Prove Schwarz's for complex numbers: (x,y)^2 <= |x|^2 |y|^2
Under what conditions does equality hold?
Hint:
| lx + my |^2 = l^2 (x,x) + 2lm (x,y) + m^2 (y,y)
is positive definite wrt l and m
The hint is really confusing me...
Prove Schwarz's for complex numbers: (x,y)^2 <= |x|^2 |y|^2
Under what conditions does equality hold?
Hint:
| lx + my |^2 = l^2 (x,x) + 2lm (x,y) + m^2 (y,y)
is positive definite wrt l and m
The hint is really confusing me...
I believe the problem is a notation one. Allow me to change inner products from $\displaystyle (y,x)$ to $\displaystyle \langle y \, , \,x \rangle$. So, we have:
$\displaystyle \langle z \, , \, x \rangle = \left\langle y - \frac{\langle y \, , \, x \rangle}{|x|^2} \, , \, x \right\rangle = \langle y \, , \, x \rangle - \frac{\langle y \, , \, x \rangle}{|x|^2} \langle x \, , \, x \rangle = $
$\displaystyle = \langle y \, , \, x \rangle - \frac{\langle y \, , \, x \rangle}{|x|^2} \cdot |x|^2 = \langle y \, , \, x \rangle - \langle y \, , \, x \rangle = 0$
I hope it's clearer now.