3
$\begingroup$

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 $\mathbb R^3$?

$\endgroup$
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.