build a matrix with 2 bases?! im stuck on this Task :/
W:= span((1,1,0,0),(1,1,1,1) subspace of $ \mathbb{R^4} $  and
$ \pi : \mathbb{R^4} \to \mathbb{R^4} /W$
$v \to [v]:= v+W $
$ B_1:= ([e_1],[e_3])$ is a Base of $ \mathbb{R^4} /W $ and $B=(e_1,e_2,e_3,e_4)$
is the Standard base.
How does the Matrix $\pi $ Looks like?
i know how to build a Matrix of two Bases..but because of $B_1$ im not sure how to do that...hopefully someone can help out :)

You have to write $\pi(e_1)$, $\pi(e_2)$, $\pi(e_3)$ and $\pi(e_4)$ as a linear combination of $[e_1]$ and $[e_3]$.
The easy part is
$$
\pi(e_1)=[e_1]=1[e_1]+0[e_3]
\qquad
\pi(e_3)=[e_3]=0[e_1]+1[e_3]
$$
Note also that, by definition,
$$
\pi(e_1+e_2)=\pi((1,1,0,0))=[0]
$$
so $\pi(e_2)=-\pi(e_1)=-1[e_1]+0[e_3]$. Finally,
$$
\pi(e_1+e_2+e_3+e_4)=[0]
$$
so $\pi(e_4)=-\pi(e_3)$.
Now grab the coefficients of the linear combinations and write them as columns:
$$
\begin{bmatrix}
1 & -1 & 0 & 0 \\
0 & 0 & 1 & -1
\end{bmatrix}
$$
