I've been given a vector space of linear polynomials in x, p ( x ) = a x + b, q ( x ) = c x + d, and the inner product is defined to be ⟨ p , q ⟩ = a c + b d. I've been able to verify all the axioms for the inner product except for the complex conjugate one, ⟨ p , q ⟩ ∗ = ⟨ q , p ⟩, 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 ( a x + i y ) ∗ = a x − i y, 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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I&#039;ve been given a vector space of linear polynomials in x,   p  (  x  )  =  a  x  +  b,   q  (  x  )  =  c  x  +  d, and the inner product is defined to be   ⟨  p  ,  q  ⟩  =  a  c  +  b  d. I&#039;ve been able to verify all the axioms for the inner product except for the complex conjugate one,   ⟨  p  ,  q      ⟩    ∗    =  ⟨  q  ,  p  ⟩, where   p and   q are vectors.The issue I&#039;m having is that I don&#039;t understand how the complex conjugate can apply if there isn&#039;t an   i in the equation. All I know is that the complex conjugate takes the form   (  a  x  +  i  y      )    ∗    =  a  x  −  i  y, but I&#039;m really confused as to what this means without   i.What is the complex conjugate of a vector that doesn&#039;t have an imaginary component?

rideonthebussp · Accepted Answer

If your base field is       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       C   as the base field, you must consider the coefficients as complex numbers: if   a  =  x  +  y  i with   x  ,    y  ∈      R  , then       a    ∗    =  x  −  y  i. In that case, you should define your inner product as   ⟨  p  ,  q  ⟩  =  a      c    ∗    +  b      d    ∗  .

I've been given a vector space of linear polynomials in x, p ( x ) = a x +

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