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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 $\mathbb R^3$?

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1 Answer 1

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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 $\text{span}\{u_{1}, u_{2} \} \subseteq \text{span}\{v_{1}, v_{2} \}$

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