1. ## norm question..

i cant understnad why they put absolute value sign there
why they put the complement sign above some coefficient
$\displaystyle <ae_1+be_2,ae_1+be_2>=$$\displaystyle \left | a \right |^2<e_1,e+1>+a\bar{b}<e_1,e_2>+b\bar{a}<e_2,e_1>+\ left | b \right |^2<e_2,e_2>\\ i whould do it like this \displaystyle <ae_1+be_2,ae_1+be_2>=<ae_1,ae_1+be_2>+<be_2,ae_1+ be_2>=$$\displaystyle a^2<e_1,e_1>+ab<e_1,e_2>$ 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
$\displaystyle <ae_1+be_2,ae_1+be_2>=$$\displaystyle \left | a \right |^2<e_1,e+1>+a\bar{b}<e_1,e_2>+b\bar{a}<e_2,e_1>+\ left | b \right |^2<e_2,e_2>\\ i whould do it like this \displaystyle <ae_1+be_2,ae_1+be_2>=<ae_1,ae_1+be_2>+<be_2,ae_1+ be_2>=$$\displaystyle a^2<e_1,e_1>+ab<e_1,e_2>$ etc..
Not "complement", "complex conjugate". The definition of an inner product on a vector space over the complex numbers requires that $\displaystyle <u, v>= \overline{<v, u>}$. To take a simple example, if u= (a+bi, c+di) is a vector in $\displaystyle C^2$, pairs of complex numbers, and v= (e+fi,g+hi) is another, then the inner product on $\displaystyle C^2$ is $\displaystyle (a+bi)(\overline{e+fi})+ (c+di)(\overline{g+hi})$$\displaystyle = (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, $\displaystyle ||v||= \sqrt{<v, v>}$ 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 $\displaystyle ||v||= \sqrt{-2}= i\sqrt{2}$. 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 $\displaystyle ||v||= \sqrt{2}$. And if v= (1, i), $\displaystyle <v, v>= 1(1)+ i(-i)= 1+ 1= 2$ and $\displaystyle ||v||= \sqrt{2}$.