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NIKOLA KAMBUROV: Hi, guys.

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Today we're going
to see how one can

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use linear algebra to
describe graphs and networks.

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In particular, we'll do
the following problem.

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We're given this very
simple graph here

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with five nodes and six edges.

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We've already labeled
them, and we've

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put directions on the edges.

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And we are asked to write
down the incidence matrix A,

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and then to compute its kernel
and the kernel of A transpose.

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And finally, we're
asked to compute

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the trace of A transpose A.

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I'll give you a few moments to
try the problem on your own.

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And then you'll
see my take on it.

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Hello again.

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OK, so let's first recall
what an incidence matrix is.

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So an incidence
matrix is supposed

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to encode how the nodes
connect to the edges.

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In particular, it
has as many rows

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as there are edges and as many
columns as there are nodes.

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And we're going to
fill in the rows.

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And we'll fill them
out as follows.

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So we're going to use
only negative 1, 1, and 0.

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And we're going to put a
negative 1 in entry i and 1

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in entry j if the corresponding
edge connects node i to node j.

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OK, let me just
do it concretely.

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So let's look at edge number 1.

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So it corresponds
to the first row.

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It connects 1 to 2.

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So we have a negative 1 and a 1.

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Then edge number 2, it connects
node 2 to 3, so negative 1, 1.

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Edge number 3 connects node
1 to 3, so negative 1, 1.

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And I believe you get
the picture, right?

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So I'm just going to fill
out the rest of the entries.

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All right, 4 is:
negative 1, to 1.

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5 is: well, 4, well,
negative 1, 1 here.

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And 6 is: negative 1 and 1.

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OK.

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So we've constructed
the matrix A. Now,

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we'll compute its null space.

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And we're going to do it
without performing any row

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operations whatsoever.

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So in order to do this,
it's helpful to look

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at the graph as an
electric circuit

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and to assign to each of the
nodes an electric potential.

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If we collect all the
electric potentials

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in a vector x, then A times x
is a vector with as many entries

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as there are edges
and gives precisely

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the potential differences
across the edges of the graph.

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OK, so then if A*x is to be
0, this means that across

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the graph, across all
the edges of the graph,

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all potential differences are 0.

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Therefore, all the
potentials at all the nodes

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need to be equal to
a constant number.

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So therefore, we conclude
that the null space of A

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is spanned by constant 1.

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OK?

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There are five 1's
here, corresponding

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to the five nodes.

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Now what about the null
space of A transpose?

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Adopt this analogy
with electric circuits.

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But this time, we're going
to look at currents flowing

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across the edges of the graph.

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Oh, and we are going to adopt
the following convention

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for the currents.

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So a current is
going to be positive

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if it flows in the direction
of the edge and negative

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if it flows in the
opposite direction.

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Right.

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So then, what is A
transpose y, where

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y is a vector, each of
whose entries is a current

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on the edge?

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Well, the entries
of A transpose y

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are precisely equal to the total
current flowing through each

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of the nodes of the graph.

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So A transpose y
being equal to 0

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means that there is a
balance in the circuit,

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that the currents that
flow into each node

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equal the currents
that flow out of it.

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Right.

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And it's fairly easy to
find such a configuration

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of currents that satisfies
this balance equation.

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We do it by flowing
around loops of the graph.

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So you see, this
graph has three loops.

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The first one is this
triangle up there.

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The second one is this square.

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And I'm just, by this
curled direction,

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I'm signifying in which way
I'm going to trace the loop.

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And there is third
loop, is along

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the outer contour of the graph.

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But in fact, the
third one can be

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thought of as a
superposition of these two,

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and I'll explain
why in a second.

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So let's figure out the
configuration of currents

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that balances these loops.

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So if we flow a
current 1 from 1 to 2

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and then flow a current of
1 along edge 2, from 2 to 3,

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and then we flow a
current of negative 1,

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mind that the
direction is opposite

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to the direction
of the loop, then

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we're going to have a balanced
configuration of currents.

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So let me write this down.

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The following configuration, so
1 along edge 1, 1 along edge 2,

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and negative 1 along
edge 3, and the rest 0,

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is a solution to A transpose y.

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Let's see what solution
we get by flowing around

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the loop in the square.

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Well, we flow a current
of 1 along edge 4,

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current of 1 along edge 5,
current of 1 along edge 6,

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and current of negative
1 along edge 2.

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So let's be careful.

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So it was 0, then along
edge 2 was negative 1.

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Along 3: 0; along 4: 1;
along 5: 1; along 6: 1.

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Now we can do the same
thing with the big loop

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and produce a vector
corresponding to it.

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And I prompt you to do it.

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But what you'll see is that
the vector that you get

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is precisely a sum
of these two vectors.

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In a way, the big loop
is a superposition

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of the small loops.

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OK, so we figured out what the
null space of A transpose is.

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And now, let's
concentrate our attention

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on finding the trace
of A transpose A. We're

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going to do it right here.

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So the trace of a matrix is the
sum of its diagonal entries.

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And we've seen this
many times already,

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that the diagonal
entries of A transpose A

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are precisely the magnitudes
squared of the columns of A.

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OK?

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So the (1, 1) entry
is the magnitude

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squared of the first column.

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The (2, 2) entry
is the magnitude

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squared of the second
column, and so on.

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Now what is the
magnitude squared

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of a column of an
incidence matrix?

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Well, each entry in a column
of an incidence matrix

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is either 1, negative 1, or 0.

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So when we square these
entries, we get 1's or 0's.

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And when we add them up,
we get precisely a number

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which is the nontrivial
entries in that column.

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OK?

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So the magnitude
squared of the column

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is the number of
nontrivial entries in it.

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But if we go back
to the matrix A,

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and we count the number
of non-zero entries,

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this is precisely the number of
edges that connect with a node.

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OK, so the number of edges
that connects with each node

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is called the
degree of the node.

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In this way, trace
of A transpose A

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will be just the
sum of the degrees

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of the graph in the picture.

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So we have there are 2
edges connecting to 1, so 2,

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plus 3 edges connecting to
2, 3 edges connecting to 3.

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And we've got a 2 for the
number of edges connecting to 4,

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and 2 for the number of
edges connecting to 5.

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So altogether, we get 12.

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So you see, in this problem, we
computed certain linear algebra

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objects without performing the
usual algebraic operations,

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but just by looking at
the graph and seeing

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how the linear algebra
is encoded in it.

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I hope it was most illuminating.

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I'll see you next time.