1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    (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

    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper
    Gold Member

    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).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Linear Algebra - Subspaces proof
  1. Linear Algebra Proof (Replies: 8)

  2. Linear Algebra proof (Replies: 21)