A proof that an orthogonal matrix with a determinant 1 is a rotation matrix Reading proof(starting on page 5) for item 1 of "Rotation Matrix Theorem" in this doc i'm stuck at understanding its last step. Matrix A being an orthogonal Matrix, at this step the conclusion that A is a rotation matrix is reached based on these facts:


*

*$u_1$ is a unit vector such that $A u_1 = u_1$

*$u_2$ is a unit vector perpendicular to $u_1$

*$u_3 = u_1 \times u_2$

*$A u_2 = \cos(\theta)u_2 + \sin(\theta)u_3$

*$A u_3 = -\sin(\theta)u_2 + \cos(\theta)u_3$


I can't realize how that conclusion is reached, although I clearly understand a matrix of the form:
\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos\theta & -\sin\theta \\
0 & \sin\theta & \cos\theta \\
\end{bmatrix}
is a rotation matrix and applying A to the unit vectors has the effect of rotating them about the axis through $u_1$ by the angle $\theta$.
 A: I understand your statement now. I will summarize my answers here.
As I mentioned in my comments, $A$ is a rotation matrix since it can be written in the "standard" form under the new basis. In other words, it rotate any vector about the axis that is in the direction of $u_1$. 
Notice that this last statement "it rotate any vector about the axis that is in the direction of $u_1$" is independent of the basis. 
Your statement "to prove that there's a rotation such that for an orthonormal basis, applying the rotation to vectors of that basis gives A1, A2 and A3" is also correct. Because it also says the columns of $A$ is an orthonormal basis, which is true. But I think this is kind of a confusing way to prove a matrix is a rotation. Also you'll have to construct another rotation to prove it. And in what form would you construct another rotation? 
A more straightforward way, as above, is to prove that the matrix $A$ rotates vectors.
Edit:
This is with regard to the matrix $A$ represented in terms of basis $u_1, u_2, u_3$. We can write a linear transformation $T$ as a matrix in terms of any basis using the following way. Given basis $e_1, e_2, e_3$, the matrix in terms of this basis is calculated by
$$[Te_1\quad Te_2 \quad Te_3]$$
where $Te_i$ is the column coordinate vector after you apply $T$ to $e_i$ in terms of this basis. 
Looking at the question in this way, we see that $Au_1=\begin{pmatrix}1\\0\\0\end{pmatrix}$, $Au_2=\begin{pmatrix}0\\ \cos\theta\\ \sin\theta\end{pmatrix}$, $Au_3=\begin{pmatrix}0\\ -\sin\theta\\ \cos\theta\end{pmatrix}$.
Then
$$\begin{bmatrix}
1 & 0 & 0 \\
0 & \cos\theta & -\sin\theta \\
0 & \sin\theta & \cos\theta \\
\end{bmatrix}$$
is exactly the matrix $A$ in terms of the basis $u_1, u_2, u_3$.
