
#1
Apr2712, 04:28 PM

P: 5

I think I am missing a subtle point of the definition of a inner product. All the texts I have seen state <v,v> >= 0
If you have say: v=(1,2i) then <v,v> = 3 (Using the definition where you do the dot product, while conjugating the first term) This is a negative number and defies the above definition of an inner product. Is it that (1,2i) is not in an inner product space and therefore doen't have an inner product? 



#2
Apr2712, 05:59 PM

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P: 26,167

hi 9k9! welcome to pf!
(1,2i).(1,2i) = 1.1 + 2i.2i = 1  4i^{2} = 5 



#3
Apr2812, 05:31 AM

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P: 6,189

Welcome to PF, 9k9!
The inner product for vectors with complex numbers is defined with a complex conjugate (as tt showed). Without it, you have shown yourself that the resulting vector product does not satisfy the axioms for an inner product. 



#4
Apr2812, 10:15 AM

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P: 6,189

Why is <v,v> >= 0 ?
It seems to me that you did not conjugate the first term.
Perhaps you can write out your calculation? 



#5
Apr2812, 10:17 AM

P: 5

Thanks for the welcomes, I see now I didn't read the definiton correctly and I was only trying to conjugate the first element in the vector, rather than the first term in each of the products.




#6
Apr2812, 11:10 AM

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#7
Apr3012, 09:29 PM

P: 11

<v,v>=0 only when v=0
the definition for innerproduct is Ʃ(v[itex]_{j}[/itex])([itex]\overline{v_{j}}[/itex]) for 1≤j≤n where n is the length of vector v note that [itex]\overline{v_{j}}[/itex] is defined as the adjoint, or conjugate transpose when dealing in ℝ, you'll never get <v,v>=0 because it is merely taking the square of each term {(a[itex]_{1}[/itex])[itex]^{2}[/itex]+...+(a[itex]_{k}[/itex])[itex]^{2}[/itex]} for all k[itex]\epsilon[/itex]dim(v) and then taking the square root of that sum √(Ʃa[itex]_k{}[/itex]) which means that the inside sum must be ≥0 else the inner product wouldn't exist because we are in ℝ, but would also never equal zero unless v=0 and then 0[itex]^{2}[/itex]=0 it's the same for ℂ since squaring terms ends up those new terms becoming positive, rather nonzero and nonnegative 



#8
May112, 03:37 AM

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Hi Krovski! Welcome to PF!
it's "<v,v> >= 0" … and the >= is because 9k9 doesn't have a Mac with a key that types "≥" !! 



#9
May112, 05:49 AM

P: 11




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