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I just want to remind you of
the main facts.

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The first thing that you have
to do is, of course,

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we are going to have to be
doing it several times today.

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That is the system we are
trying to solve.

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And the first thing you have to
do is find a characteristic

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equation which is general form,
although this is not the form

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you should use for two-by-two,
is A minus lambda I

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 equals zero.
And its roots are the

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eigenvalues.

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And then with each eigenvalue
you then have to calculate its

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eigenvector, which you do by
solving the system (A minus

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lambda1, let's say,
times I) alpha equals zero

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because
the solution is the eigenvector

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alpha 1.
And then the final solution

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that you make out of the two of
them looks like alpha 1 times e

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to the lambda 1t.

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Of course you do that for each
eigenvalue.

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You get the associated
eigenvector.

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And then the general solution
is made up out of a linear

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combination of these individual
guys with constant coefficients.

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The lecture today is devoted to
the two cases where things do

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not go as smoothly as they seem
to in the homework problems you

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have been doing up until now.
The first one will take

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probably most of the period.
It deals with what happens when

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an eigenvalue gets repeated.
But I think since the situation

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is a little more complicated
than it is where the case of a

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characteristic root gets
repeated in the case of a

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second-order equation as we saw
it, you know what to do in that

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case, here there are different
possibilities.

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And I thought the best thing to
do would be to illustrate them

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on an example.
So here is a problem.

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It came out of a mild
nightmare, but I won't bore you

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with the details.
Anyway, we have this circular

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fish tank.
It is a very modern fish tank.

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It is divided into three
compartments because one holds

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Siamese fighting fish and one
goldfish, and one-- They should

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not eat each other.
And it is going to be a simple

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temperature problem.
The three actual compartments

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have to be kept at different
temperatures because one is for

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tropical fish and one is for
arctic fish and one is for

43
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everyday garden variety fish.
But the guy forgets to turn on

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the heater so the temperatures
start out what they are supposed

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to be, tropical,
icy, and normal.

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But as the day wears on,
of course, the three

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compartments trade their heat
and sort of tend to all end up

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at the same temperature.
So we are going to let (x)i

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equal the temperature in tank i.
Now, these are separated from

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each other by glass things.
Everything is identical,

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each has the same volume,
and the same glass partition

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separates them out and no heat
can escape.

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This is very well-insulated
with very double-thick

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Thermopane glass or something
like that.

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You can see in,
but heat cannot get out very

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well.
Heat essentially is conducted

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from one of these cells to the
other.

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And let's assume that the water
in each tank is kept stirred up

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because the fish are swimming
around in it.

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That should be a pretty decent
way of stirring a fish tank.

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The question is how do each of
these, as a function of time,

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and I want to know how they
behave over time,

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so find these functions.

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Well, we are going to find them
in solutions to differential

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equations.
And the differential equations

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are not hard to set up.
They are very much like the

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diffusion equation you had for
homework or the equations we

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studied in the beginning of the
term.

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Let's do one carefully because
the others go exactly the same

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way.
What determines the flow,

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the change in temperature?
Well, it is the conductivity

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across the barriers.
But there are two barriers

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because heat can flow into this
first cell, both from this guy

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and it can flow across this
glass pane from the other cell.

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We have to take account of both
of those possibilities.

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It is like in your homework.
The little diffusion cell that

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was in the middle could get
contributions from both sides,

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whereas, the two guys on the
end could only get contribution

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from one.
But here, nobody is on the end.

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It is circular table.
Everyone is dying equally.

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Everybody can get input from
the other two cells.

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x1 prime is some
constant of conductivity times

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the temperature difference
between tank three and tank one.

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And then there is another term
which comes from tank two.

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So a times tank two minus the
temperature difference,

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tank two minus tank one.
Let's write this out.

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Remember there will be other
equations, too.

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But instead of doing this,
let's do a more careful job

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with this first equation.
When I write it out,

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remember, the important thing
is you are going to have x1,

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x2, x3 down the left,
so they have to occur in the

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same order on the right in order
to use these standard eigenvalue

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techniques.
The coefficient of x1 is going

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to be minus a x1 and then
another minus a x1.

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In other words,
it is going to be minus 2 ax1.

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And then the x2 term will be
plus a x2.

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And the x3 term will be plus a
x3.

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Well, you can see now that is
the equation for x1 prime in

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terms of the other variables.
But there is symmetry.

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There is no difference between
this tank, that tank,

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and that tank as far as the
differential equations are

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concerned.
And, therefore,

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I can get the equations for the
other two tanks by just changing

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1 to 2, just switching the
subscripts.

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When I finally do it all,
the equations are going to be,

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I will write them first out as
a system.

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Let's take a equal 1
because I am going to want to

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solve them numerically,
and I want you to be able to

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concentrate on what is
important, what is new now and

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not fuss because I don't want to
have an extra a floating around

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everywhere just contributing
nothing but a mild confusion to

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the proceedings.
So x1 prime,

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I am going to take a equal 1
and simply write it minus 2 x1

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plus x2 plus x3.

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And so now what would the
equation for x2 prime

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be?
Well, here x2 plays the role

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that x1 played before.
And the only way to tell that

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x1 was the main guy here was it
occurred with a coefficient

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negative 2, whereas,
the other guys occurred with

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coefficient 1.
That must be what happens here,

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too.
Since x2 prime is our

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main man, this is minus x2 and
this must be x1 here plus x3.

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And finally the last one is no
different, x3 prime is x1 plus

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x2.
And now it is the x3 that

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should get negative 2 for the
coefficient.

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There is a perfectly
reasonable-looking set of

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equations.
Just how reasonable they are

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depends upon what their
characteristic polynomial turns

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out to be.
And all the work in these

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problems is trying to find nice
models where you won't have to

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use Matlab to calculate the
roots, the eigenvalues,

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the roots of the characteristic
polynomial.

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So we have to now find the
characteristic polynomial.

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The matrix that we are talking
about is the matrix,

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well, let's right away write A
minus lambda I

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I cannot use the trace and

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determinant form for this
equation because it is not a

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two-by-two matrix.
It is a three-by-three matrix.

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I have to use the original form
for the characteristic equation.

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But what is this going to be?
Well, what is A?

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A is minus 2.
I am going to leave a little

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space here.
1, 1, 1 minus 2,

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1.
And finally 1,

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1, negative 2.
subtract lambda from

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the main diagonal,
minus 2 minus lambda

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minus 2 minus lambda,
minus 2 minus lambda.

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And now that equals zero is the
characteristic equation.

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The term with the most lambdas
in it is the main diagonal.

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That is always true,
notice.

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Now, each of these I would be
happier writing lambda plus 2,

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so there would be a negative
sign, negative sign,

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negative sign.
The product of three negative

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signs is still a negative sign
because three is an odd number.

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So it is minus the principle
term.

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The product of these three is
minus lambda plus 2 cubed.

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Now, the rest of the terms are

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going to be easy.
There is another term 1 times 1

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times 1, another term 1 times 1
times 1.

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So to that I add 2,
1 and 1 for those two other

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terms.
And now I have the three going

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in this direction,
but each one of them has to be

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prefaced with a minus sign.
What does each one of them come

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to?
Well, this is minus 2 minus

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lambda when I multiply those
three numbers together.

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And so are the other guys.
This is 1 times 1 times minus 2

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minus lambda,
the same thing.

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There are three of them.
Minus because they are going

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this way, minus 3 because there
are three of them,

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and what each one of them is is
negative 2 negative lambda.

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That is equal to zero,
and that is the characteristic

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equation.
Now, it doesn't look very

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promising.
On the other hand,

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I have selected it for the
lecture.

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Simple psychology should tell
you that it is going to come out

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okay.
What I am going to do is expand

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this.
First imagine changing the

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sign.
I hate to have a minus sign in

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front of a lambda cubed,

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so let's make this plus and we
will make this minus and we will

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make this plus.
I will just change all the

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signs, which is okay since it is
an equation equals zero.

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That doesn't change its roots
any.

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And now we are going to expand
it out.

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What is this?
Lambda plus 2 cubed.

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Lambda cubed plus,

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and don't get confused because
it is this 2 that will kill you

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when you use the binomial
theorem.

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If there is 1 here everybody
knows what to do.

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If there is an A there
everybody knows what to do.

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It is when that is a number not
1 that everybody makes mistakes,

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including me.
The binomial coefficients are

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1, 3, 3, 1 because it is a
cubed, I am expanding.

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So it is lambda cubed plus 3
times lambda squared times 2.

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I won't explain to you what I
am doing.

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I will just do it and hope that
you all know what I am doing.

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Plus 3 times lambda times 2
squared plus the last term,

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which is 2 cubed.

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And now we have the other term.

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All that is plus because I
changed its sign.

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The next thing is negative 2.
And then the last thing is plus

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3 times (minus 2 minus lambda).

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Let's keep it.
So what is the actual

205
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characteristic equation?
Maybe I can finish it.

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I should stay over here instead
of recopying all of it.

207
00:13:22 --> 00:13:28

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Well, there is a lot more work
to do.

209
00:13:29 --> 00:13:35
Let's see if we can at least
write down the equation.

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What is it?
It is lambda cubed.

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00:13:35 --> 00:13:41
What is the lambda squared

212
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erm?
It is six and that is all there

213
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is.
How about the lambda term?

214
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Well, we have 12 lambda minus 3
lambda which makes plus 9

215
00:13:48 --> 00:13:54
lambda.
That looks good but constant

216
00:13:51 --> 00:13:57
terms have a way of screwing
everything up.

217
00:13:54 --> 00:14:00
What is the constant term?
It is A minus 2 minus 6.

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00:14:00 --> 00:14:06
Zero.
The constant term is zero.

219
00:14:02 --> 00:14:08
That converts this from a hard
problem to an easy problem.

220
00:14:07 --> 00:14:13
Now it is a cinch to calculate
the stuff.

221
00:14:11 --> 00:14:17
Let's go to this board and
continue the work over here.

222
00:14:17 --> 00:14:23

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00:14:24 --> 00:14:30
The equation is lambda cubed
plus 6 lambda squared plus 9

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lambda is zero.

225
00:14:31 --> 00:14:37
It is very easy to calculate
the roots of that.

226
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You factor it.
Lambda is a common factor.

227
00:14:37 --> 00:14:43
And what is left?
Lambda squared plus 6 lambda

228
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plus 9.

229
00:14:43 --> 00:14:49
That is the sort of thing you
got all the time when you were

230
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studying critical damping.
It is the square of lambda plus

231
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3. Lambda squared plus 6 lambda

232
00:14:54 --> 00:15:00
plus 9 equals zero.
So the eigenvalues,

233
00:14:58 --> 00:15:04
the roots are what?
Well, they are lambda equals

234
00:15:02 --> 00:15:08
zero from this factor and then
lambda equals minus 3.

235
00:15:07 --> 00:15:13
But what is new is that the

236
00:15:10 --> 00:15:16
minus 3 is a double root.
That is a double root.

237
00:15:14 --> 00:15:20
Now, that, of course,
is what is going to cause the

238
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trouble.
Because, for each one of these,

239
00:15:21 --> 00:15:27
I am supposed to calculate the
eigenvector and make up the

240
00:15:26 --> 00:15:32
solution.
But that assumed that I had

241
00:15:29 --> 00:15:35
three things to get three
different solutions.

242
00:15:32 --> 00:15:38
Here I have only got two
things.

243
00:15:34 --> 00:15:40
It is the same trouble we ran
into when there was a repeated

244
00:15:38 --> 00:15:44
root.
We were studying second or

245
00:15:40 --> 00:15:46
third order differential
equations and the characteristic

246
00:15:43 --> 00:15:49
equation had a repeated root.
And I had to go into a song and

247
00:15:47 --> 00:15:53
dance and stand on my head and
multiply things by t and so on.

248
00:15:51 --> 00:15:57
And then talk very hard arguing
why that was a good thing to do

249
00:15:54 --> 00:16:00
to get the answer.
Now, I am not going to do the

250
00:15:57 --> 00:16:03
same thing here.
Instead, I am going to try to

251
00:16:02 --> 00:16:08
solve the problem instead.
Let's get two points by at

252
00:16:06 --> 00:16:12
least doing the easy part of it.
Lambda equals zero.

253
00:16:11 --> 00:16:17
What am I supposed to do with

254
00:16:13 --> 00:16:19
lambda equals zero?
I am looking for the alpha that

255
00:16:17 --> 00:16:23
goes with that.
And I find that eigenvector by

256
00:16:21 --> 00:16:27
solving this system of
equations.

257
00:16:24 --> 00:16:30
Let's write out what that
system of equations is.

258
00:16:29 --> 00:16:35
Well, if lambda is zero,
this isn't there.

259
00:16:32 --> 00:16:38
It is just the matrix A times
alpha equals zero.

260
00:16:38 --> 00:16:44
And the matrix A is,

261
00:16:40 --> 00:16:46
I never even wrote it anywhere.
I never wrote A.

262
00:16:45 --> 00:16:51
I thought I would get away
without having to do it,

263
00:16:49 --> 00:16:55
but you never get away with
anything.

264
00:16:53 --> 00:16:59
It's the principle of life.
That is A.

265
00:16:58 --> 00:17:04
If I subtract zero from the
main diagonal,

266
00:17:01 --> 00:17:07
that doesn't do a great deal to
A.

267
00:17:04 --> 00:17:10
And the resulting system of
equations is those same things,

268
00:17:10 --> 00:17:16
except you have the a1's there,
too.

269
00:17:13 --> 00:17:19
There is one.
a1 minus 2 a2 plus a3 equals

270
00:17:17 --> 00:17:23
zero.
I am just subtracting zero from

271
00:17:22 --> 00:17:28
the main diagonal so there is
nothing to do.

272
00:17:26 --> 00:17:32
a2 minus 2 a3 equals zero.
Now I am supposed to solve

273
00:17:33 --> 00:17:39
those.

274
00:17:35 --> 00:17:41

275
00:17:40 --> 00:17:46
Of course we could do it.
Well, how do you know how to

276
00:17:43 --> 00:17:49
solve a system of three linear
equations?

277
00:17:45 --> 00:17:51
Well, elimination.
You can always solve by

278
00:17:48 --> 00:17:54
elimination.
Now we are much more

279
00:17:50 --> 00:17:56
sophisticated than that.
You all have pocket calculators

280
00:17:54 --> 00:18:00
so you could use the inverse
matrix, right?

281
00:17:56 --> 00:18:02
No.
You cannot use the inverse

282
00:17:58 --> 00:18:04
matrix.
What will happen if you punch

283
00:18:03 --> 00:18:09
in those coefficients and then
punch in A inverse.

284
00:18:08 --> 00:18:14
What answer will it give you?
0, 0, 0.

285
00:18:12 --> 00:18:18
No, I am sorry.
It won't give you any answer.

286
00:18:16 --> 00:18:22
What will it say?
It will say I cannot calculate

287
00:18:21 --> 00:18:27
the inverse to that matrix
because the whole purpose of

288
00:18:27 --> 00:18:33
this exercise was to find a
value of lambda such that this

289
00:18:33 --> 00:18:39
system of equations is
dependent.

290
00:18:38 --> 00:18:44
The coefficient determinant is
zero and, therefore,

291
00:18:42 --> 00:18:48
the coefficient matrix does not
have an inverse matrix.

292
00:18:46 --> 00:18:52
You cannot use that method.
In other words,

293
00:18:49 --> 00:18:55
the inverse matrix will never
work in these problems because

294
00:18:54 --> 00:19:00
the system of equations you will
be trying to solve is always a

295
00:18:59 --> 00:19:05
non-independent system.
And, therefore,

296
00:19:03 --> 00:19:09
its determinant is always zero.
And, therefore,

297
00:19:06 --> 00:19:12
there is no inverse matrix
because the determinant of the

298
00:19:11 --> 00:19:17
coefficient is zero.
All you can do is use

299
00:19:14 --> 00:19:20
elimination or physical insight
and common sense.

300
00:19:18 --> 00:19:24
Now, because I teach
differential equations everybody

301
00:19:23 --> 00:19:29
assumes, mistakenly,
as I think, that I really know

302
00:19:27 --> 00:19:33
something about them.
I get now and then graduate

303
00:19:32 --> 00:19:38
students, not in mathematics,
but some obscure field of

304
00:19:36 --> 00:19:42
engineering or whatever drift
into my office and say I see you

305
00:19:40 --> 00:19:46
teach differential equations.
Do you have a minute here?

306
00:19:45 --> 00:19:51
And before I can say no they
write their differential

307
00:19:49 --> 00:19:55
equation on the board.
And almost invariably it is

308
00:19:52 --> 00:19:58
nothing I have ever seen before.
And they so look at me

309
00:19:57 --> 00:20:03
hopefully and expectantly.
So what do I ask them?

310
00:20:01 --> 00:20:07
I don't ask them what they have
tried.

311
00:20:05 --> 00:20:11
What I ask them is where did
this come from?

312
00:20:08 --> 00:20:14
What field did it come from?
Because each field has its own

313
00:20:14 --> 00:20:20
little tricks.
It gets the same differential

314
00:20:17 --> 00:20:23
equations all the time and has
its own little tricks for

315
00:20:22 --> 00:20:28
solving them.
You should do the same thing

316
00:20:26 --> 00:20:32
here.
Well, of course we can solve

317
00:20:29 --> 00:20:35
this.
And by now most of you have

318
00:20:33 --> 00:20:39
solved it just by inspection,
just by sort of psyching out

319
00:20:37 --> 00:20:43
the answer.
But a better way is to say

320
00:20:40 --> 00:20:46
look, suppose we had the
solution, what would the

321
00:20:44 --> 00:20:50
solution look like?
Well, it would look like (a1,

322
00:20:48 --> 00:20:54
a2, a3), whatever the values of
those variables were which gave

323
00:20:52 --> 00:20:58
me the solution to the equation,
times e to the 0t.

324
00:20:57 --> 00:21:03
But what is this?
e to the 0t is one

325
00:21:02 --> 00:21:08
for all time.
And, therefore,

326
00:21:04 --> 00:21:10
this is a constant solution.
What I am asking is to find a

327
00:21:09 --> 00:21:15
constant solution.
Now, can I, by inspection,

328
00:21:13 --> 00:21:19
find a constant solution to
this?

329
00:21:15 --> 00:21:21
If so it must be the one.
Well, there is an obvious

330
00:21:19 --> 00:21:25
constant solution.
All the cells have the same

331
00:21:23 --> 00:21:29
temperature.
If that is true then there is

332
00:21:26 --> 00:21:32
no reason why it should ever
change as time goes on.

333
00:21:32 --> 00:21:38
The physical problem itself
suggests what the answer must

334
00:21:35 --> 00:21:41
be.
You don't have to solve

335
00:21:37 --> 00:21:43
equations.
In other words,

336
00:21:38 --> 00:21:44
any constant like (1,
1, 1).

337
00:21:40 --> 00:21:46
Well, could it be (20,
20, 20)?

338
00:21:42 --> 00:21:48
Yeah, that is a constant
multiple of (1,

339
00:21:44 --> 00:21:50
1, 1).
That is included.

340
00:21:45 --> 00:21:51
My basic constant solution,
therefore, is simply (1,

341
00:21:48 --> 00:21:54
1, 1) times e to the 0t.

342
00:21:51 --> 00:21:57
You don't have to include e to
the 0t because it is one.

343
00:21:54 --> 00:22:00
Now, just to check,
is (1, 1, 1) a solution to

344
00:21:57 --> 00:22:03
these equations?
It certainly is.

345
00:22:01 --> 00:22:07
1 plus 1 minus 2 is zero in
every case.

346
00:22:04 --> 00:22:10
The equations are essentially
the same, except they use

347
00:22:08 --> 00:22:14
different variables.
By inspection or,

348
00:22:11 --> 00:22:17
if you like,
by elimination,

349
00:22:14 --> 00:22:20
but not by finding the inverse
matrix you solve those

350
00:22:18 --> 00:22:24
equations.
And we have our first solution.

351
00:22:21 --> 00:22:27
Now let's go onto the second
one.

352
00:22:24 --> 00:22:30
For the second one,
we are going to have to use the

353
00:22:28 --> 00:22:34
eigenvalue lambda equals
negative 3.

354
00:22:33 --> 00:22:39
And now what is the system of
equations?

355
00:22:36 --> 00:22:42
Well, now I have to take this
and I have to subtract negative

356
00:22:41 --> 00:22:47
3 from the diagonal elements.
Minus 2 minus negative 3 is

357
00:22:46 --> 00:22:52
plus 1, right?

358
00:22:49 --> 00:22:55
Got that?
Each of the diagonal elements,

359
00:22:53 --> 00:22:59
after I subtract minus 3 turns
into plus 1.

360
00:22:58 --> 00:23:04
And, therefore,
the system becomes,

361
00:23:00 --> 00:23:06
the system I have to solve is
a1 plus a2 plus a3 equals zero.

362
00:23:05 --> 00:23:11
And what is the second

363
00:23:09 --> 00:23:15
equation?
Symmetry is preserved.

364
00:23:11 --> 00:23:17
All the equations are
essentially the same,

365
00:23:14 --> 00:23:20
except for the names of the
variables so they all must give

366
00:23:19 --> 00:23:25
you the same thing after I
subtract minus 3 from the main

367
00:23:24 --> 00:23:30
diagonal.
Well, that is what we call a

368
00:23:27 --> 00:23:33
dependent system of equations.
All I have is the same equation

369
00:23:34 --> 00:23:40
repeated twice,
but I still have to solve it.

370
00:23:38 --> 00:23:44
Now, what you see is that there
are lots of solutions to this.

371
00:23:44 --> 00:23:50
Let me write down one of them.
For example,

372
00:23:49 --> 00:23:55
suppose I made a1 equal to 1
and I made a2 a 0,

373
00:23:54 --> 00:24:00
then a3 would be negative 1.

374
00:24:00 --> 00:24:06
So here is a solution.
That is the eigenvector.

375
00:24:03 --> 00:24:09
And with it,
I can make the solution by

376
00:24:06 --> 00:24:12
multiplying by e to the negative
3t.

377
00:24:10 --> 00:24:16
There is a solution.
But that is not the only alpha

378
00:24:14 --> 00:24:20
I could have chosen.
Suppose I chose this one

379
00:24:18 --> 00:24:24
instead.
Suppose I kept this 1,

380
00:24:20 --> 00:24:26
but this time made a3 zero.
Well, in that case,

381
00:24:24 --> 00:24:30
there would be a2 that had to
be minus 1.

382
00:24:30 --> 00:24:36
Now, is this essentially
different from that one?

383
00:24:32 --> 00:24:38
It would still be multiplied by
e to the minus 3t,

384
00:24:36 --> 00:24:42
but don't be fooled by the e to
the minus 3t.

385
00:24:39 --> 00:24:45
That is our scalar.
That is not what is essential.

386
00:24:42 --> 00:24:48
What is essential is the
content of these two vectors.

387
00:24:45 --> 00:24:51
Is either one a multiple of the
other?

388
00:24:47 --> 00:24:53
The answer is no.
Therefore, they are

389
00:24:49 --> 00:24:55
independent.
They are pointing in two

390
00:24:51 --> 00:24:57
different directions in three
space, these two vectors.

391
00:24:56 --> 00:25:02
And, therefore,
I have two independent

392
00:24:59 --> 00:25:05
solutions just by picking two
different vectors that solve

393
00:25:03 --> 00:25:09
those three equations.
This is also a solution.

394
00:25:07 --> 00:25:13
If I call this the eigenvector
alpha 1, then I ought to call

395
00:25:12 --> 00:25:18
this one the alpha 2.
Hey, we can keep on going

396
00:25:16 --> 00:25:22
through this.
Why not make the first one

397
00:25:19 --> 00:25:25
zero?
Well, what would happen if I

398
00:25:21 --> 00:25:27
made the first one 0,
and then 1, and minus 1?

399
00:25:25 --> 00:25:31
The answer is this one is no
longer independent of those two.

400
00:25:32 --> 00:25:38
I can get it by taking a
combination of those two.

401
00:25:34 --> 00:25:40
Do you see what combination I
should take?

402
00:25:36 --> 00:25:42

403
00:25:42 --> 00:25:48
This one minus that one.
This guy minus that guy gives

404
00:25:47 --> 00:25:53
me that guy, isn't that right?
1 minus 1, 0 minus minus 1,

405
00:25:52 --> 00:25:58
minus 1 minus 0.
This is not a new one.

406
00:25:56 --> 00:26:02
It looks new,
but it is not.

407
00:26:00 --> 00:26:06
I can get it by taking a linear
combination of these two.

408
00:26:04 --> 00:26:10
It is not independent delta.
And that would be true for any

409
00:26:08 --> 00:26:14
other possible solution you
could get for these equations.

410
00:26:12 --> 00:26:18
Once you found two solutions,
all the others will be linear

411
00:26:16 --> 00:26:22
combinations of them.
Well, I cannot use that one.

412
00:26:20 --> 00:26:26
It is not new.
And the general solution,

413
00:26:23 --> 00:26:29
therefore, will be a
combination, c1 times that one

414
00:26:26 --> 00:26:32
plus a constant times this one.
Plus the first one that I found

415
00:26:32 --> 00:26:38
c3 times (1, 1,
1) e to the 0t,

416
00:26:36 --> 00:26:42
which I don't have to write in.
That is the general solution to

417
00:26:42 --> 00:26:48
the system, (x1,x2, x3).

418
00:26:45 --> 00:26:51
What happens as time goes to
infinity?

419
00:26:48 --> 00:26:54
Regardless of what the values
of these two C's this term goes

420
00:26:54 --> 00:27:00
to zero, that term goes to zero
and what I am left with is a

421
00:27:00 --> 00:27:06
constant solution.
So all of these solutions tend

422
00:27:06 --> 00:27:12
to be the solution where all the
cells are at the same

423
00:27:11 --> 00:27:17
temperature.
Well, of course there must be

424
00:27:15 --> 00:27:21
some vocabulary word in this.
There is.

425
00:27:19 --> 00:27:25
There are two vocabulary words.
This is a good eigenvalue.

426
00:27:25 --> 00:27:31
There are also bad eigenvalues.
This is a good repeated

427
00:27:32 --> 00:27:38
eigenvalue, but good is not the
official word.

428
00:27:37 --> 00:27:43
An eigenvalue like this,
which is repeated but where you

429
00:27:44 --> 00:27:50
can find enough eigenvectors,
if lambda is a repeated

430
00:27:50 --> 00:27:56
eigenvalue, it occurs multiply
in the characteristic polynomial

431
00:27:57 --> 00:28:03
as a root.
But you can find enough

432
00:28:03 --> 00:28:09
independent eigenvectors --
Forget the "but."

433
00:28:10 --> 00:28:16

434
00:28:30 --> 00:28:36
-- to make up the needed number
of independent solutions.

435
00:28:36 --> 00:28:42
For example,
if it is repeated once,

436
00:28:40 --> 00:28:46
that is it occurs doubly then
somehow I have got to get two

437
00:28:46 --> 00:28:52
solutions out of that as I was
able to here.

438
00:28:51 --> 00:28:57
If it occurred triply,
I have got to get three

439
00:28:56 --> 00:29:02
solutions out of it.
I would look for three

440
00:29:02 --> 00:29:08
independent eigenvectors and
hope I could find them.

441
00:29:06 --> 00:29:12
That is the good case because
it tells you how to make up as

442
00:29:12 --> 00:29:18
many solutions as you need.
And this kind of eigenvalue is

443
00:29:17 --> 00:29:23
called in the literature the
complete eigenvalue.

444
00:29:23 --> 00:29:29

445
00:29:30 --> 00:29:36
Now, how about the kind in
which you cannot?

446
00:29:33 --> 00:29:39
Well, unfortunately,
all my life I have called it

447
00:29:37 --> 00:29:43
incomplete, which seems to be a
perfectly reasonable thing to

448
00:29:43 --> 00:29:49
call it.
However, terminology changes

449
00:29:46 --> 00:29:52
slowly over time.
The notes, because I wrote

450
00:29:50 --> 00:29:56
them, call it an incomplete
eigenvalue.

451
00:29:53 --> 00:29:59
But the accepted term nowadays
is defective.

452
00:29:57 --> 00:30:03
I don't like that.
It violates the "eigenvalues

453
00:30:02 --> 00:30:08
with disabilities act" or
something.

454
00:30:06 --> 00:30:12
But I have to give it to you
because that is the word I am

455
00:30:11 --> 00:30:17
going to try to use from now on,
at least if I remember to use

456
00:30:18 --> 00:30:24
it.
It would be the word,

457
00:30:20 --> 00:30:26
for example,
used in the linear algebra

458
00:30:24 --> 00:30:30
course 18.06 "plug,
plug," defective otherwise.

459
00:30:30 --> 00:30:36
A defective eigenvalue is one
where you can get one

460
00:30:33 --> 00:30:39
eigenvector.
If it is double,

461
00:30:34 --> 00:30:40
for example,
if it a double eigenvalue.

462
00:30:37 --> 00:30:43
It is defective if you can get
one eigenvector that goes with

463
00:30:41 --> 00:30:47
it, but you cannot find an
independent one.

464
00:30:43 --> 00:30:49
The only other ones you can
find are multiples of the first

465
00:30:47 --> 00:30:53
one.
Then you are really in trouble

466
00:30:49 --> 00:30:55
because you just don't have
enough solutions that you are

467
00:30:53 --> 00:30:59
supposed to get out of that,
and you have to do something.

468
00:30:58 --> 00:31:04
What you do is turn to problem
two on your problem set and

469
00:31:01 --> 00:31:07
solve it because that tells you
what to do.

470
00:31:04 --> 00:31:10
And I even give you an example
to work.

471
00:31:06 --> 00:31:12
Problem two,
that little matrix has a

472
00:31:09 --> 00:31:15
defective eigenvalue.
It doesn't look defective,

473
00:31:12 --> 00:31:18
but you cannot tell.
It is defective.

474
00:31:14 --> 00:31:20
But you, nonetheless,
will be able to find two

475
00:31:17 --> 00:31:23
solutions because you will be
following instructions.

476
00:31:22 --> 00:31:28

477
00:31:32 --> 00:31:38
Now, the only other thing I
should tell you is one of the

478
00:31:35 --> 00:31:41
most important theorems in
linear algebra,

479
00:31:37 --> 00:31:43
which is totally beyond the
scope of this course and is

480
00:31:41 --> 00:31:47
beyond the scope of most
elementary linear algebra

481
00:31:44 --> 00:31:50
courses as I have taught around
the country but,

482
00:31:47 --> 00:31:53
of course, not at MIT.
But, nonetheless,

483
00:31:49 --> 00:31:55
it is the last theorem in the
course.

484
00:31:51 --> 00:31:57
That means it is liable to use
stuff.

485
00:31:53 --> 00:31:59
The theorem goes by different
names.

486
00:31:55 --> 00:32:01
Sometimes it is called the
principle axis theorem.

487
00:32:00 --> 00:32:06
Sometimes it is called the
spectral theorem.

488
00:32:04 --> 00:32:10
But, anyway,
what it says is,

489
00:32:06 --> 00:32:12
if A is a real end-by-end
matrix which is symmetric,

490
00:32:11 --> 00:32:17
you know what a symmetric
matrix is?

491
00:32:15 --> 00:32:21
The formal definition is it is
equal to its transpose.

492
00:32:20 --> 00:32:26
What that means is if you flip
it around the main diagonal it

493
00:32:25 --> 00:32:31
looks just the same as before.
Somewhere on this board,

494
00:32:32 --> 00:32:38
right there,
in fact, is a symmetric matrix.

495
00:32:36 --> 00:32:42
What happened to it?
Right here was the symmetric

496
00:32:40 --> 00:32:46
matrix.
I erased the one thing which I

497
00:32:44 --> 00:32:50
had to have.
Minus 2, 1, 1;

498
00:32:46 --> 00:32:52
1, minus 2, 1;
1, 1 minus 2.

499
00:32:48 --> 00:32:54

500
00:32:51 --> 00:32:57
That was our matrix A.
The matrix is symmetric because

501
00:32:57 --> 00:33:03
if I flip it around the diagonal
it looks the same as it did

502
00:33:02 --> 00:33:08
before.
Well, not exactly.

503
00:33:06 --> 00:33:12
The ones are sort of lying on
their side, but you have to take

504
00:33:10 --> 00:33:16
account of that.
Is that right?

505
00:33:12 --> 00:33:18
The twos are backward.
Well, you know what I mean.

506
00:33:15 --> 00:33:21
Put that element there,
this one here,

507
00:33:18 --> 00:33:24
that one there.
Exchange these two.

508
00:33:20 --> 00:33:26
Notice the diagonal elements
don't all have to be minus 2 for

509
00:33:24 --> 00:33:30
that.
No matter what they were,

510
00:33:26 --> 00:33:32
they are the guys that aren't
moved when you do the flipping.

511
00:33:32 --> 00:33:38
Therefore, there is no
condition on them.

512
00:33:34 --> 00:33:40
It is these other guys.
Each guy here has to use the

513
00:33:38 --> 00:33:44
same guy there.
This one has to be the same as

514
00:33:42 --> 00:33:48
that one, and so on.
Then it will be real and

515
00:33:45 --> 00:33:51
symmetric.
If you have a matrix that is

516
00:33:48 --> 00:33:54
real and symmetric,
like the one we have been

517
00:33:51 --> 00:33:57
working with,
the theorem is that all its

518
00:33:54 --> 00:34:00
eigenvalues are complete.
That is a very unobvious

519
00:33:58 --> 00:34:04
theorem.
All its eigenvalues are

520
00:34:02 --> 00:34:08
automatically complete.
And it is a remarkable fact

521
00:34:07 --> 00:34:13
that you can prove that purely
generally with a certain amount

522
00:34:14 --> 00:34:20
of pure reasoning no calculation
at all.

523
00:34:18 --> 00:34:24
But it has to be true,
and it is true.

524
00:34:22 --> 00:34:28
You will find there are whole
branches of applied differential

525
00:34:28 --> 00:34:34
equations.
You know, equilibrium theory,

526
00:34:33 --> 00:34:39
all the matrices that you deal
with are always symmetric.

527
00:34:37 --> 00:34:43
And, therefore,
this repeated eigenvalues is

528
00:34:40 --> 00:34:46
not something you have to worry
about, finding extra solutions.

529
00:34:45 --> 00:34:51
Well, I guess that is the end
of the first part of the

530
00:34:49 --> 00:34:55
lecture.
I have a third of it left.

531
00:34:52 --> 00:34:58
Let's talk fast.
I would like to,

532
00:34:55 --> 00:35:01
with the remaining time,
explain to you what to do if

533
00:34:59 --> 00:35:05
you were to get complex
eigenvalues.

534
00:35:03 --> 00:35:09

535
00:35:08 --> 00:35:14
Now, actually,
the answer is follow the same

536
00:35:11 --> 00:35:17
program.
In other words,

537
00:35:12 --> 00:35:18
if you solve the characteristic
equation and you get a complex

538
00:35:17 --> 00:35:23
root, follow the program,
calculate the corresponding

539
00:35:20 --> 00:35:26
complex eigenvectors.
In other words,

540
00:35:23 --> 00:35:29
solve the equations.
Everything will be the same

541
00:35:26 --> 00:35:32
except that the eigenvectors
will turn out to be complex,

542
00:35:30 --> 00:35:36
that is will have complex
entries.

543
00:35:34 --> 00:35:40
Don't worry about it.
Then form the solutions.

544
00:35:37 --> 00:35:43
The solutions are now going to
look once again like alpha times

545
00:35:43 --> 00:35:49
e to the a plus bi to the t.

546
00:35:47 --> 00:35:53
This will be complex and that
will be complex,

547
00:35:51 --> 00:35:57
too.
This will have complex entries.

548
00:35:54 --> 00:36:00
And then, finally,
take the real and imaginary

549
00:35:58 --> 00:36:04
parts.
Those will be real and they

550
00:36:02 --> 00:36:08
will give real and two
solutions.

551
00:36:04 --> 00:36:10
In other words,
the program is exactly like

552
00:36:08 --> 00:36:14
what we did for second-order
differential equations.

553
00:36:12 --> 00:36:18
We used the complex numbers,
got complex solutions.

554
00:36:16 --> 00:36:22
And then, at the very last
step, we took the real and

555
00:36:20 --> 00:36:26
imaginary parts to get two real
solutions out of each complex

556
00:36:25 --> 00:36:31
number.
I would like to give you a

557
00:36:27 --> 00:36:33
simple example of working that
out.

558
00:36:30 --> 00:36:36
And it is the system x prime
equals x plus 2y.

559
00:36:35 --> 00:36:41
And y prime equals minus x

560
00:36:39 --> 00:36:45
minus y.
Because it is springtime,

561
00:36:45 --> 00:36:51
it doesn't feel like spring but
it will this weekend as it is

562
00:36:51 --> 00:36:57
getting warmer.
And since, when I am too tired

563
00:36:56 --> 00:37:02
to make up problem sets for you
late at night,

564
00:37:01 --> 00:37:07
I watch reruns of Seinfeld.
I am from New York.

565
00:37:06 --> 00:37:12
It is just in my bloodstream.
Of course, the most interesting

566
00:37:12 --> 00:37:18
character on Seinfeld is George.
We are going to consider Susan

567
00:37:19 --> 00:37:25
who is the girlfriend who got
killed by licking poison

568
00:37:24 --> 00:37:30
envelopes.
And George carried on their

569
00:37:28 --> 00:37:34
love affair until Susan was
disposed of by the writers by

570
00:37:33 --> 00:37:39
this strange death.
And we are going to consider x

571
00:37:39 --> 00:37:45
is modeling Susan's love for
George.

572
00:37:43 --> 00:37:49
That is x.
And George's love for Susan

573
00:37:47 --> 00:37:53
will be y.
Now, I don't mean the absolute

574
00:37:51 --> 00:37:57
love.
If x and y are zero,

575
00:37:53 --> 00:37:59
I don't mean that they don't
love each other.

576
00:37:58 --> 00:38:04
I just mean that that is the
equilibrium value of the love.

577
00:38:05 --> 00:38:11
Everything else is measured as
departures from that.

578
00:38:10 --> 00:38:16
So (0, 0) represents the normal
amount of love,

579
00:38:15 --> 00:38:21
if love is measured.
I don't know what love units

580
00:38:20 --> 00:38:26
are.
Hearts, I guess.

581
00:38:22 --> 00:38:28
Six hearts, let's say.
Now, in what sense does this

582
00:38:27 --> 00:38:33
model it?
This is a normal equation and

583
00:38:32 --> 00:38:38
this is a neurotic equation.
That is why this is George and

584
00:38:36 --> 00:38:42
this is Susan who seemed very
normal to me.

585
00:38:39 --> 00:38:45
Susan is a normal person.
When y is positive that means

586
00:38:43 --> 00:38:49
that George seems to be loving
her more today than yesterday,

587
00:38:47 --> 00:38:53
and her natural response is to
be more in love with him.

588
00:38:51 --> 00:38:57
That is what most people are.
If y is negative,

589
00:38:54 --> 00:39:00
hey, what's the matter with
George?

590
00:38:57 --> 00:39:03
He doesn't feel so good.
Maybe there is something wrong

591
00:39:02 --> 00:39:08
with him.
She gets a little mad at him

592
00:39:06 --> 00:39:12
and this goes down.
x prime is negative.

593
00:39:10 --> 00:39:16
And the same way why is this
positive?

594
00:39:13 --> 00:39:19
Well, again,
it is a psychological thing,

595
00:39:17 --> 00:39:23
but all the world loves a
lover.

596
00:39:20 --> 00:39:26
When Susan is in love,
as she feels x is high,

597
00:39:24 --> 00:39:30
that makes her feel good.
And she loves everything,

598
00:39:28 --> 00:39:34
in fact.
Not just George.

599
00:39:32 --> 00:39:38
It is one of those things.
You all know what I am talking

600
00:39:38 --> 00:39:44
about.
Now, George,

601
00:39:40 --> 00:39:46
of course, is what makes the
writers happy.

602
00:39:44 --> 00:39:50
George is neurotic and,
therefore, is exactly the

603
00:39:49 --> 00:39:55
opposite.
He sees one day that he feels

604
00:39:53 --> 00:39:59
more in love with Susan than he
was yesterday.

605
00:40:00 --> 00:40:06
Does this make him happy?
Not at all.

606
00:40:02 --> 00:40:08
Not at all.
It makes y prime more negative.

607
00:40:05 --> 00:40:11
Why?
Because all he can think of is,

608
00:40:08 --> 00:40:14
my God, suppose I am really in
love with this girl?

609
00:40:11 --> 00:40:17
Suppose I marry her.
Oh, my God, 40 years of seeing

610
00:40:15 --> 00:40:21
the same person at breakfast all
the time.

611
00:40:18 --> 00:40:24
I must be crazy.
And so it goes down.

612
00:40:20 --> 00:40:26
Here is our neurotic model.
The question for differential

613
00:40:24 --> 00:40:30
equations is,
what do the solutions to that

614
00:40:27 --> 00:40:33
look like?
In other words,

615
00:40:31 --> 00:40:37
how does, in fact,
their love affair go?

616
00:40:34 --> 00:40:40
Now, there is a reason why the
writers picked that model,

617
00:40:39 --> 00:40:45
as you will see.
It means they were able to get

618
00:40:44 --> 00:40:50
a year's worth of episodes out
of it.

619
00:40:47 --> 00:40:53
And why is that so?
Well, let's solve it.

620
00:40:50 --> 00:40:56
The characteristic equation is
lambda squared.

621
00:40:54 --> 00:41:00
The matrix that governs this
system is A equals (1,

622
00:40:59 --> 00:41:05
2; negative 1,
negative 1)

623
00:41:02 --> 00:41:08
The trace of that matrix,

624
00:41:06 --> 00:41:12
the sum of the diagonal
elements is zero.

625
00:41:10 --> 00:41:16
There is the zero lambda here.
The determinant,

626
00:41:13 --> 00:41:19
which is the constant term,
is negative 1,

627
00:41:16 --> 00:41:22
minus negative 2,
which is plus 1.

628
00:41:19 --> 00:41:25
So the characteristic equation,
by calculating the trace and

629
00:41:23 --> 00:41:29
determinant is lambda squared
plus 1 equals 0.

630
00:41:28 --> 00:41:34
The eigenvalues are plus and

631
00:41:32 --> 00:41:38
minus i.
Now, you don't have to pick

632
00:41:35 --> 00:41:41
both of them because the
negative one lead to essentially

633
00:41:40 --> 00:41:46
the same solutions but with
negative signs.

634
00:41:44 --> 00:41:50
Either one will do just as we
solved second order equations.

635
00:41:49 --> 00:41:55
The system for finding the
eigenvectors,

636
00:41:53 --> 00:41:59
well, we are going to have to
accept the complex eigenvector.

637
00:42:00 --> 00:42:06
What is the system going to be?
Well, I take the matrix and I

638
00:42:05 --> 00:42:11
subtract i.
We will use i.

639
00:42:07 --> 00:42:13
Subtract i from the main
diagonal.

640
00:42:10 --> 00:42:16
So the system is (1 minus i)
times a1 plus 2a2 is zero.

641
00:42:15 --> 00:42:21
And let's, for good measure,

642
00:42:19 --> 00:42:25
write the other one down,
too.

643
00:42:22 --> 00:42:28
It is negative a1 plus minus (1
minus i) times a2.

644
00:42:28 --> 00:42:34
Then what is the solution?

645
00:42:33 --> 00:42:39
Well, you get the solution the
usual way.

646
00:42:39 --> 00:42:45
Let's take a1 equal to 1.

647
00:42:44 --> 00:42:50
Then what is a2?
a2 is 1 minus i divided by 2

648
00:42:50 --> 00:42:56
from the first equation.

649
00:42:58 --> 00:43:04
So the complex solution is 1
minus i over 2 times 

650
00:43:02 --> 00:43:08
e to the it.
Now you have to take the real

651
00:43:05 --> 00:43:11
and imaginary parts of that.
This is the only part which

652
00:43:09 --> 00:43:15
technically I would not trust
you to do without having someone

653
00:43:13 --> 00:43:19
show you how to do it.
What do you do?

654
00:43:15 --> 00:43:21
Well, of course,
you know how to separate the

655
00:43:18 --> 00:43:24
real and imaginary parts of
that.

656
00:43:20 --> 00:43:26
It is the first thing is to
separate the vectors.

657
00:43:25 --> 00:43:31
I don't know how to explain
this.

658
00:43:29 --> 00:43:35
Just watch.
The real part of it is 1,

659
00:43:33 --> 00:43:39
one-half.
It should be negative 1,

660
00:43:37 --> 00:43:43
so minus this plus that because
I didn't put that on the right

661
00:43:45 --> 00:43:51
side.
It is minus one-half plus i

662
00:43:49 --> 00:43:55
times (0, one-half).

663
00:43:54 --> 00:44:00
Anybody want to fight?

664
00:44:00 --> 00:44:06
1 plus i times 0 minus one-half
plus one-half times i.

665
00:44:06 --> 00:44:12
You saw how I did that?
Okay.

666
00:44:10 --> 00:44:16
When you do these problem you
do it the same way,

667
00:44:16 --> 00:44:22
but don't ask me to explain
what I just did.

668
00:44:21 --> 00:44:27
Here it is cosine t plus i sine
t.

669
00:44:30 --> 00:44:36
And so the real part will give
me one solution.

670
00:44:34 --> 00:44:40
The imaginary part will give me
another.

671
00:44:39 --> 00:44:45
Since I have a limited amount
of time, let's just calculate

672
00:44:45 --> 00:44:51
the real part.
What is it?

673
00:44:48 --> 00:44:54
Well, it is (1,
minus ½) times cosine t,

674
00:44:52 --> 00:44:58
i squared is negative 1,
so minus (0,

675
00:44:56 --> 00:45:02
one-half), the negative 1 from
the i squared,

676
00:45:01 --> 00:45:07
times sine t.

677
00:45:04 --> 00:45:10
Now, what solution is that?

678
00:45:10 --> 00:45:16
This is (x, y).
Take the final step.

679
00:45:13 --> 00:45:19
It doesn't have to look like
that.

680
00:45:16 --> 00:45:22
x equals cosine t.

681
00:45:19 --> 00:45:25
Do you see that?
x equals cosine t plus 0 times

682
00:45:24 --> 00:45:30
sine t.

683
00:45:27 --> 00:45:33
What is y?
y is minus one-half times

684
00:45:31 --> 00:45:37
cosine t, minus one-half times
sine t, plus sine t.

685
00:45:37 --> 00:45:43
Now, you may have the pleasure

686
00:45:42 --> 00:45:48
of showing eliminating t.
You get a quadratic polynomial

687
00:45:47 --> 00:45:53
in x and y equals zero.
This is an ellipse.

688
00:45:51 --> 00:45:57
As t varies,
you can see this repeats its

689
00:45:55 --> 00:46:01
values at intervals of 2pi,
this gives an ellipse.

690
00:46:01 --> 00:46:07
And if you want to use a little
computer program,

691
00:46:05 --> 00:46:11
linear phase,
this is not in the assignment,

692
00:46:09 --> 00:46:15
but the ellipses look like this
and go around that way.

693
00:46:14 --> 00:46:20
And that is the model of George
and Susan's love.

694
00:46:19 --> 00:46:25
x, Susan.
y, George.

695
00:46:21 --> 00:46:27
They go round and round in this
little love circle,

696
00:46:25 --> 00:46:31
and it stretches on for 26
episodes.