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Take any $2 \times 2$ matrix $A$. Then $A$ is defind to be positive definite if for any $v\in \mathbb R^2$ except the zero vector, it is the case that $v^TAv>0$

This expression 'simplifies' to $$\forall x,y\in \mathbb R: ax^2+(b+c)xy+dy^2>0$$ where $a, b, c, d$ are the elements of $A$.

if we take first $x=0$, and then $y=0$, then we find that it must hold that $$a>0$$ $$d>0$$ i.e. the diagonal elements of $A$ must be positive.

Now if we define $k\in \mathbb R$ so that $x=-ky$, then our condition becomes: $$\forall k>0: ak^2-(b+c)k+d>0$$

I can find a number of restrictions on the parameters, simply by choosing for example $k=1, k=2$:

$$a+d>b+c$$ $$4a+d>2(b+c)$$

But I don't know how to turn this into a finite set of simple constraints on the parameters that are necessary and sufficient.

user56834
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You can use Sylvester's criterion, which says that a Hermitian matrix is positive definite if and only if it's leading principal minors are all positive. For the case $$A = \begin{pmatrix} a& b \\ \bar b & d\end{pmatrix}$$ this turns into $$a >0$$ $$ad-|b|^2>0.$$

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    Is this also true for non-symmetric matrices ? – Peter Oct 10 '17 at 14:50
  • I had left $b\neq c$ as it is possible that the matrix is not symmetric if we allow for imaginary entries. I've updated it to be more clear. –  Oct 10 '17 at 14:54
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    @Peter: $v^T A v = v^T A_s v$ where $A_s =\frac12 (A+A^T)$ is the symmetric part of $A$. So it's enough if you know how to check it for symmetric matrices. – Hans Lundmark Oct 10 '17 at 15:00