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We have a binary operation over vectors in $\mathbb{R}^2$ like this:
$\langle X,Y\rangle = 5x_1y_1 + x_1y_2 + x_2y_1 + 3x_2y_2 $ where $X=(x_1,x_2)$ and $Y$ is defined similarly.

The only remaining facts I have to show are:
$\langle X, X \rangle \ge 0$ and $\langle X,X \rangle = 0$ iff $X=0$

I've seen the answer key, which says that I only need to note that the polynomial $5z^2 + 2z +3 =0$ has no real roots. I don't see how this polynomial relates to the problem, however. How am I supposed to use this polynomial?

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It is not linear in $Y$ and is thus not an inner product. –  Will Jagy Feb 18 '13 at 1:33
    
Oh! Maybe the answer key is wrong. –  Mark Feb 18 '13 at 1:41
    
@WillJagy I fixed the question. It's linear now, but it still remains to be seen that it is an inner product. –  Mark Feb 18 '13 at 1:45
    
In that case, the polynomial they invoke should be $5 z^2 + 2 z + 3.$ –  Will Jagy Feb 18 '13 at 1:47
    
Can you elaborate on how that helps me? I see that it's similar to $5x_1^2 + 2x_1x_2 + 3x_2^2$ but you replaced $x_2$ with $1$? –  Mark Feb 18 '13 at 1:54

2 Answers 2

up vote 1 down vote accepted

Write out $\langle X,X \rangle.$ There are two cases to consider:

(A) $x_2 = 0.$

(B) $ x_1 = z x_2$ with real $z.$

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Without using the hint, here is another way of solving this. If you let $X := (x,y)$ for notational simplicity,

$\begin{align} <X,X> & = 5x^2 + 2xy + 3y^2\\ & = (x+y)^2 +4x^2 + 2y^2 \quad \begin{cases} \geq 0, & \text{for all $X$} \\ =0, & \text{iff all 3 terms are $0$}\end{cases} \end{align}$

So $<X,X> = 0 \text{ iff } X = 0.$

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Nice! Pretty simple solution. –  Mark Feb 18 '13 at 2:36
    
@Mark: Thanks! :) –  gnometorule Feb 18 '13 at 2:53

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