|^2 = a.a,  $
let the orthogonal unit vectors $e_1 = (1,0)^T, (0,1)^T$ 
the first column of $A$ is the vector $Ae_1$ should have length $1.$ so 
$$Ae_1 = (\cos \theta, \sin \theta)^T$$ similarly $$Ae_2 = \cos \phi, \sin \phi)$$
$e_1, e_2$ are orthogonal implies $Ae_1,Ae_2$ is orthogonal so that $Ae-1.Ae_2 = 0$
which gives $0= \cos \theta \cos \phi + \sin \theta \sin \phi = \cos(\theta - \phi)$
therefore $\theta - \phi = \pm \pi/2$ taking $\phi = \phi + \pi/2$ give you a rotation matrix $$\pmatrix{\cos \theta & -\sin \theta\\\sin \theta & \cos\theta}$$
taking $\phi = \phi - \pi/2$ give you a reflection matrix $$\pmatrix{\cos \theta & \sin \theta\\\sin \theta & -\cos\theta}$$
A: Your matrix is not a rotation matrix.
Its a reflection.
Use the following link. 
http://en.wikipedia.org/wiki/Rotation_matrix
As for rotation in higher dimensions, first you need to define what exactly do you mean.
This was suppose to be a comment.
A: A rotation matrix is an orthognal matrix with determinant +1. And the converse of this is also true, i.e any orthognal matrix with determinant +1 is a rotation matrix. So you can check for that
