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$\begingroup$

$x = (x_1,x_2)$, 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 $ = a$

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

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

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

= $a$

Axiom 2 $ = + $

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

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

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

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

Axiom 3 $ = $

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    Axiom $2$ should be about linearity in the first coordinate, I think you have it stated the wrong way.2012-10-14

2 Answers 2

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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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    @FRU5TR8EDD: You’re welcome!2012-10-14
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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 ?