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Check my proof for quality (Inner Product Space / Subspaces)

  1. Oct 10, 2015 #1

    RJLiberator

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    Gold Member

    1. The problem statement, all variables and given/known data
    S = a non-empty set of vecotrs in V
    S' = set of all vectors in V which are orthogonal to every vector in S
    Show S' = subspace of V

    2. Relevant equations
    Subspace requirements.
    1. 0 vector is there
    2. Closure under addition
    3. Closure under scalar multiplication

    3. The attempt at a solution

    1. 0 is held within S' by definition here
    2. Let v and v' exist within S' and let s exist within S.
    <v+v',s>=<v,s>+<v',s> = 0+0 =0 which is what we want since
    <v,s>=0 and <v',s>=0.
    3. if x is a scalar, then <xv,s> = x<v,s> =x*0 = 0

    therefore we have satisfied all the necessary requirements and have shown that S' is a subspace of V.

    Question: In my notes, the instructor lists a fourth requirement, namely If v exists within S, S a subspace, then -v exists within s.
    Must I prove this as well?
     
  2. jcsd
  3. Oct 10, 2015 #2

    Mark44

    Staff: Mentor

    I wouldn't say "by definition". The 0 vector is orthogonal to every vector in S, so ##0 \in S'##.
    Minor quibble -- "is an element of S'" (##\in S'##) is used more commonly than "is held within".
    The conclusion here is that if ##v, v' \in S'##, then ##v + v' \in S'##. This shows closure of addition.
    The above shows that if ##v \in S'## and x is a scalar, the ##xv \in S'##, showing closure of scalar multiplication.
    -v = (-1) * v, right? If you have already shown that the set is closed under scalar multiplication, then that also includes the case where the scalar is -1.
     
  4. Oct 10, 2015 #3

    RJLiberator

    User Avatar
    Gold Member

    Excellent. It looks good with a few minor touch ups.

    Thank you for your check and analysis of my writing. It is always helpful.

    I see the importance in saying "this shows closure under ____" as that is what we set out to prove.

    Thank you.
     
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