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 $\alpha(Y)\subseteq Z$. Also $\alpha$ induces $\bar{\alpha}:V/Y \rightarrow W/Z$ and $\alpha|_Y:Y \rightarrow Z$.
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 $Z$.
The Matrix of $\alpha$, regarding the bases $(v_1,…,v_n)$ and $(w_1,…,w_m)$, has the form:
$\pmatrix{A&C\\ 0&B}$.
From this block matrix, I want to determine the matrix of $\alpha|_Y$ with respect to the
Bases $(v_1,\dots, v_k)$ of $Y$ and $(w_1,\dots, w_l)$ of $Z$, but also the matrix of $\bar{\alpha}$
with respect to the bases $(v_{k+1} + Y,\dots, v_n + Y)$ of $V/Y$ and $(w_{l + 1} + Z,\dots, w_m+Z)$ from $W/Z$.
Is this already the solution?:
$$\begin{pmatrix}
a_{11} & \dots &a_{1k} & a_{1(k+1)} & \dots &a_{1n} \\
\vdots & \ddots &\vdots & \vdots & \ddots &\vdots \\
a_{l1} & \dots &a_{lk} & a_{l(k+1)} & \dots & a_{ln} \\
0 & \dots &0& a_{(l+1)(k+1)} & \dots &a_{(l+1)n} \\
\vdots & \ddots &\vdots & \vdots & \ddots &\vdots \\
0& \dots & 0 & a_{m(k+1)} & \dots &a_{mn}
\end{pmatrix}$$
with $A$ as the Basis of $Y$? I am not sure whose matrices are $B$, $C$ and the Nullmatrix for. If not then how to calculate the matrices?
Thanks in advance.
Best Answer
Restricting the domain to $Y$ corresponds to only looking at the first few columns of the matrix (since $v_1,\dots,v_k$ are a basis of $Y$). Similarly, since $\alpha(Y) \subset Z$, you can restrict to the first few rows of the matrix (the bottom several rows are zero anyways). So the matrix of $\alpha|_Y$ is just $A$.
On the other hand, the matrix of the induced map $\overline{\alpha} \colon V/Y \to W/Z$ is given by $B$. Let me explain why. You have the bases $\bar{v}_{k+1},\dots,\bar{v}_n$ of $V/Y$ and $\bar{w}_{\ell+1},\dots,\bar{w}_{m}$ of $W/Z$ (here I'm using the notation $\bar{v}_i$ instead of $v_i + Y$). The $j$-th column of the matrix tells you that
$$ \alpha(v_j) = \sum_{i=1}^m a_{ij} w_i $$
Taking equivalence classes in the quotient spaces you get
$$ \bar{\alpha}(\bar{v}_j) = \sum_{i=1}^m a_{ij} \overline{w}_i = \sum_{i=\ell+1}^m a_{ij} \overline{w}_i $$ The last equality is because $\overline{w}_i = 0$ when $i \leq \ell$ (because $w_1,\dots,w_\ell$ is a basis of $Z$). This means the $j$-th column of the matrix for $\overline{\alpha}$ is $$ \begin {pmatrix} a_{\ell+1,j} \\ a_{\ell+2,j} \\ \vdots \\ a_{m,j} \end{pmatrix} $$ This is precisely the $j$-th column of $B$.