GILBERT STRANG: OK.
I want to continue
the last video, which
was about incidence matrices,
and graphs, and networks,
and flows in the network.
So that was 5.6.
This is 5.6b.
And I'll remember
the same graph.
You remember a graph is
some nodes, four nodes here,
and some edges, and in
this case five edges.
So I have a 5 by 4 matrix,
and that's what it was.
And I'll remember
how it was created.
Every row corresponds
to an edge.
So the first edge there
goes from node 1 to node 2,
so I put a minus 1 and a
1 in columns one and two.
That tells me what that
first edge is doing
and it gives me one row
of the incidence matrix.
Five edges give me five rows.
There is the matrix.
And here I'll multiply by
v, thinking of a vector v
as voltages at the four
nodes, and I get that answer.
The 1 and minus 1 produce
this kind of answer.
OK.
Now I'm ready for questions
about the matrix A, the 5
by 4 matrix.
These matrices, these
incidence matrices,
are beautiful examples
of rectangular matrices
where we can ask all the
key questions about a matrix
and get a nice answer.
And the key questions
that I have in mind are
what are their solutions
to Av equals 0?
Are there--
That says, are there are
combinations of the columns
that give the zero column?
So it's asking, are
the columns dependent?
If the columns were dependent,
then I'll find some solutions,
and here I will.
If the columns are independent,
the only solution I will find
will be v equals 0.
But those columns are dependent.
Now, how can we see that?
Well, in this case,
we can find a solution
to Av equals 0,
because I can see
how do I get all those
differences to be 0?
Well, not hard.
v could be the
vector of all 1's.
Then the differences would all
be 1 minus 1, would all be 0.
I would be solving Av equals 0.
And of course, I can
multiply by any constant.
The voltage is--
So all I'm saying
is if all the voltages are
equal, there won't be any flow.
If all the voltages
are equal and I
don't have any batteries or
other sources in the network,
there will be no flow.
And those are all the solutions.
But the only way I
could make all those 0
would be for all the
v's to be the same.
So all the v's have
to be the same.
v is C, C, C, C.
And I learn something important.
Av equals 0 has some solutions.
And I'll just jump ahead
one electrical moment.
That's not good if we
want an invertible matrix.
In the end we would
have A transpose A
and it won't be invertible
unless we do something.
And what do we do?
We want to get rid
of that last column.
We can have three columns.
Those will be independent,
but that fourth column
is a combination of the others.
And what we do,
in reality, is we
ground a node, which means we
set one of the v's, maybe v4,
if we set that to 0, it's like
we're fixing the temperature,
we're fixing the
voltage, we often
have to do this on
a sliding scale.
If we only know
differences in temperature,
we have to say, where is 0?
And if we make that
point 0, then we
have only three unknown
voltages and a 5
by 3 matrix and all well.
OK.
So that's the discussion
of Av equals 0.
Now, what about A
transpose w equals 0?
So now I'm asking about the
transpose of that matrix.
Now, this is a 4 by 5 matrix.
Again, a beautiful
example, 4 by 5 matrix.
w, of course.
It's a 4 by 5 matrix
multiplying w, which is 5 by 1.
So 4 by 5 times 5 by 1.
And I want to get all
zeros, four zeros this time.
Right.
So first of all, if I
have a 4 by 5 matrix,
so when I transpose this is
you could say short and wide,
I think there are
automatically solutions.
There will be solutions
in a 4 by 5 matrix.
With five unknowns and
only four equations,
I'm going to have some
solutions to that system.
So there will be some solutions.
Well, the question is how
many different w's could I
find, how many
different solutions,
and what do they mean.
And that's the beauty
of this example,
that it's not just a bunch
of 20 numbers in the matrix.
The matrix has a meaning.
The incidence matrix
takes differences A to Av
is differences in v, but what's
the meaning of A transpose?
That's the key question here.
Why is this equation
very important?
OK.
So I have to tell you the
meaning of A transpose.
And maybe I have to copy
down what A transpose is.
So let me go to the next board
and copy down A transpose.
So I'm looking now
at A transpose w.
So now it will be 4 by 5.
So that first row
becomes a column.
The second row becomes another
column in the transpose.
The third row, another column,
the fourth row, is that column.
And the fifth row is that one.
And that will multiply w1, 2,
3, 4, and 5 to give 0, 0, 0, 0.
And that's called the current
law, Kirchhoff's current law.
And what is that law?
What does it mean?
It means that in the network at
a typical node, so at node 1,
you remember, there
was an edge out.
Edge 1 went out.
Three edges went out actually.
This was to node 2.
This was to node 3,
and that was to node 4.
At node 1, three
edges are going out.
And what does the
current law tell me?
It tells me that the
total flow out is 0.
The net flow, any flow in, which
would be negative w's, and any
flows out, which would
be positive w's-- w,
that came from the first edge.
This was maybe the second edge.
And I think that happened
to be the fourth edge--
flows out of w.
And that's what I see here.
A 1, a 2, and a
4 are multiplying
w1, w2, and w4 are the-- The
first equation there is minus.
w1 plus w2 plus w4 equals 0.
So that came from the first
row of A transpose w equals 0.
Right?
I just took those numbers
from the first row.
I wrote down that
first equation.
And you see it says exactly
the sum of those three flows
has to be 0.
So if there's some
positive flows going out,
there must be some negative
w's coming in to balance.
OK.
And that was at node 1, and
similarly at nodes 2, and 3,
and 4, currents balance.
It's the balance equation.
Kirchhoff's law, it's
the balance equation.
It's conservation.
A fundamental
equation in modeling
applied mathematics is if
a body is sitting there
in equilibrium, then the
forces on it are in balance.
If I have steady flow
around the network,
the currents are in balance.
Always there's a
balance equation,
so that things are not
collecting up at a node.
It's stable.
OK.
So that's the meaning of
Kirchhoff's current law.
That's the meaning of
A transpose w equals 0.
And what about solutions?
Solutions w.
Now, so now we're getting
down into the details.
Can we actually find the w's?
Well, there will be some.
There will be some.
As I said, we've got five
unknowns here and only
four equations.
So we're certainly going
to find a solution.
And let me suggest one
good way to look for it.
Suppose the flow-- Let me
put in the other two edges--
suppose the flow
goes around a loop.
Loops are the key here.
The key to the
solution is a loop.
So that's a flow that sends
a flow of 1 along that edge,
a flow of 1 going that way
along that edge, which I think
was w5, and a flow
of 1 going that way.
Pay attention.
It's going to send 1
Amp around the loop.
I go with the arrow, with the
flow, this way and this way,
but this one is
against the arrow.
So I'm thinking that a
solution is w1 equals 1.
You see I'm writing
down a solution
without doing any elimination
or other linear algebra.
I'm just understanding
the picture.
w1 is 1.
w5 is 1.
w5 is 1.
And what is w4?
Negative 1, because it's
going against the arrow.
And the other two
w's are 0, w2 and w3.
This was w3 here.
Those are not
involved in this loop.
So there is a solution
with w2 and w3 equal to 0.
And I think that
how could it fail
on Kirchhoff's current law?
Nothing is piling up at a node.
We're just sending
it around a loop.
And of course, I
put in that's a 1.
w2 is a 0.
w4 is a minus 1.
I have a 1 and a minus 1.
I get 0, just right.
And all the equations
would be solved.
In other words,
conclusion, the solutions w
come from loops in the network.
Every loop in the
network gives me a w.
Here's another loop.
I could send flow down there.
Now that would be a w4 plus 1.
This way.
This way.
Do you see that second loop?
Let me draw my
little loopy symbol.
Flow going around that loop.
That loop happens to
have four edges on it.
So I'd have four w's.
1 minus 1, 1, and minus 1,
and no flow on edge 1, and I
would have another solution.
And it would be a
different solution.
So I'm going from-- Can
I insert here two loops?
In that graph I see two
loops, two small loops.
And each of those small
loops gives me a flow, a w,
that solves the current law,
because it's just continuously
running around and around.
Now, there's another
question to ask you,
and that is what about
the big loop, w1, w3--
I think that is-- and minus w2?
What if I send flow
around the big loop?
No problem.
That gives me
another set of w's .
Those satisfy
Kirchhoff's current law.
They satisfy these equations.
They satisfy A
transpose w equals 0.
But I don't want that big loop.
I don't want to include that in
my list of w's, because I was
only looking for two w's.
I was only looking for two w's.
And linear algebra told me that
was the number to look for.
And here you're suggesting--
I'll blame you--
a third around the big loop.
So what's up?
Well, do you see it?
The flow around that big
loop does solve A transpose w
equals 0, but it's not new.
It's the sum of a
flow around that loop
plus a flow around that.
Do you see?
If I add together the flow
vector, the loop vector for w
for that loop and
for that loop, they
will cancel on the edges
that are in both loops,
and I'll just be left with the
flow there, the flow there,
and the flow there, and
that's the big loop.
In other words, that
big loop doesn't give me
a new vec-- It doesn't
give me-- It gives me
a vector w that's a combination
of what I already have.
And in linear algebra,
that's always the question.
You want the number
of independent w's,
and this big loop
is a dependent w,
because it's a combination
of the other two.
OK.
So that's the picture for
one particular example.
I'll just end with linear
algebra facts, linear algebra
facts.
OK.
So how many-- So if I
have an m by n matrix,
and suppose Av equals 0 has
how many independent solutions
shall I say?
k independent solutions.
And in my example,
the incidence matrix,
the answer was, for A equal
incidence matrix, k was 1.
So if I know the number of
solutions to that equation,
then how many
solutions do I expect
to-- This has-- So how many
solutions do I expect there?
The difference between m and n
comes in it, and then plus k.
So independent solutions.
That's a basic fact
of linear algebra
that I never wrote down before.
I never wrote it
in this notation.
I'll make that a question on
a future linear algebra exam.
What I'm saying is that if I
know how many solutions Av has,
how many combinations,
these are combinations
of the columns of
A that give 0, then
I know how many combinations
of the rows of A.
Let's just check that this
counting theorem was correct.
This was k equals 1, right?
The only solution to Av equals
0 was the constants, 1, 1, 1, 1.
Then m was 5.
n was 4.
k was 1.
5 minus 4 plus 1 is 2.
And that's the number
of loop solutions
to Kirchhoff's current law.
OK.
We have voltages.
We have currents.
And there's a lot of beautiful
linear algebra involved
with these matrices.
I'll also include a video about
RLC circuits, which are totally
an application of this.
And there I'll begin with
just one loop, one RLC loop.
But the reality of
modern electronics
is thousands of nodes,
thousands of edges,
maybe tens of thousands of
edges, and many, many loops.
Good.
Thank you.
