2
$\begingroup$

An inner product $\langle \cdot,\cdot \rangle$ satisfies the following properties:

Let $u$, $v$, and $w$ be vectors and $\alpha$ be a scalar, then:

  1. $\langle u+v,w\rangle=\langle u,w\rangle+\langle v,w\rangle$ and $\langle \alpha v,w\rangle=\alpha\langle v,w\rangle$ (Linearity in first coordinate).

  2. $\langle v,w\rangle=\langle w,v\rangle$ (Symmetry) .

  3. $\langle v,v\rangle\ge 0$ and $\langle v,v\rangle=0$ $\Leftrightarrow$ $v=0$ (Positive definiteness).

This the general definition of inner products, but in complex vector spaces the symmetry part is $\langle u,v\rangle=\overline{\langle v,u\rangle}$, how can be possible? this is supposed to be the same symmetry of the general properties above, btw what's the difinition of $\overline{\langle u,v\rangle}$?

EDIT

I know how to prove $\langle u,v\rangle=\overline{\langle v,u\rangle}$, what I wanna know is why this is called an inner product, because we don't have $\langle u,v\rangle={\langle v,u\rangle}$

Thanks in advance

$\endgroup$
  • $\begingroup$ It's the complex conjugate. And if you restrict $x\mapsto \overline{x}$ to the reals, you get the identity. $\endgroup$ – xavierm02 Apr 15 '14 at 17:04
  • $\begingroup$ Note that in the complex case you can't have both symmetry and positive definiteness. It turns out that positive definiteness is the more useful property - by taking the square root you can give a complex number a length - and this generalises. But to get a positive definite form you need to introduce the complex conjugate. $\endgroup$ – Mark Bennet Apr 15 '14 at 18:18
3
$\begingroup$

There is no way to preserve all of these 3 properties in complex field. We have to choose the least important one to sacrifice.

Linearity is the foundation of linear algebra, and positive definiteness is required to make inner product a metric. Therefore, we abandon symmetry.

$\endgroup$
  • $\begingroup$ but what's the general definition of inner products? $\endgroup$ – user42912 Apr 15 '14 at 18:17
  • $\begingroup$ An inner product space is a normed vector space satisfying the parallelogram law. $\endgroup$ – jdh8 Apr 15 '14 at 18:23
1
$\begingroup$

Since you are talking about the complex inner product (not a normal real inner product), you cannot apply the real inner product properties. If you check the property for complex inner product (the Hermitian inner product), you can find the property you are talking about. In fact, if you apply real numbers on this Hermitian inner product, you get the same result as apply them on the real inner product.

You can have a look at this document.

http://www2.math.umd.edu/~hking/Hermitian.pdf

$\endgroup$
  • 1
    $\begingroup$ Thank you for the link, but it doesn't answer my question, I know how to prove $\langle u,v\rangle=\overline{\langle v,u\rangle}$, my question is the requirement to be an inner product: this is not an inner product because we don't have $\langle u,v\rangle={\langle v,u\rangle}$. $\endgroup$ – user42912 Apr 15 '14 at 17:57
  • $\begingroup$ @user42912 Since you are talking about the complex inner product (not a normal real inner product), you cannot apply the real inner product properties. If you check the property for complex inner product (the Hermitian inner product), you can find the property you are talking about. In fact, if you apply real numbers on this Hermitian inner product, you get the same result as apply them on the real inner product. $\endgroup$ – Yilun Zhang Apr 15 '14 at 18:04
  • $\begingroup$ what's the general definition of inner products? $\endgroup$ – user42912 Apr 15 '14 at 18:07
  • $\begingroup$ @user42912 for real inner product, the properties are what you stated in your question. $\endgroup$ – Yilun Zhang Apr 15 '14 at 18:08
  • 2
    $\begingroup$ @user42912 It really depends on whether or not your scalar field is the complex numbers or the real numbers! We define the inner product differently for these two. However, notice that the more general condition $\langle u, v \rangle = \overline{\langle v, u \rangle}$ simplifies to the $\langle u, v \rangle = \langle v, u \rangle$ if $\langle v, u \rangle$ is real. $\endgroup$ – user98602 Apr 15 '14 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.