# Proving linearly independence even when the first vector is omitted

Problem:

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?

Edit:

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?

• Surely you mean they are linearly INdependent in (B)? – Alfred Yerger Feb 26 '18 at 5:16
• "are also linearly dependent"... ?! – Mark Twain Feb 26 '18 at 5:17
• Something something matroid theory – MCT Feb 26 '18 at 5:17
• Yes typo sorry. – Adagio Feb 26 '18 at 5:17
• "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. – 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 $$c_1=0$$.