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Linear Algebra. Proving differentiable functions are a vector space.

  1. Mar 4, 2012 #1
    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.
     
  2. jcsd
  3. Mar 4, 2012 #2

    tiny-tim

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    welcome to pf!

    hi datran! welcome to pf! :smile:
    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!)
     
  4. Mar 4, 2012 #3
    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!
     
  5. Mar 5, 2012 #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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