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I need to prove that if a set of vectors $S$ is linearly independent, then $f(S)$ is also independent. Assuming f is the next homomorphism of vector spaces:

$$ f : V \rightarrow W$$

Demonstration attempt

We assume S is the next set of vectors:

$$ S = \left \{ u_{1}, u_{2},...,u_{n} \right \}$$

By definition of a linearly independent set, we know that the only solution to the equation is the trivial one:

$$\alpha_{1} u_{1} + \alpha_{2} u_{2} + ... + \alpha_{n} u_{n} = 0$$

Then,

$$f(\alpha_{1} u_{1}) + f(\alpha_{2} u_{2}) + ... + f(\alpha_{n} u_{n}) = 0 \Rightarrow \\ \alpha_{1} f(u_{1}) + \alpha f(u_{2}) + ... + \alpha f(u_{n}) = 0 \Rightarrow \\ \alpha_{1} = ... = \alpha_{n} = 0$$

It is a linearly independent set. $\blacksquare$

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  • $\begingroup$ I did not downvoted you. Whomever did, must give a reason why they did this. $\endgroup$ – Will M. Nov 7 '18 at 18:03
  • $\begingroup$ By the way, @Carlos. what you are proving is that if $f$ and $S$ are such that $f(S)$ is linearly independent, then $S$ is linearly independant. $\endgroup$ – Will M. Nov 7 '18 at 18:04
  • $\begingroup$ However, if the image is linearly independent, the anti-image is also independent? $\endgroup$ – Carlos Nov 7 '18 at 18:10
  • $\begingroup$ Exactly what I wrote one comment above. $\endgroup$ – Will M. Nov 7 '18 at 18:13
  • $\begingroup$ And the demonstration would be similar to mine? $\endgroup$ – Carlos Nov 7 '18 at 18:16
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The result is false unless you have extra hypothesis on $f.$ The linear function $f(x) = 0$ already shows this.

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  • $\begingroup$ I know that by defining linear transformation that linear dependence if preserved, its demonstration would be similar to what I have tried to pose. So, based on that I have tried to do that proof. $\endgroup$ – Carlos Nov 7 '18 at 17:58
  • $\begingroup$ Can you rephrase that? What are you trying to prove? $\endgroup$ – Will M. Nov 7 '18 at 17:58
  • $\begingroup$ I'm trying to show that if a set is linearly independent, so is its image. When I studied linear transformations in Algebra, I observed that if a vector is linearly dependent, the image of the vector is also linearly dependent. Well, based on that proof, I've tried this one. $\endgroup$ – Carlos Nov 7 '18 at 18:01
  • $\begingroup$ I already showed to you that the stament "if a set is linearly independent, so is its image [under a linear transformation]" is false. So, no, you cannot. $\endgroup$ – Will M. Nov 7 '18 at 18:02
  • $\begingroup$ So, unless the linear application is $f(x)=0$, the image of an independent linear set doesn't have to be, does it? $\endgroup$ – Carlos Nov 7 '18 at 18:05

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