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I was reading the definition of inner product here and they state the following:

Let $V$ be a vector space over a field $\mathbb F$ ( $\mathbb R$ or $\mathbb C$). A map $\left< \cdot , \cdot \right>: V \times V \to \mathbb F$ is called an inner product if, for all $x,y,z \in V$ and $\alpha \in \mathbb F$:

  1. $\left< \alpha x , y \right> = \alpha \left< x , y \right>$ and $\left< x + z , y \right> = \left< x , y \right> + \left< z , y \right>$
  2. $\left< x , y \right> = \overline{\left< y , x \right>}$
  3. $\left< x , x \right> > 0$, if $x \neq 0$.

My question is about property 3. Let $V$ be a vector space over $\mathbb C$, then $\left< \cdot , \cdot \right>: V \times V \to \mathbb C$, this means that for all $x \in V$, $\left< x , x \right>$ is a complex number. So what do they mean by $\left< x , x \right> > 0$? Do they mean that the inner product of $x$ with itself is allawys a real number, so if $\left< \cdot , \cdot \right>$ is an inner product then $\Im(\left< x , x \right>) = 0$? If not, what does it mean for a complex number to be greater than $0$ in this context?

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2 Answers2

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Because of the second property, $\langle x,x \rangle = \overline{\langle x,x \rangle}$, so $\Im(\langle x,x \rangle)=-\Im(\langle x,x \rangle)=0$, i.e. $\langle x,x \rangle$ is real.

Ian
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Because of point 2.:

$$\left< x , y \right> = \overline{\left< y , x \right>}$$ applied for $x= y$, you get

$$\left< x , x \right> = \overline{\left< x , x \right>}.$$

Therefore $\left< x , x \right>$ is a real number.

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