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Homework Help: Factoring/Proving an inequality

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data
    ax^2 + 2bxy + cy^2 such that x,y are variables and a,b,c are scalars.
    Show that ax^2 + 2bxy + cy^2 ≥ 0 and =0 iff x,y=0 and iff b^2 - ac < 0 and a > 0

    2. Relevant equations

    3. The attempt at a solution

    This is actually a differential geometry problem, but I have it worked out all the way to this point. I'm proving the 4th axiom of inner products for this particular inner product and I just cannot find a factorization to help prove this inequality. If any of this doesn't make sense, just say so and I may be able to clarify.
  2. jcsd
  3. Sep 7, 2009 #2
    I see the b² - ac < 0 inequality. It is an possibility that use of http://en.wikipedia.org/wiki/Discriminant" [Broken] is relevant for the inequality.
    Last edited by a moderator: May 4, 2017
  4. Sep 7, 2009 #3
    Am I parsing the logical conditions wrong, or is that incorrect?

    If x,y=0 then a,b,c can be anything, right?
  5. Sep 8, 2009 #4
    Hint: Find the discriminant D of:

  6. Sep 8, 2009 #5


    Staff: Mentor

    I agree with PhaseShifter. I believe the problem has been stated incorrectly. If x = 0 and y = 0, then ax2 + bxy + cy2 = 0 for all values of a, b, and c.
  7. Sep 8, 2009 #6
    I think he meant:
    [itex]ax^2 + 2bxy + cy^2 = 0[/itex] if x,y=0 and [itex]ax^2 + 2bxy + cy^2 > 0 [/itex] if [itex]b^2 - ac > 0[/itex] and a > 0
  8. Sep 8, 2009 #7


    Staff: Mentor

    That's not what the OP wrote, and it doesn't make much sense. Again, if x and y are zero, then ax^2 + 2bxy + cy^2 is identically zero for all values of a, b, and c.
  9. Sep 8, 2009 #8
    That's where I'm getting hung up. "if and only if x,y=0, and if and only if b^2 - ac < 0 and a > 0"

    I think one of those ANDs should be an OR, or the IFFs should only be IF. (because it's easy to think of examples that work even when some of the "only if" parts aren't true)

    here are the conditions I'm coming up with where the inequality holds:

    if x,y=0
    or x=0 and c>=0
    or y=0 and a>=0
    or b^2-ac<=0 and a>=0
    Last edited: Sep 8, 2009
  10. Sep 8, 2009 #9

    If we can factor the equation because of [itex]\sqrt{D}[/itex], [itex]b^2-ac[/itex] must be positive, in other case it would be complex number.

    [tex]x_{1,2}=\frac{-2by \pm \sqrt{D}}{2a}[/tex]

    I guess this task is a total mess. :smile:
  11. Sep 8, 2009 #10


    Staff: Mentor

  12. Sep 8, 2009 #11
    Oh wow, that turned out poorly. My apologies to all.

    Let me first start with the original question as worded in my text.

    Show that the function
    <x,y>= (1x2)matrix[x1 x2](2x2)matrix:{a11=a, a12=a21=b, a22=c](2x1)matrix[y1 y2] is an inner product on R2 if and only if b2-ac < 0 and a > 0.

    Sorry about the poor matrix set up, have yet to take the time to read how to do that.
    This function obviously passes the <x,y>=<y,x>, <x,y+z>=<x,y>+<x,z>, and <cx,y>=c<x,y>.
    The last one is proving <x,x> > 0 if x≠0.

    Clearly <x,x> = ax12 + 2bx1x2 + cx22. My approach was to somehow factor this into only having squares and somehow relying on the "b2-ac < 0 and a > 0". I however got stumped and so my poorly original question came about.

    Hope this helps clarify.
    Last edited: Sep 9, 2009
  13. Sep 8, 2009 #12
    Other than the fact that there seem to be exceptions when a=0 or c=0, you seem to have it. The trinomial can be factored into a product of two binomials, and you can then decide on a case by case basis...each of the factors is either positive, negative, or zero, giving nine cases, and the product is greater than zero only when both factors are negative or both are positive.

    The problem is when either [tex](x_{1}=0,x_{2}\neq0)[/tex] or [tex](x_{1}\neq0,x_{2}=0)[/tex], because then the resulting product can be equal to zero depending on the values of a and c.

    Apparently you also need the constraint that a,c≠0.
  14. Sep 8, 2009 #13
    From b2-ac < 0 and a > 0 You can gather that a,c ≠0 because
    b2-ac < 0 -> 0 ≤ b2 < ac -> 0 < ac, but a >0, so c > 0.

    And also, I don't believe that this proof should require 9 cases, that seems a bit more lengthy than we would normally be asked to do.
  15. Sep 8, 2009 #14
    I'm really just looking for the factorization of that expression into squares to just show that is ≥ 0. I just could not see a suitable factorization with the given constraints to make it work.
  16. Sep 8, 2009 #15
    You mean solving for the eigenvectors of the matrix and using them as a basis?
  17. Sep 9, 2009 #16
    I'm not the most familiar with the uses of eigenvectors and have definitely not seen a connection to them at the moment. If there is a way to use them in the problem, please enlighten me.
  18. Sep 9, 2009 #17
    Actually what I was thinking about the eigenvectors wouldn't work unless the eigenvectors were orthogonal.
  19. Sep 9, 2009 #18
    I just now saw this post, my apologies. I originally tried this, but the text does in fact say b2-ac<0. Which made this useless to me. It is possible it is a typo in text. Anyone know of a good source to find out corrections to typos in texts?
  20. Sep 10, 2009 #19
    Have you tried checking the publisher's website for errata?
  21. Sep 12, 2009 #20
    Well, the book is correct. I found out that the preferred method of proof uses eigenvalues/eigenmatrix. As of now, I have not found out how to utilize this, but I haven't really had a chance to think more about it.
  22. Sep 15, 2009 #21
    So I was correct in thinking that it could be factored to prove that it is greater than 0.

    ax2 + 2bxy + cy2

    b2-ac<0 and a>0 -> c>0
    and hence, -(ac)1/2<b<(ac)1/2

    ax2 - 2(ac)1/2xy + cy2 < ax2 + 2bxy + cy2 < ax2 + 2(ac)1/2xy + cy2

    Hence, 0 ≤ (a1/2x - c1/2y)2 < ax2 + 2bxy + cy2 < (a1/2x + c1/2y)2

    Hence, ax2 + 2bxy + cy2 > 0 for all x,y≠0 and
    ax2 + 2bxy + cy2 = 0 only if x,y=0
  23. Sep 17, 2009 #22
    I have to take issue with the "only if" part.

    It's still not accounting for the cases (x=0, c=0) , (y=0,a=0) or where the discriminant is equal to zero and x=y.

    In all three cases ax^2 + 2bxy + cy^2 = 0 without meeting the requirement x,y=0.

    x,y=0 is therefore sufficient but not necessary for the expression to equal zero.
    Last edited: Sep 17, 2009
  24. Sep 17, 2009 #23
    But there is the restriction of a>0 and c>0.

    And if x=y and a=c to make the first expression equal to 0, then the fact that it is strictly less than the expression I'm dealing with gives me the desired solution.
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