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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?

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    $\begingroup$ Do you know rank + nullity? This is the theorem that says that the dimension of the image plus the dimension of the kernel equals the dimension of the domain. If you do, then there's a very clean proof which doesn't involve writing down elements. $\endgroup$
    – Cass
    Aug 8, 2014 at 13:55
  • $\begingroup$ No, this is how it's proven in my book and while it's not obligatory that I know this exact proof, I'd still like to understand it. $\endgroup$
    – Joshua
    Aug 8, 2014 at 14:04

1 Answer 1

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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 $$z=0\implies v=-w=0$$

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