Question

Please, help

Answer 1

@wio

Answer 2

I don't know how to put the condition of alpha, beta, gamma and delta for <x.x> >0
Please,

Answer 3

I can't exactly read it. What is the problem?

Answer 4

I am sorry, let me type the problem :
Given 4 complex numbers \(\alpha , \beta , \gamma , \delta\), try to defined an inner product in \(C^2\) by writing
\((x,y) = \alpha \xi_1\bar \eta_1+\beta\xi_2\bar\eta_1+\gamma\xi_1\bar\eta_2+\delta\xi_2\bar\eta_2\)
whenever x =(\(\xi_1,\xi_2\)) ; y = (\(\eta_1,\eta_2\) ). Under what conditions on \(\alpha, \beta, \gamma, \delta\), does this equation define an inner product?

Answer 5

I'm a bit confused by notation now...

Answer 6

So \(x\) and \(y\) are arbitrary vectors in \(C^2\), or basically they are two dimensional vectors, where each entry is a complex number?

Answer 7

yes, x, y are arbitrary vectors in C^2. In general, they have that form

Answer 8

So we have \(\mathbf x = \langle x_1,x_2\rangle\) and \(\mathbf y = \langle y_1,y_2\rangle\), and \(x_1,x_2,y_1,y_1 \in \mathbb C\)
Then we are saying \[
\mathbf x \cdot \mathbf y = ax_1\overline{y_1}+bx_2\overline{y_1}+cx_1\overline{y_2}+dx_2\overline{y_2}
\]

Answer 9

yes, and find the condition of a, b, c, d to have it is inner product

Answer 10

Is there a definition of inner product we must consider in this case?

Answer 11

I suppose \(a,b,c,d\in\mathbb C\)?

Answer 12

They should be scalars since the inner product should return a scalar

Answer 13

yes, it is
1/ \((x,y) =\bar{(y,x)}\)
2/ (x, ay +bz) = a(x,y) + b(x,z) (a, b are scalar, x, y ,z are vectors)
3/ (x,x) \(\geq 0\) ; =0 iff x =0

Answer 14

\[
(x,y)=\overline{(y,x)}
\]Have you checked this one out?

Answer 15

the first part on my paper, it turns out that if a, b, c, d are real, then \((x,y)=\overline{(y,x)}\)

Answer 16

That is my first condition.

Answer 17

Umm, are you sure it isn't \[
\alpha = \overline{\alpha} ,\beta = \overline{\gamma}, \gamma = \overline{\beta}, \delta = \overline{\delta}
\]?

Answer 18

so \(\alpha\) and \(\delta\) are real, and \(\gamma\) is just the conjugate of \(\beta \).

Answer 19

nope, the proof show that all are equal to its conjugate, not as you did

Answer 20

oh, yes, I am sorry, you are correct. I am wrong

Answer 21

For the last one, consider \(\mathbf x=\langle x_1,0\rangle\)

Answer 22

Then you have \[
(x,x) = \alpha x_1x_1
\]The other terms go to \(0\).

Answer 23

If \(x_1\neq 0\) then it must be \(\alpha \neq 0\). Same applies to \(\delta\) by symmetry.

Answer 24

So, in the general case \(\alpha \neq 0\) and \(\delta \neq 0\). If you multiply any non-zero numbers together, they will be non-zero.

Answer 25

unless they are complex..? Well we know \(\alpha\) and \(\delta\) are real.

Answer 26

It's so complicated. I feel dumb.