Intuition behind conjugate symmetry axiom in inner product spaces

194 Views Asked by At

In inner product spaces what is the motivation behind conjugate symmetry axiom?

Well if it is a real inner product I could make sense of symmetry, but here I couldn't make sense of conjugate symmetry

Any insight toward it or any geometric ideas would be helpful

2

There are 2 best solutions below

0
On BEST ANSWER

If we replace conjugate symmetry by symmetry, we will have $\alpha^2\langle u,u\rangle=\langle \alpha^2u,u\rangle=\langle \alpha u,\alpha u\rangle>0$ for every nonzero complex scalar $\alpha$ and nonzero vector $u$. But this is a contradiction as $\alpha^2$ is not necessarily real or positive.

0
On

Not exactly geometric but I find it makes more sense when looking at a standard inner product, like the one on $L^2(\Omega;\mathbb{C})$ $$ \langle f,g \rangle = \int_{\Omega} f(x) \bar{g}(x) dx $$ If it was defined without conjugation of the 2nd (eqv: 1st) argument then $\langle f,f\rangle$ could be negative or complex valued and so wouldn't induce a positive-definite bilinear form, which is necessary for it to induce a norm and other nice things. And a consequence of conjugating the second (or first) argument is the conjugate symmetry.