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The number of connected components in a graph is equal to the dimension of the null space of the edge-node incident matrix. The non-zero entries ...
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Chapter 2: Vector Spaces, Linear Transformations, and their applications Section 5: Applications to networks Ivan Contreras, Sergey Dyachenko and Bob Muncaster University of Illinois at Urbana-Champaign February 9 2018 Ivan Contreras, Sergey Dyachenko and Bob MuncasterUniversity of Illinois at Urbana-ChampaignApplied Linear Algebra——– February 9 2018 1 / 8
Networks arise in a variety of settings (internet structure, transportation
patterns, electrical circuits, etc.) Structurally a network is represented by a
directed graph (digraph) along which things (information, product,
electrical current) flow. Here is an example:
1 2
3 4
1
2
3
4
5
loop 1
loop 2
N Node
E Edge
an arrow indicates a direction of positive flow
no edge goes from a node to that same node
there is at most one edge pointing from one node to another
For an edge-node incident matrix, each of the four fundamental subspaces
captures some information about the structure of the network. Let us
begin with the null space. For the matrix in our example:
Ax = 0 ⇔
−x 1 + x 2 = 0
−x 1 + x 3 = 0
−x 2 + x 3 = 0
−x 2 + x 4 = 0
−x 3 + x 4 = 0
all the xi ’s
are the same
⇔ N(A) = Span{
Now consider a second example:
1 2
3 4
, N(B) = Span{
The difference in these two examples lies in “connectedness.” The network
represented by A is connected, i.e. there is a path (even if you have to go
backwards along an edge) from each node to every other node. The graph
represented by B is disconnected. But we say that it has two connected
components, nodes 1 and 3, and nodes 2 and 4.
The number of connected components in a graph is equal
to the dimension of the null space of the edge-node incident
matrix. The non-zero entries in each basis vector indicate the
nodes that represent each connected component of the graph.
Look next at the column space. We note in example A that Row 1 + Row
3 = Row 2. Also Row 3 + Row 5 = Row 4. For Ax = b this means that
b 1 + b 3 = b 2 , b 3 + b 5 = b 4 (conditions restricting b for C (A))
Note that we can read these off from the small loops in the graph!
Finally we consider the row space. We noted above that Row 2 and Row
4 were sums of the other rows in the matrix. This indicates linear
dependence in the row space and we can throw such rows away when
finding a basis of C (A
T ). But each row of A represents an edge! So what
happens if we modify the graph by throwing these edges away? We get:
1 2
3 4
1
3
5
minimal ︸ ︷︷ ︸
smallest # edges
spanning ︸ ︷︷ ︸
contains all nodes
tree ︸︷︷︸
no loops
Each basis vector of the row space of the edge-node incident matrix
represents one of the edges in a minimal spanning tree for the graph.
This is the smallest simplification of the network (by removing
redundant edges) that maintains its connectedness and flow pattern.
Euler established an interesting identity concerning “connected” graphs:
(# nodes) - (# edges) + (# small loops) = 1
With our machinery we can easily establish this. If the graph is connected
then the dimension of the null space is 1. This means the rank of A is
r = n − 1 (nullity + rank = n). And the dimension of the left null space is
m − r. Hence
(# nodes) - (# edges) + (# small loops) = n − m + (m − r )
= n − m + (m − (n − 1 ))
