Complex conjugate on an inner product

[bugatti79]

Homework Statement

Consider the set $C^2= {x=(x_1,x_2):x_1,x_2 \in C}$.

Prove that $<x,y>=x_1 \overline{y_1}+x_2 \overline{y_2}$ defines an inner product on $C^2$

Homework Equations

The Attempt at a Solution

$<,y>=\overline {<y,x>}$

$= \overline {y_1x_1} + \overline {y_2x_2}$

I think I need some other information to continue...?

Thanks

 

[CornMuffin]

Homework Statement

Consider the set $C^2= {x=(x_1,x_2):x_1,x_2 \in C}$.

Prove that $<x,y>=x_1 \overline{y_1}+x_2 \overline{y_2}$ defines an inner product on $C^2$

Homework Equations

The Attempt at a Solution

$<,y>=\overline {<y,x>}$

$= \overline {y_1x_1} + \overline {y_2x_2}$

I think I need some other information to continue...?

Thanks

verify that this satisfies the definition of an inner product:

$<x,y>=\overline{<y,x>}$
$<x+y,z>=<x,z>+<y,z>$
$<ax,z>=a<x,z>$
$<x,x> \geq 0$
$<x,x>=0 \iff x=0$

for all x,y,z in $C^2$ and for all complex scalars a

 

[bugatti79]

verify that this satisfies the definition of an inner product:

$<x,y>=\overline{<y,x>}$
$<x+y,z>=<x,z>+<y,z>$
$<ax,z>=a<x,z>$
$<x,x> \geq 0$
$<x,x>=0 \iff x=0$

for all x,y,z in $C^2$ and for all complex scalars a

I can verify the above for x,y,z in $R^2$ but don't know how to extend to the complex field...? Especially the first axiom..

THanks

 

[CornMuffin]

I can verify the above for x,y,z in $R^2$ but don't know how to extend to the complex field...? Especially the first axiom..

THanks

$\overline{x}$ means the complex conjugate of $x$, that is if $x=a+ib$ for $a,b \in \mathbb{R}$ then $\overline{x} = a-ib$

find $\overline{<y,x>}$ and show that it simplifies to $x_1\overline{y_1} + x_2\overline{y_2}=<x,y>$

 

[Fredrik]

$\overline {<y,x>}$

$= \overline {y_1x_1} + \overline {y_2x_2}$

This is wrong. Use the definition you posted, and you'll see that.

 

[bugatti79]

$\overline{x}$ means the complex conjugate of $x$, that is if $x=a+ib$ for $a,b \in \mathbb{R}$ then $\overline{x} = a-ib$

find $\overline{<y,x>}$ and show that it simplifies to $x_1\overline{y_1} + x_2\overline{y_2}=<x,y>$

This is wrong. Use the definition you posted, and you'll see that.

1) $<x,y>=\overline{<y,x>}=\overline{ y_1\overline{x_1}}+\overline { y_2\overline{x_2}}=\overline{y_1}x_1+\overline{y_2}x_2$

2) $<x+y,z>=(x_1+y_1)\overline{z_1}+(x_2+y_2) \overline{z_2}=x_1\overline {z_1}+y_1\overline {z_1}+x_2\overline {z_2}+y_2\overline {z_2}=<x,z>+<y,z>$

3)$<\alpha x,y>= \alpha x_1 \overline{y_1}+\alpha x_2 \overline{y_2}=\alpha(x_1 \overline{y_1}+x_2\overline{y_2})=\alpha<x,y>$

4)$<x,x>=x_1 \overline{x_1}+x_2 \overline{x_2}=|x_1|^2+|x_2|^2>=0$ since by definition $z \overline{z}=|z|^2>=0$ and $<x,x>=0$ iff x=0, that is x1=x2=0...

thanks

 

[Fredrik]

2-4 are fine. In part 1, the first equality you wrote down is the one you're trying to prove. This is of course not OK. Your string of equalities should end with =<x,y>, not begin with <x,y>=. The best way to do this is one step at a time, like this:
$$\overline{\langle y,x\rangle} =\overline{y_1\overline{x_1}+y_2\overline{x_2}} =\overline{y_1\overline{x_1}} +\overline{y_2\overline{x_2}} =\overline{y_1}\,\overline{\overline{x_1}} +\overline{y_2}\,\overline{\overline{x_2}}
= \overline{y_1}x_1+\overline{y_2}x_2 =x_1\overline{y_1}+x_2\overline{y_2} =\langle x,y\rangle$$.

 

[bugatti79]

Thanks Fredrik for the clarification.