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$x = (x_1,x_2)$, 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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What have you tried ? –  Belgi Oct 14 '12 at 8:48
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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Did Oct 14 '12 at 8:51
    
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
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I should rellay have domething like did's comment on a notepad by now. +1 –  Belgi Oct 14 '12 at 8:53
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@Belgi It ain't mine... :-) –  Did Oct 14 '12 at 9:00

2 Answers 2

up vote 0 down vote accepted

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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