Question on a linear algebra formula for **inner product**

Question

I am currently studying linear algebra, specifically the orthogonality of a set of vectors. I learnt that the inner product of the 2 vectors in the field $\mathbb {C}$ can be expressed by the sum of multiplication of the one's counterpart with another's conjugated one.

$$ <u,v> = \sum a_i.\bar bi $$

This formula can apply to a set of $\mathbb {R}$ too. My question is why in the field $\mathbb{C}$ we need to use the conjugated vector instead of the normal one.

Thank you in advance

Answer 1

The reason why we use conjugated vectors in complex inner products is because there is a property we need verified with inner products : Hermitian symmetry.

Hermitian symmetry is defined as such (here, $E$ is an inner product space over $\mathbb{F}$):

For all $(x,y) \in E^2, \left\langle x,y \right\rangle = \overline{\left\langle y,x \right\rangle}$

This property makes $\left\langle x,y \right\rangle_? = \sum x_i y_i$ improper because it would be symmetrical. One could argue that it would be better for an inner product to be symmetrical, but Hermitian symmetry allows us to do this :

$ \left\langle x,x \right\rangle = \overline{\left\langle x,x \right\rangle}$

Therefore $ \left\langle x,x \right\rangle \in \mathbb{R} $

And on this consequence of Hermitian symmetry and another property of inner products relies the concept of canonical norm for an inner product, and on top of a norm, we can define the notion of distance between two vectors, of angle between two vectors, and then we can follow with the bases of topology on inner product spaces.