this is about inner product.
For 2 dimensional complex linear space,
I dont see why (x.y)=conjugate of (y.x)
and (x.cy)= conjugate of c (x.y)
isn't is just same as when you do for real linear space?
ex) If I have (X.y)=(2+i,i) then (x.y) would be 2i-1 and (y.x)=(i.2+i) would be 2i-1 so they are same.
If I put conjugate sign over (y.x) wouldn't it be -2i-1 which is not equal to 2i-1
can anybody explain it?
On
If you look at the defining axioms of a skalar product you will find positive definiteness, i.e. $(x,x)\geq 0$. For this to hold you have to define the inner product in the 2-dimensional complex space as $(x,y)= x_1 \bar y_1 + x_2 \bar y_2$. If you didn't use the complex conjugation in the computation of the inner products it would never be positive definite. If you look at the definition above you will see that $(x,y)=\overline{(y,x)}$.
The positive definiteness is vital for the derivation of norms from inner products, therefore you do not want to drop that axiom in the definition.
You might also want to have a look at this.
I think the definition of inner product you are using is $<x,y> = x \overline{y}$, the product of $x$ and $\overline{y}$ under the usual definition of multiplication for complex numbers. Then you have that $$<2+i,i> = (2+i)(\overline{i}) = (2+i)(-i) = -2i - i^2 = 1 - 2i$$ $$<i,2+i> = (i)(\overline{2+i}) = (i)(2 - i) = 2i - i^2 = 1 + 2i$$
Noting that $\overline{1 - 2i} = 1 + 2i$, we see that, in this example, $<2+i,i> = \overline{<i,2+i>}$.
Furthermore, in a slightly more general case, we have, for $x = a +bi$ and $y = c+di$, that $$<x,y> = (a+bi)(\overline{c+di}) = (a+bi)(c -di) = (ac +bd) + (bc - ad)i$$ $$<y,x> = (c+di)(\overline{a +bi}) = (c+di)(a-bi) = (ac +bd) + (ad - bc)i$$
and we observe that $(ac+bd) + (bc-ad)i = \overline{(ac+bd) + (ad-bc)i}$, and therefore $<x,y> = \overline{<y,x>}$ in this case as well.