I must conclude something from this inequality, given that it must hold true for all values of $x_1$ and $x_2$. I believe it is ok to say $x_1$ and $x_2$ are the parameters and $a_{11}, a_{12},a_{21},a_{22}$ are the variables (the ones I want to find the relation between). The inequation is as follows:

$$x_1^2.a_{11} + x_2^2.a_{22} + x_1x_2.(a_{12} + a_{21}) \ge 0 ~~~\forall x_1, x_2.$$

So, I would like to know how to work with this kind of problem, what can I conclude from this inequality?


1 Answer 1


Your inequality can be written in matrix form as $(x_1,x_2)\left(\begin{array}{cc}a_{11}&a_{12}\\a_{21}& a_{22}\end{array}\right)\left(\begin{array}{c}x_1\\x_2\end{array}\right)\geq0$. This inequality is just telling you that the quadratic form associated to the matrix $\left(\begin{array}{cc}a_{11}&a_{12}\\a_{21}& a_{22}\end{array}\right)$ is positive semidefinite.

  • $\begingroup$ Hello, I know, in fact, that's where I came from! But really, I need to conclude from that inequality that the elements in the main diagonal are greater than zero and also that the determinant is greater than zero. Can I deduce these from that inequality? $\endgroup$ Commented Oct 26, 2016 at 20:02
  • $\begingroup$ The exercise I was trying asked to demonstrate it, so, although I know the conclusion, I couldn't elaborate a demonstration for it. $\endgroup$ Commented Oct 26, 2016 at 20:05
  • $\begingroup$ A matrix $A$ is positive semidefinite iff all the principal minors of A are non-negative. I think that is the theorem you're looking for. $\endgroup$
    – user378947
    Commented Oct 26, 2016 at 20:12
  • $\begingroup$ @Kurosaki_Ichigo I just found this .pdf which should be useful. $\endgroup$
    – user378947
    Commented Oct 26, 2016 at 20:16
  • 1
    $\begingroup$ Thanks for your help, I'll take a look at this theory. $\endgroup$ Commented Oct 26, 2016 at 20:19

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