I"ve been given a vector room of linear polynomials in x, $p(x)=ax+b,;;$ $q(x)=cx+d$, and the inner product is characterized to it is in $langle p,q angle=ac+bd$. I"ve been able to verify all the axioms for the inner product other than for the complicated conjugate one, $langle p,q angle^*=langle q,p angle$, where $p$ and $q$ room vectors.

The issue I"m having is that ns don"t understand how the complicated conjugate can apply if there isn"t an $i$ in the equation. All I understand is that the facility conjugate take away the type $(ax+iy)^*=ax-iy$, however I"m really confused regarding what this way without $i$.

What is the complicated conjugate the a vector that doesn"t have actually an imagine component?


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If your base field is $ptcouncil.netbbR$, so that $a$, $b$, $c$, $d$ are real, the facility conjugate that a genuine number is the number itself: $a^*=a$, and also the relation is clear satisfied.

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If you take it $ptcouncil.netbbC$ as the base field, friend must consider the coefficients as complex numbers: if $a=x+yi$ with $x,,yinptcouncil.netbbR$, then $a^*=x-yi$. In the case, friend should specify your within product together $langle p,q angle = ac^* + bd^*$.


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