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Homework Help: Linear Algebra - Subspaces proof

  1. Feb 9, 2006 #1
    Hello, just wondering if my proof is sufficient.

    Here is the question from my book:

    Show that the following sets of elements in R2 form subspaces:
    (a) The set of all (x,y) such that x = y.
    -------

    So if we call this set W, then we must show the following:
    (i) [tex]0 \in W[/tex]
    (ii) if [tex] v,w \in W[/tex], then [tex]v+w \in W[/tex]
    (iii) if [tex]c \in R[/tex] and [tex]v \in W[/tex] then [tex]cv \in W[/tex]

    Pf:
    (i) [tex]0 \in W[/tex] because we can take x = 0 = y
    (ii) if [tex]v,w \in W[/tex] then [tex](v,v) \in W[/tex] and [tex](w,w) \in W[/tex] and [tex](v,v) + (w,w) = (v + w, v + w) \in W[/tex] so [tex]v+w \in W[/tex] because v + w = x = y = v + w
    (iii) if [tex]c \in R[/tex] and [tex]v \in W[/tex], then [tex](v,v) \in W[/tex] and [tex]c(v,v) = (cv,cv) \in W[/tex] so [tex]cv \in W[/tex] because cv = x = y = cv
    Therefore W is a subspace.

    Does that look just fine? Thanks.
     
    Last edited: Feb 9, 2006
  2. jcsd
  3. Feb 9, 2006 #2
    Yah, looks good to me.
     
  4. Feb 9, 2006 #3
    Cool, thanks!
     
  5. Feb 10, 2006 #4

    matt grime

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    No, that's not right, but the idea is correct. You state v is in W but then write the vector (v,v) is in W. The v in the brackets cannot be the v outside the brackets. You need to change that, v canont simultaneously be a vector in R^2 and elements of R.
     
  6. Feb 10, 2006 #5
    Thanks matt, I think I am seeing what you are saying. So if I said V is in W, then (v,v) is in W, that would be correct? Because big V is not the same as little v. Thanks.
     
  7. Feb 10, 2006 #6

    HallsofIvy

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    Yes, that would work. Although it might be even better to say:
    V= (x,x), U= (y,y).
     
  8. Feb 10, 2006 #7

    matt grime

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    well, (v,v) is in W because of the definition of W, and is not deduced from V is in W. what you ought to say is if V in W then V=(v,v) for some v. there is no need to actually use V. W is closed under addition because (v,v)+(x,x)=(v+x,v+x).
     
  9. Feb 10, 2006 #8

    JasonRox

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    Just for future reference, you can cut down the work for the proof for a subspace, simply by showing...

    0 e W
    cx+y e W, for any scalar c (depending on the field).
     
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