Wed Nov 5 01:32:50 EST 2014

Why is a conductance matrix positive definite?

G V = I  or  V = G^-1 I

Given a network of resistors and current sources, compute the node
voltages.  The inverse problem is: given a network of resistors and
voltage sources, compute the currents drawn from the voltage sources
applied to each node.

The matrix G is positive definite if V^T G V > 0 for each V, meaning
V^T I > 0 or the total power dissipation is positive, which is clear
from physical properties.

So the this also answers a previous question: The quadratic form
associated with the symmetric conductance is the power dissipation of
the resistor network.

Does it work in the other direction?  Can a positive definite
quadratic form be written as a conductance matrix?  Probably not as
there is a structural constraint.  A more interesting question to ask
is probably: what are the eigenvectors / values?

Also, once transconductances are introduced, the symmetry breaks.