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 $$V \,\,\text{is}\,\, L.I. \iff AV=\{Av_1,Av_2,\dots,Av_r\} \,\,\text{is}\,\, L.I$$


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.

  • 1
    $\begingroup$ 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. $\endgroup$ Jun 24, 2015 at 16:39
  • $\begingroup$ Don't think about inverses. As the previous comment suggests, the rank condition should tell you that $Ax=0 \implies x=0$. Why? $\endgroup$ Jun 24, 2015 at 16:42

1 Answer 1


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.

  • $\begingroup$ Is there any way to prove this without the Rank-nullity theorem, or any interpretation of the matrix A as a linear transformation? I'm asking this because although I have studied those matters, my class haven't and so we can't use things like that. $\endgroup$ Jun 24, 2015 at 17:12
  • $\begingroup$ What can you use, because you just took all the linear algebra now. $\endgroup$ Jun 24, 2015 at 17:14
  • $\begingroup$ Hmm can't I say that $\alpha_i=0 \implies \sum \alpha_i v_i=0$ and as $\alpha_i=0$ is equivalent to $\sum A(\alpha_i v_i)=0$, $\sum A(\alpha_i v_i)=0 \implies \sum \alpha_i v_i=0$? $\endgroup$ Jul 5, 2015 at 17:47

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