Linear Algebra. Proving differentiable functions are a vector space.

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  • #1
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Question: Show the set of all differentiable functions on (-infinity, +infinity) that satisfy f′ + 2f = 0 is a vector space.

I started the problem by assuming that f and g are both differentiable functions that satisfy this vector space.

Then I ran through the ten axioms of addition and scalar multiplication and proving that each one works.

I feel like that does not answer the question though since why would I need the equation f' + 2f = 0?

How does that equation come into play?

Thanks for any help provided.
 

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  • #2
tiny-tim
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welcome to pf!

hi datran! welcome to pf! :smile:
I feel like that does not answer the question though since why would I need the equation f' + 2f = 0?

How does that equation come into play?

you have to prove eg that (f+g) satisfies that equation :wink:

(yes, i know it's obvious … but you still have to prove it!)
 
  • #3
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Oh!

So I would do (f+g) = (f+g)' + 2(f+g) = 0

and same thing over and over for the 10 axioms.

So really f and g are like variables?

Thank you so much! That actually made many more problems clearer!
 
  • #4
Fredrik
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You can start by proving that the set of all differentiable functions from ℝ to ℝ with the standard definitions of addition and scalar multiplication is a vector space. (Looks like you've done that already). Denote this space by V. Define U={f in V|f'+2f=0}. U is by definition a subset of V. If you prove that U contains the 0 function and is closed under addition and scalar multiplication, you can conclude that U is a subspace of V.
 

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