Linear indepedent functions and complex conjugation

1. Sep 12, 2008

jdstokes

Suppose $\{\varphi_i\}$ is an infinite set of linearly independent functions. Is $\{ \varphi_i^\ast \}$ linearly indepedent? How about $\{ \varphi_i \} \cup \{ \varphi_i^\ast\}$?

2. Sep 13, 2008

morphism

Well, what are your thoughts on the matter?

3. Sep 13, 2008

jdstokes

Well, if $\{ \varphi_i \}$ is LI then $\{ \varphi_i^\ast \}$ is also trivially LI, because if it wasn't then you could just take the complex conjugate violating linear indepedence of $\{ \varphi_i \}$.

It's not clear if my second claim is true, although I'd like it to be, I suspect there are counterexamples waiting to be found. I'd like to be proven wrong, however.

4. Sep 13, 2008

morphism

What if we take, say, the set $\{i f, f\}$, where f is some real-valued function?