I've been given a vector space of linear polynomials in x, $p(x)=ax+b,\;\;$ $q(x)=cx+d$, and the inner product is defined to be $\langle p,q\rangle=ac+bd$. I've been able to verify all the axioms for the inner product except for the complex conjugate one, $\langle p,q\rangle^*=\langle q,p\rangle$, where $p$ and $q$ are vectors.
The issue I'm having is that I don't understand how the complex conjugate can apply if there isn't an $i$ in the equation. All I know is that the complex conjugate takes the form $(ax+iy)^*=ax-iy$, but I'm really confused as to what this means without $i$.
What is the complex conjugate of a vector that doesn't have an imaginary component?
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$\begingroup$If your base field is $\mathbb{R}$, so that $a$, $b$, $c$, $d$ are real, the complex conjugate of a real number is the number itself: $a^*=a$, and the relation is obviously satisfied.
If you take $\mathbb{C}$ as the base field, you must consider the coefficients as complex numbers: if $a=x+yi$ with $x,\,y\in\mathbb{R}$, then $a^*=x-yi$. In that case, you should define your inner product as $\langle p,q\rangle = ac^* + bd^*$.
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