Let me first summarize the part I understand:
$\dim{W_1}=n$, $\dim{W_2}=m$, $\dim{(W_1\cap W_2)}=k$
I've defined $C$ as a basis for $W_1\cap W_2$ with $C = \{v_1, … v_k\}$.
I've expanded $C$ to a basis $B_1$ for $W_1$ and $B_2$ for $W_2$:
$B_1 = \{v_1, … v_k, w_1, …, w_{n-k}\}$
$B_2 = \{v_1, … v_k, z_1, …, z_{m-k}\}$
$B = \{v_1, … v_k, w_1, …, w_{n-k}, z_1, …, z_{m-k}\}$
I've already proven that $B$ is a generating set, now I need to prove that it's linearly independent and that contains the part I don't get:
Define the following:
$v=\alpha_1v_1+…+\alpha_kv_k\in W_1\cap W_2$
$w=\beta_1w_1+…+\beta_{n-k}w_{n-k}\in <w_1,…,w_{n-k}> \leq W_1$
$z=\gamma_1z_1+…+\gamma_{m-k}z_{m-k}\in <z_1,…,z_{m-k}> \leq W_2$
Say that $v+w+z=0$ and so $v+w=-z\in W_1\cap W2$, which means that $z\in (W_1\cap W_2)\cap <z_1, …, z_{m-k}>$
I don't understand the part right after $v+w=-z$. Why is it an element of that collection? And why is the inverse part of a different collection?
Best Answer
We have $v\in W_1$ and $w\in W_1$ so $v+w=-z\in W_1$ since $W_1$ is a subspace so it's closed by addition. But since $z\in W_2$ then $z\in W_1\cap W_2$. Finally we have $$z\in W_1\cap W_2\cap\langle z_1,\ldots,z_{m-k}\rangle=\{0\}$$ and then $$z=0\implies v=-w=0$$