Let (A) $x_1,x_2,...,x_k$ be linearly independent vectors in a vector space $V$. Prove that the vectors (B) $x_2, x_3,...,x_k$ are also linearly independent.

My thoughts:

Okay. To start off I know that since (A) is linearly independent, then $c_1x_1 + c_2x_2,+...,+c_kx_k = 0$ implies that $c_1 , c_2, ... c_k = 0$. Since only one of the vectors is omitted from (B) then since the vectors are still not linear combinations of each other then their coefficients still must be $0$ so that they can equal zero thus B is linearly independent.

Are my thoughts on the correct path to solving this?


Given (B), we have $c_2x_2 + c_3x_3 +\dots +c_kx_k = 0 $ which has an equivalent equation of $c_2x_2 + c_3x_3 +\dots +c_kx_k + 0x_{k+1} = 0 $

Since this is a zero linear combination of a linearly independent collection, all the coefficients must be zero. Is this any better?

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    $\begingroup$ Surely you mean they are linearly INdependent in (B)? $\endgroup$ – Alfred Yerger Feb 26 '18 at 5:16
  • $\begingroup$ "are also linearly dependent"... ?! $\endgroup$ – Mark Twain Feb 26 '18 at 5:17
  • $\begingroup$ Something something matroid theory $\endgroup$ – MCT Feb 26 '18 at 5:17
  • $\begingroup$ Yes typo sorry. $\endgroup$ – Adagio Feb 26 '18 at 5:17
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    $\begingroup$ "then $c_1x_1$, $c_2x_2$, $\ldots$, $c_k x_k$ implies that $c_1$, $c_2$, $\ldots$, $c_k=0$" does not make sense. Fix the beginning. $\endgroup$ – Mark Twain Feb 26 '18 at 5:19

Suppose $c_2x_2+\ldots+c_nx_n=0$, where not every $c_i$ is zero. Now extend this sum to $c_1x_1+c_2x_2+\ldots+c_nx_n$, where $c_1=0$.

I think that's the idea of your proof. Note: this is proving the contraposition.


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