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?
Best Answer
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.