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.