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Homework Help: Intro to Differential equations: Vector Spaces

  1. Oct 21, 2014 #1
    Good evening everyone, I hope everyone is having a better evening than myself thanks to this homework problem.

    Let P be the set of positive numbers. For a,b in P, define a+b=a x b; for a in P and a real number r, define r x a= a^r. Show that P is a vector space using ⊕ as addition and (circle dot) as scalar multiplication.
    For a,b in P, define ab=a x b; for a in P and a real number r, define r x a= a^r.
    My professor wants us to use these properties that are in our text,

    1. X+Y=Y+X.
    2. (X+Y)+Z=X+(Y+Z).
    3. 0+X=X+0=X.
    4. r(sX)=(rs)X.
    5. (r+s)X=rX+sX.
    6. r(X+Y)=rX+rY.
    7. 1X=X.

    My professor introduced our class to the topic of a vector space today and when he was talking about it everything made sense. Now that I am here on my own I honestly do not know where to start. Unfortunately I was unable to go to his office hours today to ask him about it.

    A brief and general description of where I should start is all I ask.
  2. jcsd
  3. Oct 22, 2014 #2


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    You mean ##r \odot a = a^r##

    Take one step at a time. Look at property 1. What is ##x\oplus y##? What is ##y\oplus x##? Are they equal? Just continue.
    Last edited: Oct 22, 2014
  4. Oct 22, 2014 #3


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    One of the standard conditions is missing from your list. It's usually stated after 3 but before 4. It goes like this: X⊕(-X)=(-X)⊕X=0.

    A typical beginning linear algebra student will find the axioms that involve 0 or -X confusing. Note that the full statement of axiom 3 goes like this: There's a Z in P such that for all X in P, we have X⊕Z=Z⊕X=X.

    Such a Z is said to be an identity element of ⊕. So axiom 3 is telling you that ⊕ has an identity element in P. It's easy to show that ⊕ has at most one identity element (see below) so axiom 3 is really telling you that ⊕ has exactly one identity element. This identity element is denoted by 0, because that's the standard notation for the identity element of a commutative binary operation that's denoted by a symbol that's similar to a plus sign. But the identity element of ⊕ is NOT the number 0. Note that (the full statement of) axiom 3 says that 0 (the identity element) is an element of P, so it's a positive real number. You will have to figure out which one.

    Similarly, the full statement of the forgotten axiom that I included at the start of this post is: For each X in P, there's a Y in P such that X⊕Y=Y⊕X=0. Here 0 denotes the identity element of ⊕, not the number 0. A positive real number Y such that X⊕Y=Y⊕X=0 is said to be an inverse of X. It's easy to prove that each element of P has at most one inverse. So the axiom is really telling us that each element of P has exactly one inverse. The standard notation for the inverse of X is -X.

    Here's the proof of the claim that there's at most one identity element in P. Suppose that there are two. Denote them by 0 and 0'. Since 0'⊕X for all X in P and 0 is in P, we have 0'⊕0=0. Since X⊕0=X for all X in P, and 0' is in P, we have 0'⊕0=0'. So 0'=0'⊕0=0.
    Last edited: Oct 22, 2014
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