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Let $B = \{v_1, v_2\}$ be a basis for the subspace $S$ of $\mathbb{R}^4$ and let $w_1, w_2, w_3 \in S$. Show that $\{w_1, w_2, w_3\}$ is linearly dependent.

We know that $\{v_1, v_2 \}$ is linearly independent. We know that $S = \text{span } \{v_1, v_2\}$ thus $w_1, w_2, w_3 \in\text{span } \{v_1, v_2\}$, so let $w_1 = a_1v_1 + a_2v_2$, $w_2 = b_1v_1 + b_2v_2, w_3 = c_1v_1 + c_2v_2$.

Any hints?

Edit: Is it enough to say that

$\alpha w_2 + Bw_3 = v_1(\alpha b_1 + Bc_1) + v_2(\alpha b_2 + Bc_2)$

And that we can let $w_3 = v_1(\alpha b_1 + Bc_1) + v_2(\alpha b_2 + Bc_2)$? Or is that not strong enough?

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  • $\begingroup$ @Aweygan, sorry $\endgroup$ – Amad27 Feb 3 '17 at 19:46
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In general, if $\{v_1,...,v_n\}$ is a linearly independent set and $\{w_1,...,w_m\}$ is a spanning set, then $n \le m$. To prove this you need an easy lemma:

Lemma. If $\{u_1,...u_N\}$ is linearly dependent, then there exists $i$ such that $u_i$ is in the span of $\{u_1,...,u_{i-1}\}$ and the span of $\{u_1, ..., u_N\} \setminus \{u_i\}$ is the same as the span of the full set $\{u_1, ..., u_N\}$.

Now to prove the claim above, you add the $v_i$ one at a time at the beginning of the list of $w_j$. Since the $w_j$ are spanning, the added vector makes the set linearly dependent, so you can remove a vector according to the lemma. You can see that the removed vector can't be one of the $v_i$, since the $v_i$ are linearly independent. Keep doing that until you've added all $n$ of the $v_i$, and conclude $n \le m$.

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