# Question on the assumption of proposition in linear independence

When talking about linear independence of a family of vectors $$(x_{1}, x_{2}, ... , x_{t})$$, where family of vectors allows repetition and order matters, it's not set of vectors, we have below proposition:

$$**Proposition**$$

Let $$V$$ be a vector space, let $$(x_{1}, x_{2}, ... , x_{t})$$ be a family of vectors in $$V$$, and suppose $$x_{1} \neq 0$$, then $$(x_{1}, x_{2}, ... , x_{t})$$ is linearly independent if and only if for each integer, $$j, j = 2, ..., s, x_{j} \notin \langle x_{1}, x_{j-1}\rangle$$

My question is that why does this proposition needs the assumption that $$x_{1} \neq 0$$, and it even emphasizes on this assumption? What if $$x_{1} = 0$$, what effect would it be?

• You dont need $x_1\neq0$ for the "$\Rightarrow$" direction but you need it for "$\Leftarrow$". Assume that $x_1 = 0$. Then for $j = 3$, $\langle x_1,x_2\rangle = \langle x_2\rangle$. So if $x_3$ is not a scalar multiple of $x_2$ we would have $x_3\not\in\langle x_2\rangle$. Now assume that this is true for $j=3,4,...,s$. We may also assume that $x_2\neq 0$. So we have that $x_j\not\in\langle x_1, x_{j-1}\rangle$ for $j=2,3,...,s$. But a set containing the zero vector is certainly not linearily independent. – Syd Amerikaner Mar 27 at 20:57
• Thanks a lot, I re-read the proposition a few times and now I see what it means and what you meant! – commentallez-vous Mar 27 at 21:08