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Let $\{v_1,v_2,...,v_n\}$ be a linear independent group of vectors, and let $\vec{0} \neq w \in\langle v_1,v_2,\ldots,v_n\rangle$. Prove that there exists a $1 \leq k \leq n$ such that $\{v_1,v_2,\ldots,v_{k-1},w,v_{k+1},\ldots,v_n\}$ is linearly independent.

I started proving with the following:

$\{w,v_1,v_2,\ldots,v_n\}$ is linearly dependent, since $w \in \{v_1,v_2,\ldots,v_n\}$. , and can be written as a linear combination: $$\vec{0} \neq w = c_1v_1 + c_2v_2 + \cdots + c_nv_n$$

So there exists a $k > 1$ such that $v_k$ can be composed as a linear combination of its preceding vectors: $$v_k = c_0w + c_1v_1 + c_2v_2 + \cdots + c_{k-1}v_{k-1}$$

I'm stuck on figuring out how to properly prove that if I now wish to remove that $v_k$ the set $\{w,v_1,v_2,\ldots,v_{k-1},v_{k+1},\ldots,v_n\}$ will become linearly independent. I am missing a final logical link between this and what I am supposed to prove, but I can't figure out which other lemmas or theorems I could use in order to prove this. Help is greatly appreciated!

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  • $\begingroup$ Indeed! thanks for pointing it out :) $\endgroup$
    – 0rka
    Feb 3, 2018 at 17:38

1 Answer 1

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Suppose that $w$ is a multiple of $v_1$, that is, $w=\lambda_1v_1$, with $\lambda_1\neq0$. Then $\{w,v_2,\ldots,v_n\}$ is linearly independent.

Now, suppose that $w=\lambda_1v_1+\lambda_2v_2$, with $\lambda_2\neq0$. Then $\{v_1,w,v_3,\ldots,v_n\}$ is linearly independent, because\begin{align}\alpha_1v_1+\alpha_2w+\alpha_3v_3+\cdots+\alpha_nv_n=0&\iff\alpha_1v_1+\alpha_2(\lambda_1v_1+\lambda_2v_2)+\alpha_3v_3+\cdots+\alpha_nv_n=0\\&\iff(\alpha_1+\alpha_2\lambda_1)v_1+\alpha_2\lambda_2v_2+\alpha_3v_3+\cdots+\alpha_nv_n=0\\&\iff\left\{\begin{array}{l}\alpha_1+\alpha_2\lambda_1=0\\\alpha_2\lambda_2=0\\\alpha_3=0\\\vdots\\\alpha_n=0\end{array}\right.\\&\iff\alpha_1=\alpha_2=\cdots=\alpha_n=0,\end{align}since $\lambda_2\neq0$. And so on: if $w=\lambda_1v_1+\cdots+\lambda_k\alpha_k$ with $\lambda_k\neq0$ and $k\in\{1,2,\ldots,n\}$, then the set $\{v_1,\ldots,v_{k-1},w,v_{k+1},\ldots,v_n\}$ is linearly independent.

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  • $\begingroup$ I see. Great explanation. I was wondering, though, if there is a way to conclude the same proof you provided by merely stating a theorem on vector set linear dependence/independence. Nevertheless, thanks for providing an answer, it cleared things up for me! $\endgroup$
    – 0rka
    Feb 3, 2018 at 17:57

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