How can you prove that a rotation is a linear transformation? (linear algebra) I am a first year college student studying linear algebra. 
I understand that all linear transformations can be represented by a matrix mapping, and more specifically, the matrix mapping can be constructed by taking the column vectors of the images of the standard basis vectors. However, if the transformation is not linear, you would not be able to construct a matrix mapping for it.
But then, my textbook proves that a rotation is a linear transformation by assuming that the transformation is linear, then constructing the corresponding matrix mapping. Then it argues that since the transformation can be represented by a matrix, it is linear. 
To me, this seems like a invalid proof, because the condition for representing a transformation with a matrix is that the transformation should be linear (also it starts off by assuming what you are trying to prove). Can someone tell me what is wrong with my reasoning? 
 A: *

*All linear transformations (from $\mathbb{R}^n$ to $\mathbb{R}^m$) can be represented as a ($m \times n$) matrix (as you described).

*Conversely, any matrix represent a linear transformation. (If $A$ is a $m \times n$ matrix, then the transformation sends a vector $v \in \mathbb{R}^n$ to $Av$.)
So, in your mind you should think of ($m \times n$) matrices and linear transformations (from $\mathbb{R}^n \to \mathbb{R}^m$) as the same thing.
So, the proof in your textbook starts with a rotation and produces a matrix $A$ such that $Av$ is the rotation of $v$. By point #2 above, the rotation must be a linear transformation.
A: By definition, 

Any linear transformation can be represented as a matrix...

Any rotation can be represented as an orthogonal matrix in the general form of 
$$\begin{bmatrix} cos\theta & -sin\theta \\ sin\theta &cos\theta \end{bmatrix}$$
And can be generalized to $\mathbb{R}^n$. You can also check that any matrix $A$ maintains linearity.
https://en.wikipedia.org/wiki/Rotation_matrix
