Prove linear independence Prove from the definition of "linearly independent" that if ${\{v_1,v_2,...,v_n\}}$ is linearly independent and if A is an invertible $n x n$ matrix, then the set ${\{Av_1,Av_2,...,Av_n\}}$ is linearly independent.
I ran in to this problem in my textbook and I am having trouble understanding how to reach the answer. My initial thought is that is has something to do with the fact that for a matrix to be invertible it's determinant must be non-zero. But that seems to contradict the definition of linearly independent which states that the only scalars that make $c_1v_1+c_2v_2+...+c_nv_n=0$ true must be $c_1=c_2=...=c_n = 0$

If $A$ is invertible then it has an inverse $A^{-1}$. (Yes its determinant is nonzero, but that won't help much.)
Suppose $c1, \dots, c_n$ are scalars such that 
$$
c_1 Av_1 + c_2 Av_2 + \dots + c_n Av_n = 0 \text{.}
$$
Then applying $A^{-1}$, we get:
$$
\begin{align}
0 &= A^{-1}(c_1 Av_1 + c_2 Av_2 + \dots + c_n Av_n) \\
&= A^{-1}(c_1 Av_1) + A^{-1}(c_2 Av_2) + \dots + A^{-1}(c_n Av_n) \\
&= c_1A^{-1} Av_1 + c_2 A^{-1} Av_2 + \dots + c_n A^{-1} Av_n \\
&= c_1 v_1 + c_2 v_2 + \dots + c_n v_n \\
\end{align}
$$
Because the $v_i$ are linearly independent, it follows that all $c_i = 0$.
So the images of the $v_i$ under $A$ are also linearly independent.
