Can Inner Product be complex?

Question

Judging from wikipedia page the result of Inner Product is a field is $V \times V \rightarrow \mathbb F$. Where $\mathbb F$ can be a Complex Number:

In this article, the field of scalars denoted $\mathbb {F}$ is either the field of real numbers $\mathbb {R}$ or the field of complex numbers $\mathbb {C}$.

But since $\langle x, x \rangle \ge 0$, I don't understand how Complex results are possible - $\mathbb C$ isn't an ordered field, so it can't be compared to $0$. So how can we both have a Complex Inner Product and have it greater than $0$?

Answer 1

You can have a complex vector space $V$ with an inner product $V\times V\to\Bbb C$ where $\langle x,x\rangle$ always happens to be real. For instance, if $V=\Bbb C^2$, the standard inner product is $$ \langle (a,b),(c,d)\rangle=a\bar c+b\bar d $$ and we see here that if $(a,b)=(c,d)$, then the result of the above multiplication is $|a|^2+|b|^2$, which is very much a real number.

I say "happens to be real", but that is actually no coincidence. Note in your Wikipedia article that inner products are required to be not symmetric, but conjugate symmetric. I.e. swapping the order of the two entries conjugates the result: $$\langle x,y\rangle=\overline{\langle y,x\rangle}$$ This gives, for any vector $x$ in the space, that $$ \langle x,x\rangle=\overline{\langle x,x\rangle} $$ which means the product has to be real. And now, since we know it's real, the requirement that this number is positive for any non-zero $x$ makes sense.

Answer 2

What that means is that, for each $x\in V$, $\langle x,x\rangle\in[0,\infty)$.

For instance, if, in $\Bbb C^2$, you define$$\bigl\langle(z_1,w_1),(z_2,w_2)\bigr\rangle=z_1\overline{z_2}+w_1\overline{w_2},$$then, if $x=(z,w)$,$$\langle x,x\rangle=\bigl\langle(z,w),(z,w)\bigr\rangle=|z|^2+|w|^2\in[0,\infty).$$