1. ## norm question..

i cant understnad why they put absolute value sign there
why they put the complement sign above some coefficient
$=$ $\left | a \right |^2+a\bar{b}+b\bar{a}+\ left | b \right |^2\\$

i whould do it like this
$=+=$ $a^2+ab$ etc..

2. Originally Posted by transgalactic
i cant understnad why they put absolute value sign there
why they put the complement sign above some coefficient
$=$ $\left | a \right |^2+a\bar{b}+b\bar{a}+\ left | b \right |^2\\$

i whould do it like this
$=+=$ $a^2+ab$ etc..
Not "complement", "complex conjugate". The definition of an inner product on a vector space over the complex numbers requires that $= \overline{}$. To take a simple example, if u= (a+bi, c+di) is a vector in $C^2$, pairs of complex numbers, and v= (e+fi,g+hi) is another, then the inner product on $C^2$ is $(a+bi)(\overline{e+fi})+ (c+di)(\overline{g+hi})$ $= (a+bi)(e-fi)+ (c+di)(g-hi)$. Inner products on vector spaces over the complex numbers are defined that way to guarentee that the norm, $||v||= \sqrt{}$ will be a positive real number. If, for example, v= (i, i) and we do NOT require the complex conjugate, we would have <v, v>= i(i)+ i(i)= -1 and then $||v||= \sqrt{-2}= i\sqrt{2}$. That's a problem since we want to use norm to "compare" the size of vectors and the complex numbers is not an ordered field. Even worse, if v= (1, i) then <v, v>= 1(1)+ i(i)= 1-1= 0 so ||v||= 0 even though v is not the 0 vector.

Using the correct inner product, if v= (i, i) then [tex]<v, v>= i(-i)+ i(-i)= 1+ 1= 2[tex] and $||v||= \sqrt{2}$. And if v= (1, i), $= 1(1)+ i(-i)= 1+ 1= 2$ and $||v||= \sqrt{2}$.