# $Ax=0 \equiv x=0 \implies \exists A^{-1}$?

Question:

Let $A\in \Bbb R^{n\times m}, \operatorname{Rank(A)}=m$ and consider $V=\{v_1,v_2,\dots,v_r\} \subset \Bbb R^m$. Prove that $$V \,\,\text{is}\,\, L.I. \iff AV=\{Av_1,Av_2,\dots,Av_r\} \,\,\text{is}\,\, L.I$$

Attempt:

I couldn't do the $\implies$ part so I was trying to do the $\impliedby$ part.

$\alpha_1 Av_1+\alpha_2 Av_2+\dots+\alpha_r Av_r =0 \iff \alpha_i=0 \forall i=1,\dots,r.$

$\alpha_1 Av_1+\alpha_2 Av_2+\dots+\alpha_r Av_r =0\iff A(\alpha_1 v_1+\alpha_2 v_2+\dots+\alpha_r v_r)=0$

So if I can multiply by the inverse of $A$ this part should be done, but I don't even know if I have a square matrix.

• Since rank of $A$ is $m$, you know that $n\ge m$. You also get an injective linear transformation, which allows you to define an inverse from the domain to the image under the linear map. So it's a well defined inverse if you restrict the range to the image of the map. This is the same thing as saying it's injective. Then your result does follow. Jun 24, 2015 at 16:39
• Don't think about inverses. As the previous comment suggests, the rank condition should tell you that $Ax=0 \implies x=0$. Why? Jun 24, 2015 at 16:42

Hint: Consider $A: \mathbb R^m \to \mathbb R^n$ the transformation associated with the matrix $A$. As the $\mathrm {rank} A = m$ then $\dim \mathcal Im (A) = m$ and $$\dim Ker (A) = m - m = 0$$
Therefore $A$ is injective and it takes L.I. sets in L.I. sets.
• Hmm can't I say that $\alpha_i=0 \implies \sum \alpha_i v_i=0$ and as $\alpha_i=0$ is equivalent to $\sum A(\alpha_i v_i)=0$, $\sum A(\alpha_i v_i)=0 \implies \sum \alpha_i v_i=0$? Jul 5, 2015 at 17:47