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Hilbert, Inner Product

  1. Mar 30, 2012 #1
    I have just realized that I accidently put it in wrong sub forum. This should be in 'calculus and beyond'.

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

    Prove the function <x,y>=x_1y_1+x_2y_2+x_3y_3 defines an inner product space on the real vector space R^3 where x=(x1,x2,x3) and y=(y1,y2,y3)

    3. The attempt at a solution

    Axiom 1 <x,y> >=0 since we have that x_n and y_n for n=1,2,3 are in R

    Axiom 2a <x,y> =x1y1+x2y2+x3y3, then <x,y>=0 iff x_n and y_n for n=1,2,3 both =0

    Axiom 2b <ax,y>=a<x,y>

    <ax,y> = ax1y1+ax2y2+ax3y3
    = a(x1y1+x2y2+x3y3)

    Axiom 3 <y,x>= complex of <x,y>

    <y,x>=(y1x1+y2x2+y3x3) but y complex =y and x complex=x in R therefore
    = (y1x1 complex+y2x2 complex +x3y3 complex)
    =<y,x> complex
    =<x,y> complex

    Axiom 4 <x+y,z>=<x,z>+<y,z>, let z=(z1,z2,z3) in R^3

    =x1z1 +x2z2+x3z3+y1z1+y2zy3z3
    Last edited: Mar 30, 2012
  2. jcsd
  3. Mar 30, 2012 #2


    Staff: Mentor

    This isn't Axiom 1. It has to do with <x, x>.
    Don't you mean "conjugate transpose"?
    Since the underlying vector space is R3, all you need to show is that <x, y> = <y, x>.
  4. Mar 30, 2012 #3
    Oh I see,

    Axiom 1 <x,x> >=0 since we have that x_n for n=1,2,3 are in R

    Yes, the bar on top of them.


    Is axiom 4 ok?

  5. Mar 30, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your statement "Axiom 1 <x,y> >=0" is NOT a property of an inner product; we can have <x,y> > 0, = 0 or < 0. However, we do have <x,x> >= 0, with <x,x> = 0 iff x is the zero vector. Can you prove that?

    Your statement "Axiom 2a <x,y> =x1y1+x2y2+x3y3, then <x,y>=0 iff x_n and y_n for n=1,2,3 both =0" is FALSE: <x,y> = 0 means only that the vectors x and y are perpendicular to one another.

    Your statement "Axiom 3 <y,x>= complex of <x,y>" is meaningless; perhaps you mean "complex conjugate". Anyway, in real space, the components of x and y are all real and <x,y> is real; the property you state is trivially true, because x1*y1 + x2*y2 + x3*y3 = y1*x1 + y2*x2 + y3*x3.

    Your statement of Axiom 4 is correct, but your proof is absolutely incorrect! Go back and read what you did.

  6. Mar 30, 2012 #5


    Staff: Mentor

    You're waving your arms here. Expand <x, x> using the definition of this inner product and it should be obvious why <x, x> >= 0.
    Yes, #4 is fine.
  7. Mar 30, 2012 #6
    <x,x>=x1x1+x2x2+x3x3>=0 where x_n>=0 for n=1,2,3 since x_n is in R

    <x,x>=0 iff x1x1+x2x2+x3x3=0 ie iff x_nx_n=0 for n=1,2,3, iff x_n=0 for n=1,2,3, ie x=x_n=0 the 0 vector.
  8. Mar 30, 2012 #7


    Staff: Mentor

    No, you're still not getting it. Any or all of the coordinates of a given vector can be negative, such as x = [1, -2, 5]. Here x2 < 0, but it's easy to show that <x, x> > 0.
    If the sum of three numbers is zero, why in this case is it necessarily true that all three numbers have to be zero?
  9. Mar 30, 2012 #8
    The product of a negative number times a negative number gives a positive therefore <x,x> will always be greater than 0..?

    Because if at least one of them is non 0 then <x,x> is non 0..?
  10. Mar 30, 2012 #9


    Staff: Mentor

    That's closer. For any real number x, x2 ≥ 0, and x2 = 0 iff x = 0.
    Now, why is x12 + x22 + x32 ≥ 0? This is what you need to establish in order to verify the <x, x> ≥ 0
  11. Mar 31, 2012 #10
    because we have that

    |x1*x1|+|x2*x2|+|x3*x3|>=0 and |x1*x1|+|x2*x2|+|x3*x3|=0 iff each |x_n*x_n|=0 for n=1,2,3
  12. Mar 31, 2012 #11


    Staff: Mentor

    Why do you have the absolute values? Your inner product is defined this way:
    <x, y> = x1y1 + x2y2 + x3y3

    So again, why is x12 + x22 + x32 ≥ 0?

    Take a closer look at what I said in post # 9.
  13. Apr 3, 2012 #12
    Because we have that x=(x1,x2,x3) in R, therefore x^2>=0 in R^3..?
  14. Apr 3, 2012 #13


    Staff: Mentor

    This makes no sense. x is a vector in R3, so x2 is not defined. Therefore you can't say that x2 ≥ 0.

    Start with <x, x> and expand it, using the definition in post #1. It is really very simple to show that <x, x> ≥ 0.
  15. Apr 3, 2012 #14
    but x1x1>=0, x2x2>=0 and x3x3>=0 since x1,x2,x3 are in R
    implies <x,x>>=0
  16. Apr 3, 2012 #15


    Staff: Mentor

  17. Apr 3, 2012 #16
    Thank you.
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