Dimension of sum and intersection of vector spaces (Strang's Linera Algebra question 43, section 3.4) The solution of the following problem is too abstract for me to understand so can anyone give me a better explanation?
This problem comes from the fifth edition of Gilbert Strang's Introduction to Linear Algebra question 43, section 3.4. The following is the description of the problem:

Intersections and sums have $\dim(V) + \dim(W) = \dim(V\cap W) + \dim(V+W)$. Start with a basis $u_1,\dots,u_r$ for the intersection $V\cap W$. Extend with $v_1,\dots,v_s$ to a basis for V, and separately with $w_1,\dots, w_t$ to a basis for W. Prove that the $u$'s, $v$'s and $w$'s together are independent. The dimensions have $(r + s) + (r + t) = (r) + (r + s + t)$.

 A: Here is a different argument which starts from the bases $\{v_1, \ldots, v_m\}$ for $V$ and $\{w_1, \ldots, w_n\}$ for $W$.
Putting them together, we get
$$\{v_1, \ldots, v_m, w_1, \ldots, w_n\}$$
and this sets certainly spans $V+W$. We wish to reduce it to a basis for $V+W$ by eliminating those $w_i$ which can be expressed as a linear combination of its predecessors.
Start with $w_1$. If you can express it as a linear combination of $v_1, \ldots, v_m$ as $$w_1 = \sum_{i=1}^m \alpha_i^{(1)}v_i$$
then discard it from the set. Otherwise keep it in the set. Then look at $w_2$. If you can express it as a linear combination of $v_1, \ldots, v_m$ (if we discarded $w_1$) or $v_1, \ldots, v_m, w_1$ (if we kept $w_1$) as
$$w_2 = \sum_{i=1}^m \alpha_i^{(2)}v_i + \beta_1w_1$$
then discard it. Otherwise keep it in the set. Moving on, try to express $w_3$ as a linear combination of $v_1, \ldots, v_m, w_1, w_2$ (the $w_1$ and $w_2$ are here if we kept them in the set, otherwise no) and so on.
Do this until you reach the end of the set. You discarded some $w_i$-s, denote them as $w_1', \ldots, w_r'$. The set which remains
$$\{v_1, \ldots, v_m, w_1'', \ldots, w_s''\}=\{v_1, \ldots, v_m, w_1, \ldots, w_n\} \setminus \{w_1', \ldots, w_r'\}, \quad r+s=n$$
is now linearly independent since no element can be expressed as a linear combination of its predecessors. Hence it is a basis for $V+W$.
Those discarded you wrote as
$$w_i' = \underbrace{\sum_{i=1}^m \alpha_iv_i}_{=u_i'} + \underbrace{\sum \beta_jw_j''}_{=g_i'}.$$
The set $\{u_1', \ldots, u_r'\}$ is then a basis for $V \cap W$. Indeed, it is contained in $V \cap W$ since
$$u_i' = \sum_{i=1}^m \alpha_iv_i \in V, \qquad u_i' = w_i' - g_i' \in W.$$
$\{u_1', \ldots, u_r'\}$ is also linearly independent. We have
$$\sum_{i=1}^r \gamma_iu_i' = \sum_{i=1}^r \gamma_iw_i' - \underbrace{\sum_{i=1}^r \gamma_ig_i'}_{\in\operatorname{span}\{w_1'', \ldots, w_s''\}}$$
so since $\{w_1, \ldots, w_n\}$ is a basis we conclude $\gamma_1 = \cdots = \gamma_n= 0$.
$\{u_1', \ldots, u_r'\}$ also spans $V \cap W$. Namely, if $x \in V \cap W$, then clearly we can write $x = \sum_{i=1}^m \alpha_iv_i$ and also
$$x = \sum_{j=1}^n \beta_j w_j = \sum_{j=1}^r \beta_j'w_j'+\sum_{j=1}^s \beta_j''w_j'' = \sum_{j=1}^r \beta_j'u_j'+\underbrace{\sum_{j=1}^r \beta_j'g_j' + \sum_{j=1}^s \beta_j''w_j''}_{\text{linear combination of $w_j''$-s}}$$
and so since $\{v_1, ldots, v_m, w_1'', \ldots, w_s''\}$ is a basis, from
$$0 = \underbrace{\sum_{i=1}^m \alpha_iv_i - \sum_{j=1}^r \beta_j'u_j'}_{\text{linear combination of $v_i$-s}} -\underbrace{\sum_{j=1}^r \beta_j'g_j' + \sum_{j=1}^s \beta_j''w_j''}_{\text{linear combination of $w_j''$-s}}$$
we get $\sum_{j=1}^r \beta_j'g_j' + \sum_{j=1}^s \beta_j''w_j'' = 0$ and hence $$x = \sum_{j=1}^r \beta_j'u_j'.$$
Therefore $\{u_1', \ldots, u_r'\}$ is indeed a basis for $V \cap W$ and hence
$$\dim (V \cap W) + \dim (V+W) = (r) + (m+s) = m + (r+s) = m+n = \dim V + \dim W.$$
A: The question is basically the building block to prove the theorem of dimension of direct sum of vector subspaces. Break it down as follows

*

*Consider the intersection of subspaces $V$, $W$. Is that a vector space? Try convincing yourself that it is using the basic definition of a vector space. Since it is a subspace, it should have a dimension and a basis. For now, we assume that set to be $u = \{u_1, u_2...u_r\}$


*What does the question mean by extend? Since $V\cap W$ is a subspace of both $V$ and $W$, one could take it's basis, and add elements to it to form a new basis for either $V$ or $W$. You have to essentially show that there is a way of doing this extension such that  all the additional vectors you add to form each basis are also mutually independent - this will allow to claim that the full set is a basis for the sum of the spaces
