# Inner Product, definite positive?

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

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 $w_1\overline{z_1}\geq 0$?

• Your definition of "positive definite" is wrong. It should be that for all $v$, $\langle v,v \rangle \ge 0$, with equality iff $v=0$. Mar 4, 2011 at 1:47
• 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 $u=0$. Mar 4, 2011 at 1:48
• Put so much effort into that question too, please delete it.. Mar 4, 2011 at 1:52
• 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. Mar 4, 2011 at 1:55

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.