Relation between standard inner products on $R^n$ and $C^n$

Question

I was looking at this problem from Hoffman's book. It states that given a conjugation $J$ on a complex vector space $V$ and $f$ a inner product on the set $W = \{x \in V: Jx = x\}$ there is a unique inner product $g$ on $V$ such that $g(a,b) = f(a,b)$ for $a,b \in W$. (a conjugation is a function such that $J(a+b) = J(a) + J(b)$ and $J(ca) = \overline {c}J(a)$ )
I understand that given $z \in V$ we can represent $z$ as $z = x+iy$ with $x,y \in W$. But, how should I proceed with this problem? Should I be thinking on Matrix representation of inner products?

Moreover, once that unique inner product is found, how that relates the standard inner product on $\mathbb R^n$ and $\mathbb C^n$? Thanks in advance.

Answer 1

Say $z = x+iy$ and $z' = x'+iy'$ where $x,y,x',y' \in W$. For $g$ to be an inner product we had better have $$ \begin{split} g(z,z') &= g(x+iy,x'+iy') \\ &= g(x,x')+g(x,iy')+g(iy,x')+g(iy,iy') \\ &= g(x,x')+ig(x,y')+ig(y,x')+i^2g(y,y') \\ &= (f(x,x')-f(y,y'))+i(f(x,y')+f(y,x')) \end{split} $$ which determines $g$ and shows it's unique. I leave it to you to show $g$ defined thus is bilinear. (Here I have presumed that by "inner product" you mean a linear inner product, as opposed to, say, a Hermitian inner product. If you wanted Hermitian, then modify the above with appropriate $\overline{i}=-i$ where needed.)

The standard inner product on $\mathbb{C}^n$ is this, with $J$ given by the standard conjugation. I don't know what to say for $\mathbb{R}^n$.