Prove two sets span the same subspace

I've found here that in order for two sets to span the same subspace, the following must be true:

• Each vector in S1 can be written as a linear combination of the vectors in S2; and
• Each vector in S2 can be written as a linear combination of the vectors in S1.

I don't know why we need the second one, because suppose a vector $$u$$ from $$S_1$$ can be written as a linear combination of vectors from $$S_2$$: $$u = \alpha_1a_1+\cdots+\alpha_na_n$$. Then shouldn't it mean that $$a_i = \frac{u -\alpha_1a_1-\cdots-\alpha_na_n}{\alpha_i}$$, and therefore, this vector in $$S_2$$ can be written as linear combination of $$S_1$$? PS: while writing this I noticed that $$a_i$$ is not written in terms of vectors in $$S_2$$, but in terms of vector in $$S_1$$ and $$S_2$$, so the converse isn't imediately true.

Ok, but,can someone give me na intuition on why these criteria are necessary?

In order to verify that

$$\{(1,-1,2),(3,0,1)\}, \{(-1,-2,3), (3,3,-4)\}$$

generate the same subspace of $$\mathbb R^3$$, I should try to solve

$$(1,-1,2) = \alpha_1(-1,-2,3)+\alpha_2(3,3,4)$$ $$(3,0,1) = \alpha_3(-1,-2,3)+\alpha_4(3,3,4)$$

and

$$(-1,-2,3) = \beta_1(1,-2,3)+\beta_2(3,0,1)$$ $$(3,3,-4) = \beta_3(1,-2,3)+\beta_4(3,0,1)$$

if there is such $$\alpha_1,\alpha_2,\alpha_3,\alpha_4,\beta_1,\beta_2,\beta_3,\beta_4$$, then they generate the same subspace of $$\mathbb R^3$$?

To show $\text{span}\{v_{1}, v_{2} \} = \text{span}\{u_{1}, u_{2} \}$, you need to show these two sets are subsets of each other. But if you show $v_{1}, v_{2} \in \text{span}\{u_{1}, u_{2} \}$, then since the span is a subspace and hence closed under addition and scalar multiplication, it follows that all possible linear combinations of $v_{1}$ and $v_{2}$ are in $\text{span}\{u_{1}, u_{2} \}$, and hence $\text{span}\{v_{1}, v_{2} \} \subseteq \text{span}\{u_{1}, u_{2} \}$.
Similarly, if you show $u_{1}, u_{2} \in \text{span}\{v_{1}, v_{2} \}$, then since the span is closed under addition and scalar multiplication, it follows that all possible linear combinations of $u_{1}$ and $u_{2}$ are in $\text{span}\{v_{1}, v_{2}\}$, and thus $\text{span}\{u_{1}, u_{2} \} \subseteq \text{span}\{v_{1}, v_{2} \}$