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Consider the vector space $\mathbb{R}^2$ with bases

$B_1 = \{ e_1, e_2 \}$, and

$B_2 = \{ e_1+e_2, e_1-e_2 \}$.

Because the vectors in $B_2$ are linear combinations of vectors in $B_1$, the transformation matrix $B_1 \rightarrow B_2$

$\begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & - \frac{1}{2} \end{bmatrix}$

can be created.

Now, consider the vector space $\mathbb{Q}(\sqrt{-1})$ with bases

$B_1=\{1, \sqrt{-1} \}$, and

$B2=\{\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \}$

then, how would I (formally) formulate the transformation matrix $B_1 \rightarrow B_2$ and $B_2 \rightarrow B_1$?

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closed as unclear what you're asking by Namaste, Trevor Gunn, Shailesh, Leucippus, Daniel W. Farlow Jul 18 '17 at 0:46

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$B_2$ is not a subset of $\mathbb Q(i)$, so it is not a basis for the vector space $\mathbb Q(i)$. Now, maybe you are thinking of a certain isomorphism between a vector space of $2 \times 2$ matrices and $\mathbb Q(i)$, and maybe this isomorphism maps $B_2$ to the set $\tilde B_2 = \{1, -i\}$, which is in fact a basis for $\mathbb Q(i)$. It makes sense to ask what is the change of basis matrix from $B_1$ to $\tilde B_2$. And that is a question you probably know how to answer.

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  • $\begingroup$ I understand, so because the matrices are not elements of Q(i) they can't be part of a sec basis of Q(i). - So the "transformation" map need to include the map describing the isomorphism. $\endgroup$ – nilo de roock Jun 28 '17 at 10:30

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