How exactly do you solve this problem?
$\varphi : V \longrightarrow W$, $M \longmapsto M \cdot A$ with $A = \left( \begin{array}{cc} 1&-3 \\ 2 & -2 \\ 3 & -1 \end{array} \right) \in \mathbb{Q}^{3 \times 2}$
We want to calculate the representation Matrix $M_{\mathcal{C},\mathcal{B}}(\varphi)$.
The two Bases are the following:
$\mathcal{B} \! := \! ( \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$, $\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right)$, $\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right)$, $\left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \end{array} \right)$, $\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right)$, $\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) )$ Base of $V$.
$\mathcal{C} := ( \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right)$, $\left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right)$, $\left( \begin{array}{cc} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{array} \right)$, $\left( \begin{array}{cc} 0 & \frac{1}{2} \\ -\frac{1}{2} & 0 \end{array} \right) )$ Base of $W$.
To calculate the representation Matrix $M_{\mathcal{C},\mathcal{B}}(\varphi)$ we first of all have calculate the following:
$\varphi( \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) \left( \begin{array}{cc} 1&-3 \\ 2 & -2 \\ 3 & -1 \end{array} \right) = \left(\begin{array}{cc} 1 & -3 \\ 0 & 0 \end{array}\right) $ $\varphi( \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right)) = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right) \left( \begin{array}{cc} 1&-3 \\ 2 & -2 \\ 3 & -1 \end{array} \right) = \left(\begin{array}{cc} 2 & -2 \\ 0 & 0 \end{array}\right) $ $\varphi( \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right)) = \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right) \left( \begin{array}{cc} 1&-3 \\ 2 & -2 \\ 3 & -1 \end{array} \right) = \left(\begin{array}{cc} 3 & -1 \\ 0 & 0 \end{array}\right) $ $\varphi( \left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \end{array} \right)) = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \end{array} \right) \left( \begin{array}{cc} 1&-3 \\ 2 & -2 \\ 3 & -1 \end{array} \right) = \left(\begin{array}{cc} 0 & 0 \\ 1 & -3 \end{array}\right) $ $\varphi( \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right)) = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right) \left( \begin{array}{cc} 1&-3 \\ 2 & -2 \\ 3 & -1 \end{array} \right) = \left(\begin{array}{cc} 0 & 0 \\ 2 & -2 \end{array}\right) $ $\varphi( \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)) = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{cc} 1&-3 \\ 2 & -2 \\ 3 & -1 \end{array} \right) = \left(\begin{array}{cc} 0 & 0 \\ 3 & -1 \end{array}\right) $
To get the representation Matrix, we have to calculate the linear combinations for C. Now here is the problem. Let's say we want to calculate the linear combination for $ \left(\begin{array}{cc} 1 & -3 \\ 0 & 0 \end{array}\right) = \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)*a + \left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right)*b + \left(\begin{array}{cc} 0 & 1/2 \\ 1/2 & 0 \end{array}\right)*c + \left(\begin{array}{cc} 0 & 1/2 \\ -1/2 & 0 \end{array}\right)*d $
How exactly do I find out, what values $a,b,c,d$ have to be, since those are Entrys in $M_{\mathcal{C},\mathcal{B}}(\varphi)$? I've tried the Gauß Algorithm but I didn't come far. Can anyone calculate $M_{\mathcal{C},\mathcal{B}}(\varphi)$ for me or at least give me some help so I can do it myself?