Let $\{v_1, v_2, v_3, v_4\}$ be a linearly independent set. Prove that the set of vectors $\{v_1, v_2, v_3\}$ form a linearly independent set.

Question

The question asks to use a proof by contradiction. I know that the coefficients of $\{v_1,v_2,v_3,v_4\}$ are all zero, but then I don't know what to do next-how to translate that into the $\Bbb R^3$ vector. Any help would be appreciated, thanks!

Answer 1

So to prove by contradiction, assume $\{v_1,v_2,v_3\}$ is linearly dependent. Then there is a nontrivial solution to $a_1v_1+a_2v_2+a_3v_3 = 0$, so at least one of $a_1,a_2,a_3$ is nonzero. Then $a_1v_1+a_2v_2+a_3v_3 + 0\cdot v_4 = 0$. But by linear independence of $\{v_1,\dots,v_4\}$ we must have $a_1=a_2=a_3 = 0$, contradicting that one of the $a_1,a_2,a_3$ is nonzero.

Answer 2

So you want to show that $$a_1v_1+a_2v_2+a_3v_3 =0 \Leftrightarrow a_1 =a_2=a_3=0$$ knowing that $$a_1v_1+a_2v_2+a_3v_3+a_4v_4 =0 \Leftrightarrow a_1 =a_2=a_3=a_4=0$$

So let's assume $\exists b_1, b_2, b_3$, with any $b_i\neq 0$ such that $b_1v_1+b_2v_2+b_3v_3 =0 $, then for $b_4 = 0$, what can you say about $b_1v_1+b_2v_2+b_3v_3+b_4v_4 =0$?

Can you go from here?