How to find the transformation matrix for rotation? 
I have a basis B in $R^n$ made of orthogonal vectors with the norm 1 (ON-Basis). Lets call them $v_1$, $v_2$ $v_3$. Find the transformation matrix for when $v_1$ rotates counterclockwise 45 degrees.
$v_1= \frac1{\sqrt2}\begin{bmatrix} 1\\ 1\\ 0 \end {bmatrix}, v_2 = \frac1{\sqrt2} \begin{bmatrix} -1\\ 1\\ 0 \end {bmatrix}, v_3 = \begin{bmatrix} 0\\ 0\\ 1 \end {bmatrix} $
If it was in the standard basis it would be totally different and I could do it, here I am struggling. The reason being I don't know how to place the order of vectors, is it $B=[v_1, v_2, v_3]$ or $B= [v_2, v_1, v_3]$. Maybe I should mention that I found these basis myself from an earlier question so I numbered them 1, 2, 3 and when I checked the answers they had the vectors in different places and thus got different matrix, or should I say a matrix where the column vectors don't have the same orders as mine? Will this cause an issue? Also the third part of the quesion is to simple give this matrix in the standard basis by $A_s = PBP^{-1}$
 A: So first of all, your answer to part (a) is correct.
The phrasing of the question seems to imply that $v_1$ should be the first element of your basis, which still allows for some degree of confusion here. However, the order of the basis will not affect the final answer that we get for part (c). To ensure that we end up with the correct final answer, it is important that your matrix $[R]_{\mathcal B}$ (of the rotation relative to your basis) is consistent with the order that you have chosen.
Let's start with the order $v_1,v_2,v_3$. In order to find the matrix $R$, we note that we're looking for a rotation about the first vector $v_1$. Conveniently, you have chosen a right-handed basis orthonormal basis $\mathcal B$, which is to say that we have $v_3 = v_1 \times v_2$ rather than $v_3 = -(v_1 \times v_2)$, where $\times$ denotes a cross-product. Equivalently, you have chosen a basis such that the matrix $B = [v_1 \ \ v_2 \ \ v_3]$ has determinant $\det(B) = 1$ instead of $\det(B) = -1$. So, we may treat $v_1,v_2,v_3$ as the $x,y,z$ axis (without flipping the direction of rotation).
The rotation about the first axis (the $x$-axis in the standard coordinates) by angle $\theta = 45^\circ$ has the matrix
$$
M = \pmatrix{
1&0&0\\
0&\cos \theta & - \sin \theta \\
0&\sin \theta & \cos \theta
} = 
\pmatrix{
1&0&0\\
0 & 1/\sqrt{2} & - 1/\sqrt{2} \\
0 & 1/\sqrt{2} & 1/\sqrt{2}
},
$$
and this matrix $M$ is our answer to part (b). That is, $[R]_{\mathcal B} = M$.
In order to convert this matrix into the matrix of the transformation relative to the standard coordinates, we need to apply a change of basis. Let $\mathcal A = \{e_1,e_2,e_3\}$ denote the standard basis, and let $B$ denote the matrix constructed from $\mathcal B$ (as defined above). We have
\begin{align}
[R]_{\mathcal A} &= 
[I]_{\mathcal B \to \mathcal A}[R]_{\mathcal B} [I]_{\mathcal A \to \mathcal B} 
\\ & = B [R]_{\mathcal B} B^{-1} = B [R]_{\mathcal B} B^\top 
\\ & = \pmatrix{
1/\sqrt{2} & -1/\sqrt{2} & 0\\
1/\sqrt{2} & 1/\sqrt{2} & 0\\
0 & 0 & 1}
\pmatrix{
1&0&0\\
0 & 1/\sqrt{2} & - 1/\sqrt{2} \\
0 & 1/\sqrt{2} & 1/\sqrt{2}
}
\pmatrix{
1/\sqrt{2} & 1/\sqrt{2} & 0\\
-1/\sqrt{2} & 1/\sqrt{2} & 0\\
0 & 0 & 1}
\\ & = 
\pmatrix{
\frac{2 + \sqrt{2}}{4} & \frac{2-\sqrt{2}}{4} & \frac{1}{2}\\
\frac{2-\sqrt{2}}{4} & \frac{2+\sqrt{2}}{4} & - \frac{1}{2}\\
- \frac{1}{2} & \frac{1}{2} & \frac{\sqrt{2}}{2}}
=
\frac 14 \pmatrix{
2 + \sqrt{2} & 2 - \sqrt{2} & 2\\
2 - \sqrt{2} & 2 + \sqrt{2} & -2\\
-2 & 2 & 2 \sqrt{2}
}
\end{align}
