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I need some help here. It's hard to me to tackle this kind of question and I'm not used to write math proofs. I need to find values of $t$ that make $$\langle(x_1, x_2), (y_1, y_2)\rangle = x_1y_1 + tx_2y_2$$ an internal product in $\Bbb R^2$.

I have showed that for $3$ of the $4$ properties, t does not matter at all. But for the property that $$\langle u,u\rangle \gt 0, u \neq 0$$ I have $$x_1^2 +tx_2^2 \gt 0$$ and them $$t \gt -((x_1/x_2)^2)$$ The answer seems to be $t \gt 0$ and I'm lost in this. Could anyone give me the right direction to complete the proof?

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  • $\begingroup$ I obtain $t> -((x_1/x_2)^2)$, not $<$. This must be true for all $(x_1,x_2)\neq 0$. $\endgroup$ – paf Aug 15 '16 at 16:26
  • $\begingroup$ If $t\ge0$ then this rule is satisfied. However it it is possible that $t<0$ then you do not have a valid inner product rule. $\endgroup$ – Doug M Aug 15 '16 at 16:26
  • $\begingroup$ Thank you for the fix. I already corrected in the original question. $\endgroup$ – daniel.franzini Aug 15 '16 at 17:10
  • $\begingroup$ I can't see why $t<0$ makes it not a valid inner product. $\endgroup$ – daniel.franzini Aug 15 '16 at 18:51
  • $\begingroup$ Try to write correct English. "It's kinda of hard to me" is not. $\endgroup$ – Jean Marie Aug 15 '16 at 22:51
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You have verified the symmetry and the bilinearity of this new product. When it comes to positive definiteness we require that $$x_1^2 + t\>x_2^2>0\qquad\forall (x_1,x_2)\ne(0,0)\ .\tag{1}$$ That's how far you got, but now you are entangled in the logic. Note that the value of $t$ is given, and we have to check whether this given $t$ is fine, i.e., satisfies $(1)$. If $t\leq0$ then it is easy to produce vectors $(x_1,x_2)\ne(0,0)$ such that $x_1^2+x_2^2\leq0$. It follows that $t\leq0$ is forbidden. On the other hand, if $t>0$ then $\tau:=\min\{t,1\}>0$, and $$x_1^2+tx_2^2\geq\tau(x_1^2+x_2^2)>0$$ whenever $(x_1,x_2)\ne(0,0)$.

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  • $\begingroup$ I got if (finally): I have to find a value of $t$ that makes the property to hold for any $(x_1, x_2)$. If, e.g., $t<0$, I can find some $(x_1, x_2)$ take makes the property not to hold anymore. $\endgroup$ – daniel.franzini Aug 22 '16 at 16:32
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You committed a silly mistake.$$x_1^2+tx_2^2>0\implies tx_2^2>-x_1^2\implies t\gt-{x_1^2\over x_2^2}$$

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  • $\begingroup$ @daniel.franzini Please accept my answer if it satisfied you! (Check the tick mark) $\endgroup$ – Qwerty Aug 15 '16 at 17:16
  • $\begingroup$ I need further insight on how to analyse this last case. If $t=0$, the properties still hold, so why it can't be included in the answer? $\endgroup$ – daniel.franzini Aug 15 '16 at 18:48
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When t must satisfy the following for all $(x_1, x_2) \in R^2$ $$t\gt-{x_1^2\over x_2^2}$$ $t$ should be meet this condition, $t \gt 0$

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  • $\begingroup$ Let's assume $x_1 = 2$, $x_2 = 1$. Then $t > -4$. This way, why $t = -2$ isn't valid? $\endgroup$ – daniel.franzini Aug 15 '16 at 19:21
  • $\begingroup$ It is true for this point. But you need to pay attention that t is for all points $(x_1, x_2) \in R^2$ including for example (0,1). $\endgroup$ – user115350 Aug 15 '16 at 19:25
  • $\begingroup$ The question is that we have a valid condition on t for some $t<0$. Why we should have $t>0$ always? $\endgroup$ – daniel.franzini Aug 15 '16 at 20:34
  • $\begingroup$ you need t to satisfy the inequality for all $(x_1, x_2) \in R^2$ but not a single point. $\endgroup$ – user115350 Aug 16 '16 at 6:03
  • $\begingroup$ I figured it out (after lots of thinking): I have to find $t$ that makes the property to hold for every $(x_1, x_2)$. Clearly $t = 0$ and $t<0$ are forbidden because in those cases it is possible to find some $(x_1, x_2)$ that makes the property not to hold anymore. $\endgroup$ – daniel.franzini Aug 22 '16 at 16:50
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To demonstrate that a function, say $f$, taking two vectors to a scalar is an internal product (a.k.a. and more commonly referred to as an inner product) is to demonstrate that it satisfies the axioms; that is, $f$ satisfies

  • linearity (i.e. $f(\alpha x+ \beta y, x) = \alpha f(x,z) + \beta f(y,z)$ for vectors $x,y,z$ and scalars $\alpha, \beta$)
  • symmetry (i.e. $f(x,y) = f(y,x)$)
  • positive definite (i.e. $f(x,x)\geq 0$ for all vectors $x$ with equality if and only if $x = 0$)

you'll find that your condition requiring $t > 0$ comes from the third axiom about positive definiteness. More specific to your question, don't forget the additional part of positive definite regarding "with equality if and only if $x = 0$".

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The matrix associate to the bilinear form $$\langle(x_1, x_2), (y_1, y_2)\rangle = x_1y_1 + tx_2y_2$$

is

$$ \begin{pmatrix} 1&0\\ 0&t \end{pmatrix}, $$

then you have an internal product if and only if is positive define if and only if the determinant is strictly positive, and all principal minor have strictly positive determinant. In your specific case this condition become $t>0$.

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