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:

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


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

It is a linearly independent set. $\blacksquare$

  • $\begingroup$ 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

The result is false unless you have extra hypothesis on $f.$ The linear function $f(x) = 0$ already shows this.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.