Proof of Sum of subspaces I am trying to do this:

Prove that the sum $S=L_1+L_2$ of two linear subspaces is equal to the intersection of all linear subspaces that contain both $L_1$ and $L_2$

I know what linear subspace means, I how is defined the sum of subspaces, in this case what I think is
 $L_1+L_2$ $=$ {$s=l_1+l_2  :  l_1 \in L_1 \land l_2 \in L_2$}, where $L_1$ and $L_2$ are subspaces of $S$.
Still I don't know how to prove it. How do I work with "the intersection of all linear subspaces that contain both $L_1$ and $L_2$". What else do I need to know in order to be able to prove it?
 A: In general, for problems like these, it's easiest to proceed by proving containment in both directions. Let $W =\bigcap_{L_1,L_2\subset V} V$. Then, you want to show $W \subset L_1 + L_2$ and $L_1 + L_2\subset W$.
For the first direction, let $w \in W$. Since $w$ is contained in all linear subspaces that contain $L_1$ and $L_2$, in particular, $w \in L_1 + L_2$ since $L_1, L_2 \subset L_1 + L_2$. Thus, $W\subset L_1 + L_2$.
For the other direction, let $l_1 + l_2\in L_1 + L_2$ and let $V$ be a linear subspace such that $L_1, L_2 \subset V$. Using the closure property of linear subspaces, $L_1 + L_2 \subset V$ as well. Thus, $l_1+l_2 \in V$ by definition and so $L_1 + L_2 \subset W$. 
Thus, $W = L_1 + L_2$ as desired. 
A: You have to prove two things :


*

*All subspace that contains $L_1$ and $L_2$ contains $L_1 + L_2$, which implies that $L_1 + L_2$ is included in the intersaction of all such subspaces.

*There is nothing else that is contained in the intersection of all subspaces containing $L_1$ and $L_2$, i.e. the intersection of all such subspaces is included in $L_1 + L_2$. This is straightforward since $L_1 + L_2$ itself is part of the intersection.
Once you have these two reciprocal inclusion, you yave the equality.
A: Consider vector $\vec{x}$ that is an element of the sum of the subspace $L_1 + L_2 = S$... then $$\exists \vec{x_1} \in L_1, \vec{x_2} \in L_2, \vec{x} = \vec{x_1} + \vec{x_2}$$
From this... consider the intersection of all subspaces that contain $L_1$ and $L_2$ and denote this as $\bigcap_{L_1, L_2}$, then it follows that if it contains both $L_1$ and $L_2$, and since $L_1$ and $L_2$ are subspaces and the encapsulating subspace of them is also a subspace, then it follows that $$\exists y, z \in \bigcap_{L_1, L_2}, y \in L_1, z \in L_2$$
Because by definition the intersection of all subspaces that contain both $L_1$ and $L_2$ will also contain both $L_1$ and $L_2$. Now look at the definition above for sum of subspaces. The equivalence of $y = x_1$ and $z = x_2$ is shown, and we have proved that $S \subset \bigcap_{L_1, L_2}$. 
The converse follows immediately... intersection of all subspaces that contain $L_1$ and $L_2$ must be a subset of $L_1 + L_2$ since the basis for $L_1 + L_2$ spans both $L_1, L_2$, this has proved that $\bigcap_{L_1, L_2} \subset S$
A: It is known that $L_1+L_2$ is the smallest subspace containg $L_1,L_2$. Also every subspace $V$ that contains $L_1,L_2$ contains $L_1+L_2$ for $V$ is closed under addition and scalar multiplication implying that it contains $L_1+L_2$. It then follows that $\bigcap _{i=1}A_i$=$L_1+L_2$ where $L_1,L_2\subset A_i$. Since $( x \in L_1+L_2) \implies (x\in A_i)$ and $\nexists A_i$ such that $A_i\subseteq L_1+L_2$. 
