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Let $\{u_1, u_2, u_3\}$ be a set of linearly independent vectors in $V$. Assume that

(1): $$v_1 = a_1u_1 + a_2u_2 + a_3u_3$$ (2):$$v_2 = a_4u_1 + a_5u_2 + a_6u_3$$ (3): $$v_3 = a_7u_1 + a_8u_2 + a_9u_3$$

Denote A = $$ \left[ \begin{array}{} a_1&a_2&a_3\\ a_4&a_5&a_6\\ a_7&a_8&a_9 \end{array} \right] $$ Show that $\{v_1, v_2, v_3\}$ is linearly independent if and only if $A$ is invertible.

What I have done is: For the vectors ${v_1, v_2, v_3}$ to be linearly independent,
$$c_1(v_1) + c_2(v_2) + c_3(v_3) = 0$$where $c_1$, $c_2$ and $c_3$ are real constants. By substituting equation (1), (2) and (3) into the above, I get

$$c_1(a_1(u_1) + a_2(u_2) + a_3(u_3)) + c_2(a_4(u_1) + a_5(u_2) + a_6(u_3)) + c_3(a_7(u_1) + a_8(u_2) + a_9(u_3)) = 0$$ and after further simplification, I arrive at:

$$(c_1a_1 + c_2a_4 + c_3a_7)u_1 + (c_1a_2 + c_2a_5 + c_3a_8)u_2 + (c_1a_3 + c_2a_6 + c_3a_9)u_3 = 0$$

From here on, how do I link it such that it is linearly independent only when $A$ is invertible?

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    $\begingroup$ math.meta.stackexchange.com/questions/5020/… $\endgroup$ – 5xum Oct 15 '18 at 12:20
  • $\begingroup$ From that you arrived at, draw the conclusion that $u_i$ independent implies $c^TA=0$. Now how can you connect that to $c=0$? $\endgroup$ – A.Γ. Oct 15 '18 at 12:49
  • $\begingroup$ Is it right to conclude that if A is invertible, A does not equals to 0, hence Ac^T = 0 can only be obtained if c = 0? $\endgroup$ – Cheryl Oct 15 '18 at 14:20
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    $\begingroup$ If $A$ is invertible then $c^TA=0$ $\Rightarrow$ $c^TAA^{-1}=0\cdot A^{-1}=0$ $\Rightarrow$ $c^T=0$. $\endgroup$ – A.Γ. Oct 16 '18 at 8:38
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Since $\{u_1, u_2, u_3 \}$ is linearly independent, we have

$$c_1a_1 + c_2a_4 + c_3a_7 = 0$$

$$c_1a_2 + c_2a_5 + c_3a_8 = 0$$ $$c_1a_3 + c_2a_6 + c_3a_9 = 0$$

which is just $$A^Tc=0$$

Hopefully you can finish up from here.

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Try to carefully understand the basic concepts. By definition, ''$\{v_1,v_2,v_3\}$ is linearly independent'' means that the equation $c_1v_1+c_2v_2+c_3v_3=0$ is satisfied only if $c_1=c_2=c_3=0$. (Of course the equation is satisfied no matter what $v_1,v_2,v_3$ are, if $c_1=c_2=c_3=0$. Linear independence of $\{v_1,v_2,v_3\}$ means that this is the only way how the equation can be satisfied.)

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