# Need help showing the following defines an inner product

$x = (x_1,x_2)$, each is an element of $\Bbb R$

$y = (y_1,y_2)$, each is an element of $\Bbb R$

Show the following defines an inner product in $\Bbb R^2$ or show otherwise:

a) $\langle x,y\rangle = 3x_1x_2-x_1y_2-x_2y_1+3x_2y_2$

b) $\langle x,y\rangle = 3x_1x_2-x_1y_2-x_2y_1-3x_2y_2$

What I have tried so far:

a) Axiom 1 $<ax,y> = a<x,y>$

= $3x_1(ax_2)-(ax_1)y_2-(ax_2)y_1+3(ax_2)y_2$

= $3ax_1x_2-ax_1y_2-ax_2y_1+3ax_2y_2$

= $a(3x_1x_2-x_1y_2-x_2y_1+3x_2y_2)$

= $a<x,y>$

Axiom 2 $<x, y+z> = <x,y> + <x,z>$

= $3x_1x_2-x_1(y_2+z)-x_2(y_1+z)+3x_2(y_2+z)$

= $3x_1x_2-x_1y_2-x_1z-x_2y_1-x_2z+3x_2y_2+3x_2z$

= $(3x_1x_2-x_1y_2-x_2y_1+3x_2y_2)+(-x_1z-x_2z+3x_2z)$

= $(3x_1x_2-x_1y_2-x_2y_1+3x_2y_2)+(-x_1z+2x_2z)$ What's going on here?

Axiom 3 $<x,y> = <y,x>$

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To expand on Belgi's question --- what do you know about inner products? What properties must something have to be an inner product? Have you checked any of them? What did you find? –  Gerry Myerson Oct 14 '12 at 8:51
I should rellay have domething like did's comment on a notepad by now. +1 –  Belgi Oct 14 '12 at 8:53
@Belgi It ain't mine... :-) –  Did Oct 14 '12 at 9:00
@did - thank you for the link! –  Belgi Oct 14 '12 at 9:02

HINTS:

For (a): Check the defining properties of an inner product. Is it true that $\langle x,y\rangle=\langle y,x\rangle$ for all $x,y\in\Bbb R^2$? Is it true that $\langle \alpha x+\beta y,z\rangle=\alpha\langle x,z\rangle+\beta\langle y,z\rangle$ for all $x,y,z\in\Bbb R^2$ and all $\alpha,\beta\in\Bbb R$? Is it true that $\langle x,x\rangle\ge 0$ for all $x\in\Bbb R^2$, and $\langle x,x\rangle=1$ if and only if $x=(0,0)$?

For (b): If $x=(1,1)$, what is $\langle x,x\rangle$? What does this tell you?

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b) if x = (1,1) -> 3(1)(1)-(1)y2-(1)y1-3(1)y2 = 3-y1-4y2 So how do I do <x,x> –  FRU5TR8EDD Oct 14 '12 at 9:09
@FRU5TR8EDD: For $\langle x,x\rangle$ you have $x=y$ in your formula, so $x_1=x_2=y_1-y_2$; just finish substituting these values, and do the arithmetic. –  Brian M. Scott Oct 14 '12 at 9:14
So 3-x-4x= 3-5(1)=-2, therefore doesn't satisfy $<x,x> >=0$ Is that right? –  FRU5TR8EDD Oct 14 '12 at 9:18
@FRU5TR8EDD: That’s exactly right. –  Brian M. Scott Oct 14 '12 at 9:19
Thanks Brian! :) –  FRU5TR8EDD Oct 14 '12 at 9:32

Hint: Try to write it as $\langle x,y\rangle=y^{t}Ax$ where $A\in M_{2}(\mathbb{R})$. When does this define an inner product ?

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