$$f(x) =ax_1^2 +2x_1x_2+2x_1x_3$$ under which condition of a, f(x) 0?
Please

Question

$$f(x) =ax_1^2 +2x_1x_2+2x_1x_3$$ under which condition of a, f(x) >0?
Please

kinggeorge · Answer

What are $x_1,x_2,x_3$?

loser66 · Answer

I am sorry, I edited my question :) I missed 2 before the last term

loser66 · Answer

@KingGeorge  actually the problem comes from positive definite matrix. I "translate" it into the simplest form :)

loser66 · Answer

$$\left[\begin{matrix}\alpha&1&1\1&0&0\1&0&0\end{matrix}\right]$$
question: under which condition of $\alpha$ the matrix is positive matrix?

loser66 · Answer

and this is translated to $$\alpha x_1^2+2x_1x_2+2x_1x_3$$

loser66 · Answer

if the expression >0 for all x1, x2, x3, then the matrix is positive definite.

ganeshie8 · Answer

$\large    x_i$ are positive integers ?

loser66 · Answer

or another way to put it in logic is $x^tAx>0$

loser66 · Answer

x is an arbitrary vector on vector space

loser66 · Answer

can we put it in logic of x1>0 and the (ax1+2x2 +2x3) > 0 simultaneously to argue for it x1<0 and the ( )<0 x1=0 --> no solution for a is it ok?

ganeshie8 · Answer

looks okay to me, you want to make $\large a$ depend on ur vector is it...

kinggeorge · Answer

We don't want to make $\alpha$ dependent on the vector.

kinggeorge · Answer

It seems to me, that there is no $\alpha$ that will make everything work nicely. For example, if $\alpha>0$ let $x_1=1/\sqrt{\alpha}$, $x_2=0$, and $x_3=-\sqrt\alpha$. Then$$\alpha x_1^2+2x_1x_2+2x_1x_3=1+0-2=-1$$

loser66 · Answer

A is a Hermitian matrix --> det A1 = alpha > o when alpha>0
det A2 = -1
we can stop here because det A2 <0 for all alpha --> with any value of alpha, A cannot be positive.

kinggeorge · Answer

If $\alpha<0$ instead choose $x_1=1/\sqrt{-\alpha}$, $x_2=0$, and $x_3=-\sqrt{-\alpha}$.

If $\alpha=0$, simply choose $x_1=-1$, and $x_2=x_3=1$.

loser66 · Answer

|dw:1397784607401:dw|