2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Indeed! thanks for pointing it out :) $\endgroup$ – 0rka Feb 3 '18 at 17:38
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '18 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.