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Let $$\langle x,y \rangle = x_1y_1 + 3x_2y_2 + 4x_3y_3 + x_1y_2 + x_2y_1 + x_1y_3 + x_3y_1 + x_2 y_3+x_3y_2$$, prove that $\langle x,y \rangle$ is positive definite.

I have simplified this to the following term:

$$(x_1+x_2)^2+ (x_2 + x_3)^2 + x_2^2 + x_3^2 +2x_1x_2$$

How can I show that this inequality is greater than 0 for all $x \neq 0$?

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    $\begingroup$ One way is to write down the symmetric matrix $A$ such that your form is $x^tAy$, and then show that all the eigenvalues of $A$ are positive. $\endgroup$ – Gerry Myerson Jun 10 '14 at 12:21
  • $\begingroup$ Thanks, that's a great idea. I'll try that. $\endgroup$ – eager2learn Jun 10 '14 at 12:23
  • $\begingroup$ @eager2learn, how did you "simplify" that to that term?? The original expression has no quadratic terms in one variable, i.e.: it has no $\;x_1^2\;,\;\;x_2^2\;$ and etc.... $\endgroup$ – DonAntonio Jun 10 '14 at 12:36
  • $\begingroup$ For that $\langle x,y \rangle$ to be pos. definite, we need $\langle x,x \rangle > 0, \forall x \neq 0$. If I use $\langle x,x \rangle$ I get $x_1^2+3x_2^2+4x_3^2 + 2x_1x_2+2x_1x_3+2x_2x_3$, or did I do something wrong? $\endgroup$ – eager2learn Jun 10 '14 at 12:44
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We have to look whether the quadratic form $$q(x):=\langle x,x\rangle=x_1^2+3x_2^2+4x_3^2+2(x_1x_2+x_2x_3+x_3x_1)$$ is positive definite. To this end we complete the squares and write $q$ in the form $$q(x)=(x_1+x_2+x_3)^2+2x_2^2+3x_3^2\ .$$ It is obvious that $q(x)\geq0$ for all $x$. Furthermore $q(x)=0$ enforces $x_2=x_3=0$ and finally $x_1=0$. It follows that $q$ is indeed positive definite.

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