# Why is <v,v> >= 0 ?

by 9k9
Tags: <v, v>
 P: 5 I think I am missing a subtle point of the definition of a inner product. All the texts I have seen state >= 0 If you have say: v=(1,2i) then = -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?
 Sci Advisor HW Helper Thanks P: 26,148 hi 9k9! welcome to pf! (1,2i).(1,-2i) = 1.1 + 2i.-2i = 1 - 4i2 = 5
 HW Helper P: 6,187 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.
 HW Helper P: 6,187 Why is >= 0 ? It seems to me that you did not conjugate the first term. Perhaps you can write out your calculation?
 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.
P: 3,297
 Quote by 9k9 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.
 P: 11 =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 =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