# Linear independence in linear applications

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. $$\blacksquare$$

• I did not downvoted you. Whomever did, must give a reason why they did this. – Will M. Nov 7 '18 at 18:03
• 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. – Will M. Nov 7 '18 at 18:04
• However, if the image is linearly independent, the anti-image is also independent? – Carlos Nov 7 '18 at 18:10
• Exactly what I wrote one comment above. – Will M. Nov 7 '18 at 18:13
• And the demonstration would be similar to mine? – 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.
• So, unless the linear application is $f(x)=0$, the image of an independent linear set doesn't have to be, does it? – Carlos Nov 7 '18 at 18:05