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:


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?

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    $\begingroup$ 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. $\endgroup$ – Syd Amerikaner Mar 27 at 20:57
  • $\begingroup$ Thanks a lot, I re-read the proposition a few times and now I see what it means and what you meant! $\endgroup$ – commentallez-vous Mar 27 at 21:08

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