# The system $x'=Ax$ is an attractor if and only if there is a positive quadratic form q such that $Dq(x)\cdot A(x)<0$ for all x

I need to show this result:

Given the system of ODEs $x'=Ax$, the origin, $0$, is an attractor (equivalently, all the eigenvalues of the real matrix $A$ are negative) if and only if there exists a positive definite quadratic form $q$ such that $Dq(x)\cdot Ax<0$ for all $x\neq 0$ (D is the differential operator, $\cdot$ is the usual inner product).

I have no idea how to start. I tried to do it in the 2x2 case, but expanding the expression gave me no clues. I believe I have to find an expression for $q$ which will imply in the inequality I want.

If anyone could give me some hint, I would be grateful; I ask not to solve it fully, since I could use any development on Linear Algebra I could get.

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Hint: Let $Q$ be a positive-definite matrix and define the function $J(x)=x'Qx$. This function is becomes zero if and only if $x=0$. If along system trajectories you have $dJ/dt<0$, then the function is strictly decreasing and bounded below by zero, thus it will eventually go to zero. Hence the state will go to zero. This is true for any initial conditions. Hence the system will be stable, i.e. all eigenvalues of $A$ will have negative real part. Start from $dJ/dt<0$.
Added: A system $dx/dt=Ax$ is an attractor if and only if $A$ has eigenvalues with negative real part if and only if given any positive-definite matrix $P$ there exists a positive definite matrix $Q$ such that $A'Q+QA=-P$. The latter is a Lyapunov equation.
Suppose that the system is an attractor. Then all eigenvalues of $A$ have negative real part. Take any positive definite matrix $P$ and define $Q=\int_{0}^{\infty}e^{At}Pe^{A't}dt$. Then verify that $dJ/dt<0$ and $Q>0$.
I believe that $x$ is a solution for the system, right? – Marra Sep 21 '12 at 14:43
Assuming it is, I couldn't find a way to justify why $\dfrac{dJ}{dt}<0$... I made some particular cases to to try and it seems to be OK, but I don't know why this derivative is lesser than zero! I have tried putting it in the diagonal form of Jordan (assuming, in a first case, that all eigenvalues are distinct) to no result also. – Marra Sep 21 '12 at 15:49
Yes, $x$ satisfies the dynamics relation. – Manos Sep 21 '12 at 16:14
Is $A'$ the conjugate transpose? – Marra Sep 21 '12 at 16:15