$\displaystyle b_1=\left\lbrace\frac{12}{9},\frac{12}{9},2\right\rbrace^T,b_2=\{-18,-18,21\}^T$ and $\displaystyle v_1=\{-1,-1,2\}^T,v_2=\{3,3,-3\}^T$. $b_1 \in \operatorname{Span}\{v_1,v_2\} \text{ and } b_2 \in \operatorname{Span}\{v_1,v_2\}$. Can we conclude that (Without performing further verification) $\operatorname{Span}\{b_1,b_2\} \subseteq \operatorname{Span}\{v_1,v_2\}$? What about $ \operatorname{Span}\{v_1,v_2\} \subseteq \operatorname{Span}\{b_1,b_2\}$?
As the question has provided, both $b_1,b_2$ belongs to the span of $v_1,v_2$. However, the answer key only pointed out that we can only conclude $\operatorname{Span}\{b_1,b_2\} \subset \operatorname{Span}\{v_1,v_2\}$ but not $ \operatorname{Span}\{v_1,v_2\} \subset \operatorname{Span}\{b_1,b_2\}$ without any further explanation.
Is the fact that both $b_1 \in \operatorname{Span}\{v_1,v_2\} \text{ and } b_2 \in \operatorname{Span}\{v_1,v_2\}$ implies that $\operatorname{Span}\{b_1,b_2\} \subseteq \operatorname{Span}\{v_1,v_2\}$ ?
More generally, as my question's title suggests, how do I verify if a Span of vectors is a subset of another span of vectors? (I am capable of verifying if a single vector belongs to a span of vectors) but I can't make any connections between the two
Best Answer
If $b_1 \in \text{Span}(v_1, v_2) $ and $b_2 \in \text{Span}(v_1, v_2)$, then $\text{Span}(b_1, b_2) \subset \text{Span}(v_1, v_2)$.
Proof:
$b_1 \in \text{Span}(v_1, v_2) $ $\implies $ $b_1 = k_1 v_1 + k_2 v_2$. Similarly, $b_2 \in \text{Span}(v_1, v_2)$ $\implies $ $b_2 = c_1 v_1 + c_2 v_2$.
Hence, an arbitrary vector $w \in \text{Span}(b_1, b_2)$ satisfies
\begin{align} w &= m_1 b_1 + m_2 b_2\\ &= (m_1 k_1 + m_2 c_1) v_1 + (m_1 k_2 + m_2 c_2 ) v_2 \end{align}
This implies $w \in \text{Span}(v_1, v_2)$. Hence, $\text{Span}(b_1, b_2) \subset \text{Span}(v_1, v_2)$.
End Proof
If $b_1 \in \text{Span}(v_1, v_2) $ and $b_2 \in \text{Span}(v_1, v_2)$, then $\text{Span}(v_1, v_2) \subset \text{Span}(b_1, b_2)$ is in general False.
Counterexample
Take $b_1 = b_2 = v_1$. Then, Span$(b_1, b_2)$ = Span$(v_1)$. Hence, $\text{Span}(v_1, v_2) \nsubseteq \text{Span}(b_1, b_2)$
Response to comments:
If $b_1 \in \text{Span}(v_1, v_2) $ and $b_2 \in \text{Span}(v_1, v_2)$, then $\text{Span}(b_1, b_2) \subset \text{Span}(v_1, v_2)$.
Now if $\text{Span}(b_1, b_2) \subset \text{Span}(v_1, v_2)$ and dim(Span($b_1, b_2$)) $=$ dim(Span($v_1, v_2$)), then indeed Span($b_1, b_2$) $=$ Span($v_1, v_2$). This follows from a general rule that any subspace $A$ of a vector space $V$ satisfying dim$(A) = $ dim($V$) implies that $A = V$.