I'm trying to check my calculations for a problem I'm working on.
Let $\alpha=(0,i-1,i+1)$ be a vector $W$.
Let $\mathcal{A}=\{\alpha_1,\alpha_2\}=\{(1,0,i),(1+i,1,-1)\}$ be a basis for a vector space $W$.
Let $\mathcal{B}=\{\beta_1,\beta_2\}=\{(1,1,0),(1,i,1+i)\}$ be another basis for a vector space $W$.
(a) Let $[\alpha]_\mathcal{A}$ and $[\alpha]_\mathcal{B}$ be the coordinates of $\alpha$ in $\mathcal{A}$, $\mathcal{B}$ respectively.
This I calculated as follows:
$\begin{pmatrix}0\\i-1\\i+1\end{pmatrix}=c_1\begin{pmatrix}1\\0\\i\end{pmatrix}+c_2\begin{pmatrix}1+i\\1\\-1\end{pmatrix}$
Which yields:
$[\alpha]_\mathcal{A}=\begin{pmatrix}2\\i-1\end{pmatrix}$
$\begin{pmatrix}0\\i-1\\i+1\end{pmatrix}=c_1\begin{pmatrix}1\\1\\0\end{pmatrix}+c_2\begin{pmatrix}1\\i\\1+i\end{pmatrix}$
Which yields: $[\alpha]_\mathcal{B}=\begin{pmatrix}1\\-1\end{pmatrix}$
(b) Find a $2\times 2$ matrix such that $U[\alpha]_\mathcal{A}=[\alpha]_\mathcal{B}$
I'm assuming that this wants us to consider the following:
$\begin{pmatrix}1\\1\\0\end{pmatrix}=c_1\begin{pmatrix}1\\0\\i\end{pmatrix}+c_2\begin{pmatrix}1+i\\1\\-1\end{pmatrix}$
and
$\begin{pmatrix}1\\i\\1+i\end{pmatrix}=c_1\begin{pmatrix}1\\0\\i\end{pmatrix}+c_2\begin{pmatrix}1+i\\1\\-1\end{pmatrix}$
This yields transition matrix:
$U=\begin{pmatrix}-i & 2-i\\1 &i\end{pmatrix}.$
However I tested the above matrix and yielded $U[\alpha]_\mathcal{A}\neq[\alpha]_\mathcal{B}$
I worked this backwards. Since I want to move from $\mathcal{A}$ to $\mathcal{B}$ I need to be computing a matrix $U$ as follows $\alpha_1=c_1\beta_1+c_2\beta_2$ and $\alpha_2=d_1\beta_1+d_2\beta_2$. Then my $U=\begin{bmatrix}c_1 & d_1\\c_2 & d_2\end{bmatrix}$