Is the following theorem correct? I will especially appreciate any comments on how to improve the writing of the proof.

NOTE: In the proof below $\mathbf{2.21-(a)}$ is the result that given a linearly dependent list $v_1,v_2,...,v_m$ $$\exists j\in\{1,2,...,m\}(v_j\in span(v_1,v_2,...,v_{j-1}))$$

Theorem. Given that $v_1,v_2,...,v_m$ is a linearly independent in $V$ and $w\in V$. Show that $v_1,v_2,...,v_m,w$ is linearly independent if and only if $$w\not\in span(v_1,v_2,...,v_m).$$ Proof. $(\Rightarrow).$ Assume for purpose of contradiction that $v_1,v_2,...,v_m,w$ is a linearly independent list in $V$, $w\in V$ and $w\in span(v_1,v_2,...,v_m)$. It follows that $$w=\sum_{i=1}^{m}a_iv_i,\ \forall i\in\{1,2,3,..,m\}(a_i\in\mathbf{F})\tag{1}$$ but this implies that $v_1,v_2,...,v_m,w$ is not linearly independent resulting in a contradiction.

$(\Leftarrow).$ We prove the contrapositive, assume that $u_1,u_2,...,u_{m},u_{m+1}=v_1,v_2,...,v_m,w$ is a linearly dependent list, then by $\mathbf{2.21-(a)}$ it follows that for some $j\in I=\{1,2,...,m,m+1\}$, $u_j\in span(u_1,u_2,...,u_{j-1})$. Consider now the following Lemma.

Lemma $j=m+1$

Proof. Assume on the contrary that $j\neq m+1$ it then follows that $j=k$ for some $k\in I\backslash\{m+1\}$ consequently $u_j=v_k\in span(v_1,v_2,...,v_{k-1})$, moreover $v_1,v_2,...,v_m$ is a linearly independent on $V$ consequently $v_1,v_2,...,v_k$ is linearly independent on $V$ but then $v_k\not\in span(v_1,v_2,...,v_{k-1})$ resulting in a contradiction.


From the above lemma it follows that $u_j=u_{m+1}=w$ implying that $w\in span(v_1,v_2,...,v_m)$.


  • $\begingroup$ I think you mean to say that given a linearly dependent list, there is a $j$ such that... $\endgroup$ – Alex Ortiz Jul 26 '17 at 20:09
  • $\begingroup$ Yes my apologies correction has been made $\endgroup$ – Atif Farooq Jul 26 '17 at 20:12

The proof is correct as written, but you can make it an $\epsilon$ clearer.

For instance, when you write $w\in\operatorname{span}(v_1,\dots,v_m)$, you note correctly that there are scalars $a_i$ such that $$ w = \sum a_iv_i. $$ Now note that this implies directly that $$ w - \sum a_iv_i = 0, $$ which is a nontrivial dependence relation (note the invisible $1$ in front of $w$), contradicting what you assumed that the list $(v_1,\dots,v_m,w)$ is independent.

For your other direction, you are given the key lemma that in any ordered dependent list $(v_1,\dots,v_m)$ there is an index $i$ such that $v_i$ is in the span of the preceding vectors. As a matter of style, you don't need this extra lemma that $j = m+1$; just go ahead and show it with your key lemma as follows:

If $(v_1,\dots,v_m,w)$ is a dependent list, then by the key lemma, there is a vector in the list which is in the span of the preceding vectors. As $(v_1,\dots,v_m)$ is known to be independent, we deduce that $w$ is in the span of $(v_1,\dots,v_m)$.

This is easier to follow because it doesn't weigh down your readers' eyes with superfluous information.

  • 1
    $\begingroup$ I see what you did there with the $\epsilon$ ! $\endgroup$ – Atif Farooq Jul 26 '17 at 22:04

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