Question

a,b,c,d \(\in R\) ; x, y \(\in C^2\) find the condition of a, b, c, d to get the > 0

Answer 1

a,b,c,d are real
\(x = (x_1,x_2)\); \(y =(y_1,y_2)\) ; the inner product defined
\[<x,y>= ax_1\bar{y_1}+bx_2\bar{y_1}+cx_1\bar{y_2}+dx_2\bar{y_2}\]
how to prove \(<x,x>\geq 0\) = iff x =0
Please, help

Answer 2

so far, I have
\[<x,x>= ax_1\bar{x_1}+bx_2\bar{x_1}+cx_1\bar{x_2}+dx_2\bar{x_2}\]
\[=a|x_1|^2 +bx_2\bar{x_1}+cx_1\bar{x_2} +d|x_2|^2\]
and stuck

Answer 3

which are you doing ?
find the condition of a, b, c, d to get the <x,x>> 0

or
how to prove <x,x>≥0 = iff x =0

Answer 4

the first one

Answer 5

the condition of a, b, c, d

Answer 6

do we need <x,x> to be real ?

Answer 7

we need it complex to have conjugate of the term \(\bar x_1\) for example

Answer 8

I mean what does it mean for a complex number C to be ≥ 0 ?
as opposed to | C | ≥ 0

Answer 9

I don't get your question

Answer 10

\[ =a|x_1|^2 +bx_2\bar{x_1}+cx_1\bar{x_2} +d|x_2|^2 \]

this will be >0 if a and d >0 and b=c=0

that might be to constraining, but it works.

Answer 11

Is there any other option else?

Answer 12

Wouldn't \(\bar{y_1}=y_1\)? If \(y\in\mathbb{C}\), then \(y_1,y_2\in\mathbb{R}\). Just wondering if you really mean to use the conjugate there.

Answer 13

Also, @loser66 you mentioned that you had to prove \(\langle x,x\rangle \ge0\) iff \(x=0\). Do you mean \(\langle x,x\rangle=0\) iff \(x=0\)?

Answer 14

we can write this problem as

\[ u^T M u ≥ 0\]
where M= [a b ; c d]
that means we want M to be positive definite

Answer 15

No, no @SithsAndGiggles, x, y in C^2, it means x1, x2, y1, y2 are complex

Answer 16

This is the whole thing and what I post here is the last part where I stuck

Answer 17

Are you sure b and c have to be real?
I am thinking M has to be hermetian, i.e. M = conj( transpose (M))

Answer 18

when proving the defined equation is an inner product, in part 1, I have a, b,c, d must be real. That's why in part 3, I use that condition to figure out the other condition.
brb

Answer 19

I think you can show that a and d are real, and b = conj( c )

define
<x,y> = y^H M x
where H means complex conjugate transpose, and M = M^H
we can show that
<y*, x*> = <x,y>

Answer 20

let me try, thanks for the help :)

Answer 21

something like this

Answer 22

yes, that 's what the first part I did and get the first condition is a, b, c, d are real

Answer 23

yes for a and d real. b = conjugate of c (not necessarily real)