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Given is $\mathbb{R}^2$ with standard basis $B^2_0$

A "new" basis $B=\left\{\vec{b_1}; \vec{b_2}\right\}$ arises from $B^2_0$ by rotation of $30°$ (clockwise). Determine the change of basis $T^{B}_{B^2_0}$ and the basis vectors $\vec{b_1}$ and $\vec{b_2}$.

So we are in $\mathbb{R}^2$ and we have a standard basis $B^2_0$. Standard matrix is another word for unit matrix I think, so we have $B^2_0 =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$.

Rotate this by $30°$, we have $B= \begin{pmatrix} \cos(30°) & -\sin(30°)\\ \sin(30°) & \cos(30°) \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix}$

So $\vec{b_1} = \begin{pmatrix} \frac{\sqrt{3}}{2}\\ \frac{1}{2} \end{pmatrix}$ and $\vec{b_2} = \begin{pmatrix} -\frac{1}{2}\\ \frac{\sqrt{3}}{2} \end{pmatrix}$

And $T^{B}_{B^2_0}$ we can solve with Gaussian elimination, right? But I'm more interested to know if I calculated the rotation matrix and the resulting bases correctly?

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    $\begingroup$ If you have posted the whole question, $g$ has nothing to do here. $\endgroup$ – Bernard May 2 '18 at 8:29
  • $\begingroup$ @Bernard Thank you, I forgot another part of the example was on the next page (back). Just editted my question :) $\endgroup$ – roblind May 2 '18 at 8:33
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    $\begingroup$ More in general with respect to a given basis of vectors $\vec v_i$ it suffice to determine/know the transformed vectors $\vec v_i \to \vec w_i$, then the matrix $[w_1\,w_2\,...\,w_n]$ represents the transformation in the basis of vectors $\vec v_i$. $\endgroup$ – user May 2 '18 at 8:49
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You seem to misunderstand what is the change of basis matrix from one basis to another: its columns are just the coordinates of the vectors in the new basis, expressed in the initial basis. So it is what you've denoted $B$.

It has the property that it lets you express the initial coordinates of a vector in function of its new coordinates.

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To calculate the rotation matrix, with reference to the standard basis, it suffices to consider what are the transformed vectors for $\vec e_1$ and $\vec e_2$ after the rotation, that is

$$\vec{e_1} = \begin{pmatrix} 1\\ 0 \end{pmatrix}\to \vec{b_1} = \begin{pmatrix} \frac{\sqrt{3}}{2}\\ \frac{1}{2} \end{pmatrix} $$

$$\vec{e_2} = \begin{pmatrix} 0\\ 1 \end{pmatrix}\to \vec{b_2} = \begin{pmatrix} -\frac{1}{2}\\ \frac{\sqrt{3}}{2} \end{pmatrix}$$

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