Let $V$ and $W$ be both K-linear vector spaces. Now $Y$ is a linear subspace of $V$ and $Z$ is a linear subspace of $W$. Let $\alpha: V \rightarrow W$ be a $K$-linear function with $\alpha(Y)\subseteq Z$. Let $(v_1,...,v_n)$ be a Basis of $V$ that contains a Basis $(v_1,...,v_k)$ of $Y$ and let $(w_1,...,w_m)$ be a Basis of $W$ that contains a Basis $(w_1,...,w_l)$ of $Z$.
Now the task is to show that the Matrix of $\alpha$, regarding the bases $(v_1,...,v_n)$ and $(w_1,...,w_m)$, has the form: $\pmatrix{A&C\\ 0&B}$. I have no idea how to approach this. I guess the Matrix has to look like this:
\begin{pmatrix} a_{11} & \dots &a_{1k} & c_{11} & \dots &c_{1n} \\ \vdots & \ddots &\vdots & \vdots & \ddots &\vdots \\ a_{m1} & \dots &a_{mk} & c_{11} & \dots &c_{1n} \\ 0 & \dots &0& b_{11} & \dots &b_{1n} \\ \vdots & \ddots &\vdots & \vdots & \ddots &\vdots \\ 0& \dots & 0 & b_{11} & \dots &b_{1n} \end{pmatrix}
But why and how to proof it? I know that $A$ is the Matrix of $Y$ and $B$ is the Matrix of $V/Y$.
First recall that if $\beta : V \to W$ is a $K$-linear function, the matrix of $\beta$ with respect to the bases $(v_1,\dots,v_n)$ and $(w_1,\dots,w_m)$ is the $m \times n$ matrix $B = (b_{ij})$ such that $$(\forall j \in \{1,\dots,n\}) \quad \beta(v_j) = b_{1j}w_1+b_{2j}w_2+\cdots+b_{mj}w_m.$$
Now, actually the matrix of $\alpha$ with respect to the bases $(v_1,\dots,v_n)$ and $(w_1,\dots,w_m)$ looks like $$A = \begin{pmatrix} a_{11} & \dots &a_{1k} & a_{1(k+1)} & \dots &a_{1n} \\ \vdots & \ddots &\vdots & \vdots & \ddots &\vdots \\ a_{\color{red}l1} & \dots &a_{\color{red}lk} & a_{l(k+1)} & \dots & a_{ln} \\ 0 & \dots &0& a_{(l+1)(k+1)} & \dots &a_{(l+1)n} \\ \vdots & \ddots &\vdots & \vdots & \ddots &\vdots \\ 0& \dots & 0 & a_{m(k+1)} & \dots &a_{mn} \end{pmatrix}$$ and this is because we have that $$(\forall j \in \{1,\dots,k\}) \quad \alpha(v_j) = a_{1j}w_1+\cdots+a_{lj}w_l+0w_{l+1}+\cdots+0w_m \in Z$$ (since $(v_1,\dots,v_k)$ is a basis for $Y$, $(w_1,\dots,w_l)$ is a basis for $Z$, and $\alpha(Y) \subseteq Z$).