Positive definite (inner product)

Question

In my linear algebra course, we defined the positive definite of the inner product where $\langle z,z\rangle \ge 0$. My professor stated that because of this $\langle z,z\rangle \notin\mathbb{C}$?

What is the definition of a positive so that they do not exist for complex numbers? Could you not simply form a definition of magnitude? Why is this definition not useful so that the value of the inner product is restricted to $\mathbb{R}$?

Answer 1

The ordering on $\mathbb{R}$ is "compatible with the operations."

This means the following: there is a subset $P$ of $\mathbb{R}$, called the "positive number", such that:

1. $\mathbb{R}$ obeys a trichotomy law: every $r\in \mathbb{R}$ satisfies one and only one of $r\in P$, $-r\in P$, and $r=0$.
2. If $a,b\in P$, then $a+b\in P$ and $ab\in P$.

Then one defines $a\lt b$ if and only if $b-a\in P$, and $a\leq b$ if and only if $a=b$ or $a\lt b$.

With these definitions, we obtain the "usual" properties : if $a\leq b$ then for all $c$ we have $a+c\leq b+c$; if $a\leq b$ and $c\leq d$ then $a+c\leq b+d$; if $a\leq b$ and $c\gt 0$, then $ac\leq bc$. Etc. That is, obtain an "ordered field".

However, there is no way to define an order on $\mathbb{C}$ that is compatible with the operations. That is, there is no subset $P$ of $\mathbb{C}$ that will satisfy properties 1 and 2 above.

To see why, note that if we had a class of "positive complex numbers", then we must have $1$ positive: because either $1\in P$ and we are done, or else $-1\in P$, and then $1=(-1)(-1)\in P$. So, either way, $1\in P$. But because $1\in P$, we must have $-1\notin P$.

Now, if $i\in P$, then $-1=(i)(i)\in P$, which contradicts trichotomy. And if $i\notin P$, then $-i\in P$, so $(-i)(-i)=-1\in P$... again contradicting trichotomy. Thus, there is no class of "positive complex numbers" satisfying 1 and 2 above, and hence no order on $\mathbb{C}$ that is compatible with the operations.

Because of that, when dealing with the algebraic side of $\mathbb{C}$, such as when working with vector spaces over $\mathbb{C}$, we do not put an order on all of $\mathbb{C}$. We still have the order on $\mathbb{R}$, which we can use to compare magnitudes of complex numbers, or real numbers. So when we have a complex number $z\in\mathbb{C}$, and a real number $r$, we only write $z\geq r$ if $z$ is real, and $z\geq r$ as real numbers. In particular, if we write $\langle z,z\rangle\geq 0$, then we are saying that $\langle z,z\rangle$ is real and nonnegative.

(You can define all sorts of orders on $\mathbb{C}$; for example, the lexicographic order on the complex plane gives $a+bi\lt c+di$ if and only if $a\lt c$, or $a=c$ and $b\lt d$; but this order does not play well with addition and multiplication; and no ordering on $\mathbb{C}$ plays well with addition and multiplication, so when you care about the algebraic structure on $\mathbb{C}$, you do not put an order on it except perhaps for ad hoc arguments.)

Answer 2

Arturo Magidin's answer is brilliant. I did however want to give you a different perspective on what we should do with complex vector spaces.

The important geometrical perspective is that $\langle z,z\rangle$ should be the norm sqaured $|z|^2$. Using this and the cosine law, one is then able to recover a notion of angle between vectors.

When we talk about a complex vector space $V$, the analogous form to consider is not an inner product but a Hermitian form.

For vector spaces over $\mathbb{R}$, a symmetric matrix $A$ defines an inner product and vice versa. i.e. $\langle v,w\rangle :=v^T A w$ defines an inner product. Since $A$ is symmetric and $v^T A w$ is a scalar (or more precisely a $1\times 1$ matrix), $\langle v,w\rangle=(v^T Aw)^T=w^T A^T v=w^T A v=\langle w,v\rangle$. We have shown that this gives you a symmetric form. You can in a similar fashion check other properties of being an inner product (i.e. bilinearity).

If we want to consider vector spaces over $\mathbb{C}$, a Hermitian matrix defines a Hermitian form. A Hermitian matrix $A$ is a matrix such that $A^{\dagger}=A$ where the dagger operation is the conjugate transpose, $A^{\dagger}=\overline{(A^T)}$.

A Hermitian matrix gives a Hermitian form defined by $v^{\dagger} A w$. This gives a form that is linear in the second component and conjugate linear in the first component.