Jtdcftul

Background

The equations formed when finding the nodal voltages of a circuit can be expressed using nodal analysis as a square system matrix $mathbf{S}$ (lets say $mtimes m$) which describes the connections and values of the conductances that correspond to these connections, and can express a whole circuit as
$$
mathbf{Sv} = mathbf{i}
$$
where $mathbf{v}$ is the collection of nodal voltages and $mathbf{i}$ are the input current sources.

Super useful matrix decomposition

In this paper, I have seen this decomposed into (for a single impedance type, e.g. resistance)
$$
mathbf{S} = mathbf{N G N}^mathrm{T}
$$
where $mathbf{N}$ specifies the connections, and is an $mtimes m$ incidence matrix which contain only values of 1, 0 and -1, and $mathbf{G}$ is an $mtimes m$ diagonal matrix containing the conductance values.

This is a ridiculously useful property as it separates the conductances from the connections making them both easily readable. No matrix decompositions I've read up on have made it clear how this works or how you'd intuitively think to apply this decomposition. Could someone explain this?

Notes

The paper actually uses modified nodal analysis but this doesn't change the application as the decomposition is only used on the nodal aspects of the circuit, not the voltage sources.

According to the document you linked, it appears to me that $mathbf{N}$ isn't $mtimes m$. Instead, it has one row per two-terminal circuit element (from a quick reading) and one column for each circuit node.

This technique has been used for decades in computing to create connections. I've used them for finding Hamiltonian cycles in graphs, for example. It's a really simple way of expressing connections.

For example, here's a 35-year old piece of code I wrote to test out a method for finding the existence of such cycles:

#include <stdio.h>

#include <stdlib.h>

typedef enum { false= 0, true= 1 } bool_t;

void hamPrint( int n, int *path ) {

    int i;

        for ( i= 0; i < n; ++i )

            printf( " %d ", path[i] );

        printf( " %dn", path[0] );

    return;

}

bool_t hamOkay( int n, int v, bool_t *graph, int *path, int pos ) {

    int i;

        if ( graph[ path[pos-1]*n + v ] == false ) return false;

        for ( i= 0; i < pos; ++i ) if ( path[i] == v ) return false;

     return true;

}

bool_t hamCycleSolver( int n, bool_t *graph, int *path, int pos ) {

    int v;

        if ( pos == n )

            return graph[ path[pos-1]*n + path[0] ];

        for ( v= 1; v < n; ++v )

            if ( hamOkay( n, v, graph, path, pos ) ) {

                path[pos]= v;

                if ( hamCycleSolver( n, graph, path, pos+1 ) == true )

                    return true;

                path[pos]= -1;

            }

    return false;

}

bool_t hamCycleExist( int n, bool_t *graph ) {

    bool_t stat;

    int i, *path= (int *) malloc( sizeof(int) * n );

        if ( path == NULL ) return false;

        for ( i= 0; i < n; ++i )

            path[i]= -1;

        path[0]= 0;

        stat= hamCycleSolver( n, graph, path, 1 );

        if ( stat == true ) hamPrint( n, path );

        free( path );

    return stat;

}

bool_t graph[5]= {    /* Create the following graph */

   { 0, 1, 0, 1, 0 },       /*      (0)     (2)     */

   { 1, 0, 1, 1, 1 },       /*       |    / |      */

   { 0, 1, 0, 0, 1 },       /*       |  (1)  |      */

   { 1, 1, 0, 0, 1 },       /*       | /    |      */

   { 0, 1, 1, 1, 0 },       /*      (3)-----(4)     */

};

int main( void ) {

        if ( hamCycleExist( sizeof(graph)/sizeof(graph[0]), (bool_t *) graph ) )

            printf( "Graph is Hamiltoniann" );

        else

            printf( "Graph is not Hamiltoniann" );

    return 0;

}

Take note of the use of a connection matrix in the matrix graph. In this case, the connections must be specified in both directions. So there are "1"s specified to connect, for example, node 0 to node 1 and also node 1 to node 0. So it's easy to change this matrix to specify a path from node 0 to node 1 without specifying a path from node 1 to node 0, here. I just didn't do that, in the above case. All connections there are explicitly arranged to work in both directions.

If interested, you can simply multiply such a matrix by an appropriate vector to get a vector of connections for each entry in the appropriate vector, too.

In any case, here is a web page I readily found on google that may also help demonstrate that these ideas have been around for a long time and are in regular use:
Graph representations.

I had simply borrowed the idea, myself. I didn't invent it. So it pre-dates my use. And that means it is practically ancient. ;) I wouldn't be the least bit surprised to hear it dates into the 1800's.

I'm no mathematician but I feel strongly this is related to the singular value decomposition (SVD) or eigendecomposition.

I first came across SVD in the context of modelling MIMO communication systems, particularly those using spatial multiplexing. I'll try to detail this to explain why I think it relates to your problem which I am not able to answer directly.

Consider a time-invariant, noiseless MIMO channel. This can be represented as.

$
mathbf{y} = H(omega)mathbf{x}
$

Where H is a matrix of transfer functions between the various parallel channels. Ideally, H would be diagonal and there would be no coupling between each channel. The presence of off-diagonal entries means that equalization will be required to prevent the channels interfering.

The SVD decomposes H into

$
H = ULambda V^*
$

Where $U$ and $V$ can be thought of as rotations and $Lambda $ is a diagonal matrix that simply scales each channel individually. $U$ and $V$ are both unitary matrices, so their inverses are their conjugate transposes. The columns of U and V also form orthonormal basis, so they can be thought of as the natural 'coordinate system' for solving the problem.

Intuitively, it takes the input channels, which are not orthogonal, and applies a transformation at the input and output that makes the behavior of the channel very simple, just attenuation (the matrix $Lambda $).

This has application to equalization, if we pre-multiply our input signals with $V$, pass it through the channel, and apply $U^*$ to the output. We get,

$
mathbf{y} = U^*ULambda V^*Vmathbf{x} \
mathbf{y} = Lambda mathbf{x}
$

Which gives us completely orthogonal channels that do not interfere. This reminds me very much of your problem, the connection matrices being the natural orthogonal basis to use, and the conductances simply scaling these.

The SVD also has some interesting applications in image processing

Edit: The decomposition in question is definitely an eigenvalue decomposition, of which the SVD can be thought of as a generalization.