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Prove Sp{(a,b),(c,d)} = R^2 if and only if ad-bc≠0

  1. Nov 26, 2012 #1
    Hi,

    1. The problem statement, all variables and given/known data

    I am trying to prove that Sp{(a,b),(c,d)} = R^2 if and only if ad-bc≠0.
    I am wondering whether the proof below would be considered rigorous.
    NB. I am not permitted to make use of number of dimentions.

    2. Relevant equations



    3. The attempt at a solution

    First direction:
    Let Sp{(a,b),(c,d)} = R^2
    Hence, (x,y) = alpha(a,b) + beta (c,d)
    Hence, x=a*alpha + c*beta; y=b*alpha + d*beta
    Hence, beta(ad-bc) = xb-ya
    In order that Sp{(a,b),(c,d)} = R^2 and any vector in R^2 could be represented as a linear combination of ((a,b),(c,d)) there has to be a solution for beta, i.e. ad-bc cannot be zero.

    Second direction: I am going to try to show that provided that ad-bc≠0, Sp{(a,b),(c,d)}=R^2.
    Supposing ad=bc, then (c,d) = (c/a)(a,b) for a≠0, or (c,d) = (d/b)(a,b) for b≠0.
    Hence, {(a,b),(c,d)} is linearly dependent over R^2.
    Hence, Sp{(a,b),(c,d)} = Sp{(a,b)} whilst a≠0, which is not equal to R^2, OR, Sp{(a,b)} whilst b≠0, which is also not equal to R^2.
    Hence, ad-bc≠0 => Sp{(a,b),(c,d)} = R^2

    Is the above sufficient? Is the proof correct?
     
  2. jcsd
  3. Nov 26, 2012 #2
    I'm assuming Sp{x,y} means span of.
    Is invertibility allowed? Since given [x y] invertible, you can show that (0,1) and (1,0) are in the span of {x,y} and thus all combinations of (0,1) and (1,0) are in the span (R2)
     
  4. Nov 26, 2012 #3
    Invertibility is allowed.
    Nevertheless, could you please comment on the proof suggested above?
     
    Last edited: Nov 26, 2012
  5. Nov 26, 2012 #4

    Mark44

    Staff: Mentor

    For the above, you should add something like
    "Every vector (x, y) in R2 can be written as a linear combination of (a, b) and (c, d)."
    Hence (x, y) = α(a, b) + β(c, d) for some constants α and β.
    It might be helpful here to explain what you're doing, which is to multiply the equation for x by b, and multiply the equation for y by a, and then subtracting the two equations.

    Then you could solve for β, getting
    β = (ax - by)/(ad - bc)
     
  6. Nov 26, 2012 #5
    Thank you very much! Yet what about the second part of the proof?
     
  7. Nov 26, 2012 #6

    Mark44

    Staff: Mentor

    The second part looks OK.
     
  8. Nov 26, 2012 #7
    That's the wrong way to start. You need to assume that ad-bc≠0, not =0.

    Actually it might be better to assume that Sp{(a,b),(c,d)} = R^2 and deduce that ad-bc can't have been ≠0 after all, i.e. instead of proving

    ad-bc ≠ 0 [itex]\Rightarrow[/itex] Sp{(a,b),(c,d)} = R^2

    you prove

    Sp{(a,b),(c,d)} ≠ R^2 [itex]\Rightarrow[/itex] ad-bc = 0
     
  9. Nov 26, 2012 #8
    But wouldn't then be exactly like the first direction?
     
  10. Nov 26, 2012 #9

    Mark44

    Staff: Mentor

    Peripatain is doing a proof by contradiction, which is a valid way to do things.
     
  11. Nov 26, 2012 #10
    True, but Peripatain has still proven the same thing twice, once
    Sp{(a,b),(c,d)}=R^2 [itex]\Rightarrow[/itex] ad-bc≠0
    and then
    ad-bc=0 [itex]\Rightarrow[/itex] Sp{(a,b),(c,d)}≠R^2.

    Adding "Hence, ad-bc≠0 => Sp{(a,b),(c,d)} = R^2" was wrong, this does not follow from either proof.

    This situation is like first proving
    "if you read good books [itex]\Rightarrow[/itex] you will grow wise"
    and then
    "if you're not wise [itex]\Rightarrow[/itex] you obviously didn't read good books".

    These are not equivalent though, since you could be wise without reading books at all (for instance by listening to advice from others).
     
  12. Nov 26, 2012 #11
    Wouldn't proving it the way you suggested be the same proof as the in first direction?
     
  13. Nov 26, 2012 #12

    Mark44

    Staff: Mentor

    You are correct. I confess that I didn't look at the 2nd part that carefully.

    Peripatain, you need to start by assuming that ad - bc ≠ 0, and then showing that the two vectors span R2.
     
    Last edited: Nov 26, 2012
  14. Nov 26, 2012 #13
    Assume ad-bc≠0. Then you can solve any equation of the kind you mentioned in the first part, that is anything like β(ad-bc) = xb-ya, no matter how you choose x and y. And that means that every vector (x,y) lies in Sp{(a,b),(c,d)}. Actually the two proofs are very similar, only the wording differs a bit:

    (1) R^2=Sp{...} => every point (x,y) fulfils the equations... => β(...) = ... must have a solution => ad-bc≠0.

    (2) ad-bc≠0 => β(...)=... always has a solution, no matter how x,y are chosen => any point (x,y) lies in Sp{...} => Sp{...}=R^2.
     
    Last edited by a moderator: Nov 26, 2012
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