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Linear Algebra- Subspaces

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine whether the following sets form subspaces of R^2 :

    a) {(x1,x2)T | x1*x2=0}

    b) {(x1,x2)T | x12=x22}

    c) {(x1,x2)T | |x1|=|x2| }


    2. Relevant equations



    3. The attempt at a solution

    My problem here is that I don't think I understand how the vectors look.
    for instance,
    {(x1,x2)T | x1*x2=0}

    I say that each vectors looks like this:
    [c,0]T
    why zero ? because x2=0/x1
    and x1=c.
    Now, if this is correct, it must be a subspace. because we can multiply by a constant and still be in the subspace, or add two vectors and be in the subspace. But, its not a subspace according to the answers.

    The problem with the other two is that I don't even know how the vectors in the subspaces look.

    Would appreciate any help.
    Thanks,
    Roni.
     
  2. jcsd
  3. Mar 30, 2010 #2
    Yes. (ab=0) <=> ((a=0) or (b=0)) is true.

    Note the "or".
    Can you think of another general form of vectors in this set?
    Then try add them (this isn't an uncommon course of action, ie, to guess two things which add and end up outside your set).
     
  4. Mar 30, 2010 #3
    I don't think I understand you.
    How do you set the vectors ? this is the part that I don't understand.
    I know how to test whether it forms a subspace or not once I have the general form of the vectors in the subspace.
     
  5. Mar 30, 2010 #4
    still need help :\

    thanks.
     
  6. Mar 30, 2010 #5
    After reading your post again, I understand that I have to do something with the 'OR' but I don't really know how to set a matrix with OR .

    I just came across another similar question:

    {(x1,x2,x3)T | x3=x1 OR x3=x2 }

    Don't really know what to do here.
     
  7. Mar 30, 2010 #6

    Mark44

    Staff: Mentor

    Neither problem has anything to do with matrices. What's more, your question about "set(ting) a matrix with OR" doesn't make any sense.

    The set in the first problem describes vectors in R2.
    {(x1,x2)T | x1*x2=0}

    The point the other poster was making was that if x1 * x2 = 0, then x1 = 0 or x2 = 0 (which includes the possibililty that both are zero). What do these vectors look like in R2? If you add any two vectors in this set, do you get another vector in this set? If you multiply any vector in this set by a scalar, do you get a vector in this set?

    For the second problem, the set consists of vectors in R3. To get you started with this one, see if you can find some vectors in this set.
     
  8. Mar 30, 2010 #7
    YEA I MEANT VECTORS, SORRY...
    I HAVE THESE TYPES OF VECTORS [0,B]^T 0OR [C,0]^T
    NOW IF I ADD ANY VECTOR OF TYPE [0,B]^T +[C,O]^T, I GET A VECTOR OUTSIDE OF THE SUBSPACE, I GET [C,B]^T....
    THE THING IS THAT I WAS TRYING TO ADD [B,0]+[C,O]... AND IT DIDN'T WORK..

    THANKS FOR THE HELP...

    NOW HOW DO THE VECTORS LOOK HERE:
    b) {(x1,x2)T | x12=x22}

    HERE THEY MUST BE SQUARED ?
    SAY
    [C2,D2]^T ?

    THANKS FOR YOUR HELP.
    (SORRY FOR THE CAPS LOCK )
     
    Last edited: Mar 30, 2010
  9. Mar 30, 2010 #8

    Mark44

    Staff: Mentor

    So what does that tell you about addition in this set?
    <B, 0> + <C, 0> = <B + C, 0>, which is a vector in this set.
    The set is {(x1,x2)T | x12=x22}
    Can you find some specific vectors that belong to this set?
    Maybe, or maybe not. Look for some specific vectors in this set.
     
  10. Mar 30, 2010 #9
    i got this one, it shows that it doesn't form a subspace, which is the correct answer

    right, i was doing only this type of vectors and was getting an answer that this is a subspace, which is incorrect.

    the thing that confuses me here is that, the vectors are of this form ? :
    <c^2,d^2 > ?
    We square both terms ? I think we do...

    Say
    <2^2,2^2> would be in the set ?
     
  11. Mar 30, 2010 #10

    Mark44

    Staff: Mentor

    The thing is, all vectors in the set have to add to another vector in the set. In your first example, <0, b> + <c, 0> don't add to a vector in the set, so even though some selected vectors add to a vector in the set, there are some that don't.
    Yes, this vector is in the set. A vector <x1, x2> is in the set iff x12 = x12.

    Here are some other vectors. Can you tell which of these are in the set?
    <1, 2>
    <2, 2>
    <3, -3>
    <9, 16>
    <-3, -3>
    <0, 0>

    Keep in mind that you aren't to the point where you're deciding whether the set is a subspace. You're at the point where you're trying to find out which vectors are in the set.
     
  12. Mar 30, 2010 #11
    Oh I see...
    then,
    <2, 2>
    <3, -3>
    <-3, -3>
    <0, 0>

    are in the set....
    but if we add 2 vectors
    say
    <3, -3>
    <-3, -3>
    we get a vector outside of the set.Therefore, it doesn't form a subspace.

    Thanks very much for your help.
    take care....
     
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