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Suppose $\{v_1,v_2,v_3\}$ $\subseteq$ $\mathbb{R}^{10}$ is a set of linearly independent vectors, and $w\in\mathbb{R}^{10}$ is not in span $\{v_1,v_2,v_3\}$. Prove $\{v_1,v_2,v_3,w\}$ is linearly independent.

Proof by Contradiction:

Let $U=\{v_1,v_2,v_3,w\}$

Assume $U$ is linearly dependent.

Then if $x_1v_1+x_2v_2+x_3v_3+x_4w=0$ , $x_1,x_2,x_3,x_4\in \mathbb{R}$ has a solution, then at leat one of the $x$'s must not equal to zero. Without loss of generality, assume $x_4\neq0$.

Then,

$w= -(\frac{x_1}{x_4}v_1+\frac{x_2}{x_4}v_2+\frac{x_3}{x_4}v_3)$

Therefore, $w$ is a linear combination of $\{v_1,v_2,v_3\}$. Then $w$ is in span $\{v_1,v_2,v_3\}$. This is a contradiction to the given statement.

Hence, $U=\{v_1,v_2,v_3,w\}$ is linearly independent.

Does this make sense?

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    $\begingroup$ I demur at your "without loss of generality" since $w$ plays a different role to the $v_i$. I suggest you consider the possibility that $x_4=0$. $\endgroup$ – Lord Shark the Unknown Apr 5 '18 at 4:00
  • $\begingroup$ u mean i should assume one of the other x's to be 0 $\endgroup$ – HAC Apr 5 '18 at 4:01
  • $\begingroup$ Since $v_1,v_2,v_3$ are linearly independent, we know that there is no non-trivial $x_1,x_2,x_3$ such that $x_1v_1 + x_2v_2 + x_3v_3 = 0.$ It is not possible for $x_1v_1 + x_2v_2 + x_3v_3 + x_4w = 0$ and $x_4 = 0$ $\endgroup$ – Doug M Apr 5 '18 at 4:03
  • $\begingroup$ The theorem above is clearly false if $v_1, v_2, v_3$ are dependent, so you have to use their independence somewhere in your proof. $\endgroup$ – Kaynex Apr 5 '18 at 4:04
  • $\begingroup$ @Kaynex $v_1, v_2, v_3$ are linearly independent $\endgroup$ – HAC Apr 5 '18 at 4:06
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Yes it is correct, the only observation is that we are allowed assume $x_4\neq 0$ since $v_i$ are linearly independent and then at least two of the x coefficient ($x_4$ and one other) must be $\neq 0$.

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