# Outward vectors to an Ellipsoid and Euclidean metrics

I'm reading Arnold's proof of the topologically equivalence of the equations $\dot{x}=Ax$ and $\dot{x}=x$ when all the eigenvalues of the $n \times n$-matrix $A$ have positive real part. The proof is based in the construction of a Lyapunov function and the equivalence of these statements:

1. There exists and Euclidean metric $g$ on $R^n$ such that $g(Ax,x)>0$ (for $x \neq 0$).
2. There exists a positive-definite quadratic form $F$ on $R^n$ whose directional derivative in the direction of $Ax$ is positive.
3. There exists an ellipsoid in $R^n$ with center at $0$ such that at each point $x$ the vector $Ax$ is directed outward.

I can see the equivalence of the first two but I don't really understand the third one, I guess the ellipsoid is defined by $F(x)=g(x,x)=$constant, but I can't see why it is equivalent to the first statement for example.

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An ellipsoid centered at the origin in $\mathbb{R}^n$ is described by $g(x,x)=1$, where $g$ is some positive definite quadratic form. If $G$ is the matrix representation of $g$ w.r.t. the standard basis, then the outward pointing normal to the ellipsoid at $x\in\mathbb{R}^n$ is given by $\mathbf{n}=Gx$. Therefore statement 3 means $\langle Ax, \mathbf{n}\rangle>0$ (where $\langle x, y\rangle\equiv y^Tx$ is the usual inner product), which is equivalent to $x^TGAx>0$ and in turn $g(Ax,x)>0$.
Thanks, I guess the normal vector can be checked that is $n$ simply because it is in the direction of the gradient of $g$. – inquisitor Jan 21 '13 at 11:23