
#1
Mar3012, 01:32 PM

P: 652

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) =a<x,y> 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 <x+y,z>=(x1+y1+x2+y2+x3+y3)(z1+z2+z3) =x1z1 +x2z2+x3z3+y1z1+y2zy3z3 =<x,z>+<y,z> ...? 



#2
Mar3012, 02:02 PM

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P: 20,962

Since the underlying vector space is R^{3}, all you need to show is that <x, y> = <y, x>. 



#3
Mar3012, 02:09 PM

P: 652

Axiom 1 <x,x> >=0 since we have that x_n for n=1,2,3 are in R Anyway, <y,x>=y1x1+y2x2+y3x3 =x1y1+x2y2+x3y3 =<x,y> Is axiom 4 ok? Thanks 



#4
Mar3012, 02:20 PM

HW Helper
Thanks
P: 4,670

Hilbert, Inner ProductYour 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. RGV 



#5
Mar3012, 02:30 PM

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P: 20,962





#6
Mar3012, 02:58 PM

P: 652

<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. 



#7
Mar3012, 03:14 PM

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P: 20,962





#8
Mar3012, 03:49 PM

P: 652





#9
Mar3012, 04:53 PM

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P: 20,962





#10
Mar3112, 04:06 AM

P: 652

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 



#11
Mar3112, 12:55 PM

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P: 20,962

Why do you have the absolute values? Your inner product is defined this way:
<x, y> = x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3} So again, why is x_{1}^{2} + x_{2}^{2} + x_{3}^{2} ≥ 0? Take a closer look at what I said in post # 9. 



#12
Apr312, 12:53 PM

P: 652





#13
Apr312, 01:16 PM

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P: 20,962

Start with <x, x> and expand it, using the definition in post #1. It is really very simple to show that <x, x> ≥ 0. 



#14
Apr312, 02:51 PM

P: 652

=x1^2+x2^2+x3^3 but x1x1>=0, x2x2>=0 and x3x3>=0 since x1,x2,x3 are in R implies <x,x>>=0 



#15
Apr312, 02:59 PM

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P: 20,962

Yes.




#16
Apr312, 03:22 PM

P: 652




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