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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

  • For real inner product spaces the norm describes length. If we simply use the dot product for $\Bbb C^n$, though, we don't get a positive-definite form. Consider the $n=1$ case, i.e. the complex plane. Here, the length satisfies $|z|^2=\bar{z}$z, which suggests we ought to use $\langle w,z\rangle=\bar{w}z$ instead of $wz$. Indeed, if we turn any orthonormal basis $\cal B$ for $\Bbb C^n$ into one ${\cal B}\sqcup i{\cal B}$ for $\Bbb R^{2n}$, the induced norm on $\Bbb R^{2n}$ matches this one on $\Bbb C^n$. – anon Jun 09 '21 at 01:01

2 Answers2

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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.

cohomonoid
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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.

user1551
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