Why do we require "conjugate symmetry" in the definition of an "inner product" instead of "symmetry"? i.e. I don't understand the motivation of $<a,b>=\overline{<b,a>}$ in the definition of the inner product. I think $<a,b>=<b,a>$ is more natural. I think the axiomatic definition of inner product on a vector space $V$ came from an elementary model. What is the model? I tried to think of the product between complex numbers, but it satisfies "symmetry" instead of "conjugate symmetry".
Asked
Active
Viewed 5,719 times
1
-
otherwise, $\langle x, x \rangle$ will not be real in general, let alone non negative. – user251257 Aug 10 '15 at 23:26
-
@user251257 So the product between complex numbers is not an inner product, but $z$ times $\bar{w}$ is an inner product? – ZTD Aug 10 '15 at 23:41
-
1yes. for illustration, notice $|z|^2 = z\bar z \ne z^2$ for a complex (not real) number $z$. – user251257 Aug 10 '15 at 23:44
-
@user251257 OK. Then I guess this is the basic model of inner product. But it seems that we don't use $z\bar{w}$ very often unless $z=w$ – ZTD Aug 11 '15 at 00:00
-
the standard inner product on $\mathbb C^n$ is defined by $\langle x, y \rangle = \sum_{i=1}^n x_i \bar y_i$! – user251257 Aug 11 '15 at 00:01
1 Answers
1
An inner-product space may be defined over both real complex planes. Remember for a real vector space $V$, the conjugates of vectors $\mathbf a, \mathbf b$ in $V$ are just $\mathbf a, \mathbf b$ themselves. So if you are using $V$ to define your inner-product space conjugate symmetry is just symmetry $\langle \mathbf a,\mathbf b\rangle = \langle \mathbf b,\mathbf a\rangle$.
As for the complex portion, I think you are getting tripped up with Drac notation. Note that $\langle \mathbf a| = \mathbf a^*$ and $|a\rangle = \mathbf a$
As an example, define $\mathbf a = e^{-i2\phi}$ and $\mathbf b = e^{i\phi}$ then $$\langle a|b\rangle =e^{i3\phi}$$ while $$\langle b|a\rangle=e^{-i3\phi}$$
Martins Bruveris
- 1,620
amo
- 11