# How to create a basis transformation matrix between “heterogenous” (…) bases? [closed]

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 $B_2 \rightarrow B_1$?

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

$B_2$ 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.