# 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 $c_1=c_2=...=c_n = 0$

• Recall that the linear equation $Av=0$ has only zero solution and you will easily figure it out. Oct 29, 2015 at 5:43

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.
Suppose that $$\lambda_1\, Av_1+\lambda_2\, Av_2+\dotsb+\lambda_n\, Av_n=\vec 0$$ Then $$A\bigl( \lambda_1v_1+\lambda_2v_2+\dotsb+\lambda_nv_n \bigr) = \vec 0\tag{1}$$ Since $A$ is invertible we know that $A^{-1}$ exists. What happens when we multiply both sides of (1) on the left by $A^{-1}$?