# $\lambda$ times linearly independent list

Is the following Proof Correct?

NOTE:$(2.21)$ in the proof below refers to the fact 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. If $v_1,v_2,v_3,...,v_m$ is a linearly independent list of vectors in $V$ and $\lambda\in\mathbf{F}$ with $\lambda\neq 0$ then $\lambda v_1,\lambda v_2,\lambda v_3,...,\lambda v_m$ is linearly independent.

Proof. Assume for purpose of contradiction that $v_1,v_2,v_3,...,v_m$ is linearly independent and the list $\lambda v_1,\lambda v_2,\lambda v_3,...,\lambda v_m$ is linearly dependent where $\lambda\neq 0$.

Since $\lambda v_1,\lambda v_2,\lambda v_3,...,\lambda v_m$ is linearly dependent $\mathbf{(2.21)}$ implies that for some $j\in\{1,2,3,...,m\}$, $\lambda v_j\in span(\lambda v_1,\lambda v_2,\lambda v_3,...,\lambda v_{j-1})$ consequently $$\lambda v_j=\sum_{i=1}^{j-1}a_i(\lambda v_i),\ \forall i\in\{1,2,3,..,j-1\}(a_i\in\mathbf{F})\tag{1}$$ Since $\lambda\neq0$ it follows that $\frac{1}{\lambda}\neq0$, multiplying both sides of equation in $(1)$ by $\frac{1}{\lambda}$ yields $$v_j=\sum_{i=1}^{j-1}a_iv_i\tag{2}$$ The result in $(2)$ implies that $v_1,v_2,v_3,...,v_m$ is not linearly dependent but we assumed that this list is linearly independent thus we have the required contradiction.

$\blacksquare$

Yes, it is correct. There is only a small problem. That's when you write “Since $\lambda\neq0$ it follows that $\frac1\lambda\neq0$”. No; since $\lambda\neq0$, it follows that $\frac1\lambda$ exists. That's all that matters.
• Yes quite right for some reason i forgot the $\lambda$ could be negative – Atif Farooq Jul 26 '17 at 13:37
• @AtifFarooq I thought that $\mathbf F$ was an arbitrary field. – José Carlos Santos Jul 26 '17 at 14:16
• It is either $\mathbf{R}$ or $\mathbf{C}$ – Atif Farooq Jul 26 '17 at 14:24
• @AtifFarooq In $\mathbb C$, “negative” makes no sense. – José Carlos Santos Jul 26 '17 at 14:24