# Linear independence of a set of vectors

Let $\{v_1, v_2\}$ be a basis for a subspace $S$ of $\Bbb R^3$ . If $\mathcal B = \{w_1, w_2, w_3\}$ is a set of vectors in $S$, then $\mathcal B$ cannot be linearly independent.

I'm not sure how to solve this, I tried coming up with a counter example where $w_1 = v_1$, $w_2 = v_2$ but that doesn't work as $w_3$ must be a linear combination of $v_1$ and $v_2$. Any hints?

• Would you know how to tell if the columns/rows of a matrix are linearly (in)dependent? – user137731 May 27 '15 at 2:01
• Yes, but I don't see how that would help here. – Filip May 27 '15 at 2:11

Have you already seen that the number of elements in a basis is called de dimension of your vector space?

What knowledge can we use?

• That doesn't sound familiar. The most recent thing we learned was solving system of equations using matrices. – Filip May 27 '15 at 2:00
• Can you comment what theorems about basis do you know please? – HeMan May 27 '15 at 2:13
• This way, I can help you. – HeMan May 27 '15 at 2:13
• For example, there is a theorem that says that if a vector space have a basis with n elements, then all basis of such vector space have n elements, but I dont know if I can use that. – HeMan May 27 '15 at 2:14

since $V = \{v_1, v_2\}$ is a basis, we can write each of the $w$'s as a linear combination of $V.$ let $$(w_1, w_2, w_3) = (v_1, v_2)\pmatrix{a_{11}&a_{12} & a_{13}\\a_{21}&a_{22}&a_{23}} \tag 1$$

suppose $$x_1w_1 + x_2w_2 + x_3w_3 = 0\tag 2$$ this can be rewritten as $$(w_1, w_2, w_3)(x_1, x_2, x_3)^\top =(v_1, v_2)(a_{11}x_1 + a_{12}x_2 + a_{13}x_3,\, a_{21}x_1 + a_{22}x_2 + a_{23}x_3)^\top = 0$$ by linear independence of $V,$ we have \begin{align} a_{11}x_1 + a_{12}x_2 + a_{13}x_3 &= 0\\ a_{21}x_1 + a_{22}x_2 + a_{23}x_3 &= 0\end{align} we have two equations and three unknowns, therefore this has a nonzero solution of $(2)$ for $x_1, x_2, x_3.$ therefore $$\{w_1, w_2, w_3\} \text{ is linearly dependent.}$$

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 $n \le m$.