While reading through my textbook it says "the most important example of an inner-product space is $F^n$", where $F$ denotes $\mathbb{C}$ or $\mathbb{R}$ .

Our definition of an inner product on a vector space $V$ is as follows:
1) Positive definite: $\langle v,v \rangle \ge 0$ with equality if and only if $v=0$
2) Linearity in the first arguement: $\langle a_1v_1+a_2v_2,w \rangle = a_1 \langle v_1,w \rangle + a_2\langle v_2,w \rangle$
3) Conjugate symmetric: $\langle u,v\rangle = \overline{\langle v,u\rangle}$

Let $$\displaystyle w=(w_1\ldots,w_n) , z=(z_1,\ldots,z_n)$$
$$\displaystyle \langle w,z\rangle =w_1\overline{z_1}+\cdots+w_n\overline{z_n}$$

I'm trying to verify that this is indeed true. So first I want to check that $\langle w,z\rangle$ satisfies condition (1).

Say that $w,z\in \mathbb{C}$.
Just looking at say $w_1=a+bi$ and $z_1=c+di$, how can we guarantee that $w_1\overline{z_1}\geq 0$?
If we can observe this, it would need to hold true for the other coordinates as well. So my question is, how do we know that $w_1\overline{z_1}\geq 0$?

  • 3
    $\begingroup$ Your definition of "positive definite" is wrong. It should be that for all $v$, $\langle v,v \rangle \ge 0$, with equality iff $v=0$. $\endgroup$ Mar 4, 2011 at 1:47
  • $\begingroup$ I think you misread the definition. $w_1\bar z_1$ can be anything, and need not be positive. The definition (1) should say that $\langle u,u\rangle\ge0$ with equality iff $u=0$. $\endgroup$ Mar 4, 2011 at 1:48
  • $\begingroup$ Put so much effort into that question too, please delete it.. $\endgroup$
    – Justin
    Mar 4, 2011 at 1:52
  • 1
    $\begingroup$ I may as well add, the angle brackets around the inner product are correctly displayed in latex with \langle u,v\rangle. You used less-than and greater-than signs, which doesn't look quite the same. $\endgroup$ Mar 4, 2011 at 1:55

1 Answer 1


Item 1 in your definition of an inner product is incorrect. A simple counter example from $\mathbb{R}^2$ is

$$(1,0).(-1,0) = -1.$$

It should read $$\langle v, v \rangle \ge 0,$$ this guarantees that all vectors in your space have a non-negative length.


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.