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?

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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