# Solution for Linear Transformation question seems wrong.

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 $$Av_1,\ldots,Av_n$$:

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

Thus,

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

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?

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

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.

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