Show that stable spriral has index of 1. How do you go about generally showing that the index at the origin for different linear systems is equal to +1? For instance, the center of a stable spiral.

For a specific system, I could show how the vector directions change as we traverse the unit circle around the point. But how do I show that its true in general for a type of linear system? Is there a theorem I can use about how the index won't change for a closed curve for systems of the same type?
 A: There is a useful and well-known result which applies to this situation:  the index of a vector field $V$ at a point $p$ where $V(p) = 0$ and $J_V(p)$, the Jacobean of $V$ at $p$, is non-singular, is equal to the sign of $\det J_V(p)$.  A nice explanation of this, with proof, may be found here.  For a non-degenerate stable spiral, the eigenvalues of $J_V(p)$ are a complex conjugate pair $\sigma \pm i\omega$, with $\sigma < 0, \omega \ne 0$; thus 
$\det J_V(p) = (\sigma + i\omega)(\sigma - i\omega) = \sigma^2 + \omega^2 > 0, \tag{1}$
whence
$\text{index}V(p) = +1. \tag{2}$
It may be seen from (1) that the index of $V(p)$ at any zero $p$ with a complex conjugate pair of eigenvalues is also $+1$; thus unstable spirals (with $\sigma > 0, \omega \ne 0$) and elliptic points ($\sigma = 0, \omega \ne 0$) are also of index $+1$; if the eigenvalues of $J_V(p)$ are both real, say $0 \ne \lambda, \mu \in \Bbb R$, then we have
$\det J_V(p) = \lambda \mu, \tag{3}$
which shows that both stable ($\lambda, \mu < 0$) and unstable ($\lambda, \mu > 0$) nodes are of index $1$ as well; the only case (for two-dimensional vector fields at least) with $\det J_V(p) < 0$ occurs when $\lambda < 0 < \mu$ or $\mu < 0 < \lambda$; only saddles have index $-1$.
This technique is especially easy to apply to linear systems of the form $V(x) = Ax$, so that $V(0) = 0$, since then $J_V(0) = A$, as may be readily seen. 
It is usually a lot easier to calculate $\det J_V(p)$ than to try and perform line integrals on circles around $p$ or otherwise analyze the way $V/\Vert V \Vert$ rotates as we traverse a path enclosing $p$.  Of course at a degenerate $0$ of $V$, where $\det J_V(p) = 0$, this method breaks down and other devices must be employed to arrive at $\text{index}V(p)$.
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
