View Full Version : Linear indepedent functions and complex conjugation
jdstokes
Sep12-08, 10:48 PM
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\}?
morphism
Sep13-08, 12:44 AM
Well, what are your thoughts on the matter?
jdstokes
Sep13-08, 01:04 AM
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.
morphism
Sep13-08, 01:32 AM
What if we take, say, the set \{i f, f\}, where f is some real-valued function?
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