I'm studying with a Linear Algebra book which presents the following question:

Be $A:E\rightarrow F$ a linear transformation. If the vectors $Av_1,\ldots,Av_n \in F$ are LI, then prove that $v_1,\ldots,v_n \in E$ are also linear.

Below is the official solution:

Assume $c_1v_1+\ldots+c_nv_n=0$. Applying $A$ (which is linear), yields $c_1Av_1+\ldots+c_nAv_n=0$. Since $Av_1,\ldots,Av_n$ are LI, then $c_1=c_2=\ldots=c_n=0$. Thus, $v_1,\ldots,v_n$ are LI.

It seems to me that this solution does not take me anywhere. It already assumed a LI set $v_1,\ldots,v_n$ and applied $A$ to it, to conclude that the same set is LI. But what if $A$ is singular?

I think it's more reasonable to linearly combine $Av_1,\ldots,Av_n$:

$$ c_1Av_1+\ldots+c_nAv_n=0 $$


$$ A\left(c_1v_1+\ldots+c_nv_n\right)=0 $$

And then to conclude that, iff $A$ is nonsingular, then $v_1,\ldots,v_n$ is LI. Otherwise, we can't be sure.

Am I missing something?

Thanks in advance.

  • $\begingroup$ How singula nature of $A$ affects the linear independency?The proof seems fine to me $\endgroup$
    – SOUL
    Dec 10, 2018 at 12:56

1 Answer 1


We want to check if $v_i$ are linearly independent, hence we form the equation that we use to check linearly independence.

$$\sum_{i=1}^n c_i v_i = 0$$

At this moment, I still do not know if all the $c_i$ must be $0$.

Now, let's multiply by $A$.

$$\sum_{i=1}^n c_i (Av_i) = 0$$

Now, I know that all the $c_i$'s are zero since we are told that $\{Av_1, \ldots, Av_n\}$ is linearly independent. Hence $\{ v_1, \ldots, v_n\}$ is linearly independent.

  • $\begingroup$ Thanks a lot. Your explanation helped me see the problem with my reasoning. $\endgroup$ Dec 10, 2018 at 13:16

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