Why <v,v> >= 0? Understanding Inner Product Definition

[9k9]

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?

 

[tiny-tim]

welcome to pf!

hi 9k9! welcome to pf!

(1,2i).(1,-2i) = 1.1 + 2i.-2i = 1 - 4i² = 5

 

[I like Serena]

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.

 

[I like Serena]

It seems to me that you did not conjugate the first term.
Perhaps you can write out your calculation?

 

[9k9]

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

 

[Stephen Tashi]

Using the definition where you do the dot product, while conjugating the first term

What do you mean by "the first term"? State the definition you are using.

 

[Krovski]

<v,v>=0 only when v=0

the definition for inner-product is Ʃ(v$_{j}$)($\overline{v_{j}}$) for 1≤j≤n where n is the length of vector v
note that $\overline{v_{j}}$ 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$_{1}$)$^{2}$+...+(a$_{k}$)$^{2}$}
for all k$\epsilon$dim(v) and then taking the square root of that sum
√(Ʃa$_k{}$) 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$^{2}$=0

it's the same for ℂ since squaring terms ends up those new terms becoming positive, rather non-zero and non-negative

 

[tiny-tim]

Welcome to PF!

Hi Krovski! Welcome to PF!

<v,v>=0 only when v=0

Ah, no … that fooled me at first, too …

it's "<v,v> >= 0" …

and the >= is because 9k9 doesn't have a Mac with a key that types "≥" ! ​

 

[Krovski]

Hi Krovski! Welcome to PF!


Ah, no … that fooled me at first, too …

it's "<v,v> >= 0" …

and the >= is because 9k9 doesn't have a Mac with a key that types "≥" ! ​

good catch and thank you