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00:00:13,000 --> 00:00:19,000
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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00:00:19,000 --> 00:00:25,000
we are going to have to be
doing it several times today.

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00:00:23,000 --> 00:00:29,000
That is the system we are
trying to solve.

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00:00:26,000 --> 00:00:32,000
And the first thing you have to
do is find a characteristic

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00:00:31,000 --> 00:00:37,000
equation which is general form,
although this is not the form

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00:00:35,000 --> 00:00:41,000
you should use for two-by-two,
is A minus lambda I

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00:00:39,000 --> 00:00:45,000
equals zero.
And its roots are the

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00:00:42,000 --> 00:00:48,000
eigenvalues.

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00:00:50,000 --> 00:00:56,000
And then with each eigenvalue
you then have to calculate its

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00:00:55,000 --> 00:01:01,000
eigenvector, which you do by
solving the system (A minus

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00:00:59,000 --> 00:01:05,000
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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00:01:18,000 --> 00:01:24,000
to the lambda 1t.

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00:01:21,000 --> 00:01:27,000
Of course you do that for each
eigenvalue.

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00:01:25,000 --> 00:01:31,000
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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00:01:51,000 --> 00:01:57,000
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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00:02:02,000 --> 00:02:08,000
an eigenvalue gets repeated.
But I think since the situation

26
00:02:06,000 --> 00:02:12,000
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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00:02:26,000 --> 00:02:32,000
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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00:02:34,000 --> 00:02:40,000
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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00:02:52,000 --> 00:02:58,000
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

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

42
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the heater so the temperatures
start out what they are supposed

43
00:03:19,000 --> 00:03:25,000
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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00:03:30,000 --> 00:03:36,000
at the same temperature.
So we are going to let (x)i

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00:03:35,000 --> 00:03:41,000
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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00:03:44,000 --> 00:03:50,000
each has the same volume,
and the same glass partition

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00:03:48,000 --> 00:03:54,000
separates them out and no heat
can escape.

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00:03:51,000 --> 00:03:57,000
This is very well-insulated
with very double-thick

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Thermopane glass or something
like that.

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00:04:00,000 --> 00:04:06,000
You can see in,
but heat cannot get out very

54
00:04:03,000 --> 00:04:09,000
well.
Heat essentially is conducted

55
00:04:06,000 --> 00:04:12,000
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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00:04:13,000 --> 00:04:19,000
because the fish are swimming
around in it.

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00:04:17,000 --> 00:04:23,000
That should be a pretty decent
way of stirring a fish tank.

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00:04:21,000 --> 00:04:27,000
The question is how do each of
these, as a function of time,

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00:04:26,000 --> 00:04:32,000
and I want to know how they
behave over time,

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00:04:29,000 --> 00:04:35,000
so find these functions.

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00:04:40,000 --> 00:04:46,000
Well, we are going to find them
in solutions to differential

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equations.
And the differential equations

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00:04:45,000 --> 00:04:51,000
are not hard to set up.
They are very much like the

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00:04:48,000 --> 00:04:54,000
diffusion equation you had for
homework or the equations we

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00:04:51,000 --> 00:04:57,000
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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00:04:57,000 --> 00:05:03,000
way.
What determines the flow,

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00:04:59,000 --> 00:05:05,000
the change in temperature?
Well, it is the conductivity

70
00:05:03,000 --> 00:05:09,000
across the barriers.
But there are two barriers

71
00:05:06,000 --> 00:05:12,000
because heat can flow into this
first cell, both from this guy

72
00:05:10,000 --> 00:05:16,000
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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00:05:21,000 --> 00:05:27,000
was in the middle could get
contributions from both sides,

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00:05:25,000 --> 00:05:31,000
whereas, the two guys on the
end could only get contribution

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00:05:29,000 --> 00:05:35,000
from one.
But here, nobody is on the end.

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00:05:34,000 --> 00:05:40,000
It is circular table.
Everyone is dying equally.

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00:05:38,000 --> 00:05:44,000
Everybody can get input from
the other two cells.

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00:05:43,000 --> 00:05:49,000
x1 prime is some
constant of conductivity times

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00:05:48,000 --> 00:05:54,000
the temperature difference
between tank three and tank one.

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00:05:54,000 --> 00:06:00,000
And then there is another term
which comes from tank two.

83
00:06:01,000 --> 00:06:07,000
So a times tank two minus the
temperature difference,

84
00:06:04,000 --> 00:06:10,000
tank two minus tank one.
Let's write this out.

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00:06:07,000 --> 00:06:13,000
Remember there will be other
equations, too.

86
00:06:10,000 --> 00:06:16,000
But instead of doing this,
let's do a more careful job

87
00:06:14,000 --> 00:06:20,000
with this first equation.
When I write it out,

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00:06:17,000 --> 00:06:23,000
remember, the important thing
is you are going to have x1,

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00:06:21,000 --> 00:06:27,000
x2, x3 down the left,
so they have to occur in the

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00:06:24,000 --> 00:06:30,000
same order on the right in order
to use these standard eigenvalue

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00:06:29,000 --> 00:06:35,000
techniques.
The coefficient of x1 is going

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

93
00:06:38,000 --> 00:06:44,000
In other words,
it is going to be minus 2 ax1.

94
00:06:42,000 --> 00:06:48,000
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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00:06:55,000 --> 00:07:01,000
terms of the other variables.
But there is symmetry.

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00:07:00,000 --> 00:07:06,000
There is no difference between
this tank, that tank,

99
00:07:04,000 --> 00:07:10,000
and that tank as far as the
differential equations are

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00:07:08,000 --> 00:07:14,000
concerned.
And, therefore,

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00:07:10,000 --> 00:07:16,000
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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00:07:22,000 --> 00:07:28,000
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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00:07:51,000 --> 00:07:57,000
plus x2 plus x3.

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00:07:54,000 --> 00:08:00,000
And so now what would the
equation for x2 prime

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00:07:57,000 --> 00:08:03,000
be?
Well, here x2 plays the role

115
00:08:01,000 --> 00:08:07,000
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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00:08:26,000 --> 00:08:32,000
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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00:09:06,000 --> 00:09:12,000
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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00:10:02,000 --> 00:10:08,000
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.

152
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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.

165
00:11:11,000 --> 00:11:17,000
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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00:11:22,000 --> 00:11:28,000
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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00:12:00,000 --> 00:12:06,000
And now we are going to expand
it out.

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00:12:02,000 --> 00:12:08,000
What is this?
Lambda plus 2 cubed.

183
00:12:05,000 --> 00:12:11,000
Lambda cubed plus,

184
00:12:06,000 --> 00:12:12,000
and don't get confused because
it is this 2 that will kill you

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00:12:10,000 --> 00:12:16,000
when you use the binomial
theorem.

186
00:12:12,000 --> 00:12:18,000
If there is 1 here everybody
knows what to do.

187
00:12:15,000 --> 00:12:21,000
If there is an A there
everybody knows what to do.

188
00:12:18,000 --> 00:12:24,000
It is when that is a number not
1 that everybody makes mistakes,

189
00:12:22,000 --> 00:12:28,000
including me.
The binomial coefficients are

190
00:12:25,000 --> 00:12:31,000
1, 3, 3, 1 because it is a
cubed, I am expanding.

191
00:12:30,000 --> 00:12:36,000
So it is lambda cubed plus 3
times lambda squared times 2.

192
00:12:34,000 --> 00:12:40,000
I won't explain to you what I
am doing.

193
00:12:37,000 --> 00:12:43,000
I will just do it and hope that
you all know what I am doing.

194
00:12:41,000 --> 00:12:47,000
Plus 3 times lambda times 2
squared plus the last term,

195
00:12:45,000 --> 00:12:51,000
which is 2 cubed.

196
00:12:49,000 --> 00:12:55,000
And now we have the other term.

197
00:12:53,000 --> 00:12:59,000
All that is plus because I
changed its sign.

198
00:12:56,000 --> 00:13:02,000
The next thing is negative 2.
And then the last thing is plus

199
00:13:03,000 --> 00:13:09,000
3 times (minus 2 minus lambda).

200
00:13:08,000 --> 00:13:14,000
Let's keep it.
So what is the actual

201
00:13:11,000 --> 00:13:17,000
characteristic equation?
Maybe I can finish it.

202
00:13:15,000 --> 00:13:21,000
I should stay over here instead
of recopying all of it.

203
00:13:27,000 --> 00:13:33,000
Well, there is a lot more work
to do.

204
00:13:29,000 --> 00:13:35,000
Let's see if we can at least
write down the equation.

205
00:13:33,000 --> 00:13:39,000
What is it?
It is lambda cubed.

206
00:13:35,000 --> 00:13:41,000
What is the lambda squared

207
00:13:38,000 --> 00:13:44,000
erm?
It is six and that is all there

208
00:13:42,000 --> 00:13:48,000
is.
How about the lambda term?

209
00:13:44,000 --> 00:13:50,000
Well, we have 12 lambda minus 3
lambda which makes plus 9

210
00:13:48,000 --> 00:13:54,000
lambda.
That looks good but constant

211
00:13:51,000 --> 00:13:57,000
terms have a way of screwing
everything up.

212
00:13:54,000 --> 00:14:00,000
What is the constant term?
It is A minus 2 minus 6.

213
00:14:00,000 --> 00:14:06,000
Zero.
The constant term is zero.

214
00:14:02,000 --> 00:14:08,000
That converts this from a hard
problem to an easy problem.

215
00:14:07,000 --> 00:14:13,000
Now it is a cinch to calculate
the stuff.

216
00:14:11,000 --> 00:14:17,000
Let's go to this board and
continue the work over here.

217
00:14:24,000 --> 00:14:30,000
The equation is lambda cubed
plus 6 lambda squared plus 9

218
00:14:27,000 --> 00:14:33,000
lambda is zero.

219
00:14:31,000 --> 00:14:37,000
It is very easy to calculate
the roots of that.

220
00:14:34,000 --> 00:14:40,000
You factor it.
Lambda is a common factor.

221
00:14:37,000 --> 00:14:43,000
And what is left?
Lambda squared plus 6 lambda

222
00:14:40,000 --> 00:14:46,000
plus 9.

223
00:14:43,000 --> 00:14:49,000
That is the sort of thing you
got all the time when you were

224
00:14:47,000 --> 00:14:53,000
studying critical damping.
It is the square of lambda plus

225
00:14:51,000 --> 00:14:57,000
3. Lambda squared plus 6 lambda

226
00:14:54,000 --> 00:15:00,000
plus 9 equals zero.
So the eigenvalues,

227
00:14:58,000 --> 00:15:04,000
the roots are what?
Well, they are lambda equals

228
00:15:02,000 --> 00:15:08,000
zero from this factor and then
lambda equals minus 3.

229
00:15:07,000 --> 00:15:13,000
But what is new is that the

230
00:15:10,000 --> 00:15:16,000
minus 3 is a double root.
That is a double root.

231
00:15:14,000 --> 00:15:20,000
Now, that, of course,
is what is going to cause the

232
00:15:18,000 --> 00:15:24,000
trouble.
Because, for each one of these,

233
00:15:21,000 --> 00:15:27,000
I am supposed to calculate the
eigenvector and make up the

234
00:15:26,000 --> 00:15:32,000
solution.
But that assumed that I had

235
00:15:29,000 --> 00:15:35,000
three things to get three
different solutions.

236
00:15:32,000 --> 00:15:38,000
Here I have only got two
things.

237
00:15:34,000 --> 00:15:40,000
It is the same trouble we ran
into when there was a repeated

238
00:15:38,000 --> 00:15:44,000
root.
We were studying second or

239
00:15:40,000 --> 00:15:46,000
third order differential
equations and the characteristic

240
00:15:43,000 --> 00:15:49,000
equation had a repeated root.
And I had to go into a song and

241
00:15:47,000 --> 00:15:53,000
dance and stand on my head and
multiply things by t and so on.

242
00:15:51,000 --> 00:15:57,000
And then talk very hard arguing
why that was a good thing to do

243
00:15:54,000 --> 00:16:00,000
to get the answer.
Now, I am not going to do the

244
00:15:57,000 --> 00:16:03,000
same thing here.
Instead, I am going to try to

245
00:16:02,000 --> 00:16:08,000
solve the problem instead.
Let's get two points by at

246
00:16:06,000 --> 00:16:12,000
least doing the easy part of it.
Lambda equals zero.

247
00:16:11,000 --> 00:16:17,000
What am I supposed to do with

248
00:16:13,000 --> 00:16:19,000
lambda equals zero?
I am looking for the alpha that

249
00:16:17,000 --> 00:16:23,000
goes with that.
And I find that eigenvector by

250
00:16:21,000 --> 00:16:27,000
solving this system of
equations.

251
00:16:24,000 --> 00:16:30,000
Let's write out what that
system of equations is.

252
00:16:29,000 --> 00:16:35,000
Well, if lambda is zero,
this isn't there.

253
00:16:32,000 --> 00:16:38,000
It is just the matrix A times
alpha equals zero.

254
00:16:38,000 --> 00:16:44,000
And the matrix A is,

255
00:16:40,000 --> 00:16:46,000
I never even wrote it anywhere.
I never wrote A.

256
00:16:45,000 --> 00:16:51,000
I thought I would get away
without having to do it,

257
00:16:49,000 --> 00:16:55,000
but you never get away with
anything.

258
00:16:53,000 --> 00:16:59,000
It's the principle of life.
That is A.

259
00:16:58,000 --> 00:17:04,000
If I subtract zero from the
main diagonal,

260
00:17:01,000 --> 00:17:07,000
that doesn't do a great deal to
A.

261
00:17:04,000 --> 00:17:10,000
And the resulting system of
equations is those same things,

262
00:17:10,000 --> 00:17:16,000
except you have the a1's there,
too.

263
00:17:13,000 --> 00:17:19,000
There is one.
a1 minus 2 a2 plus a3 equals

264
00:17:17,000 --> 00:17:23,000
zero.
I am just subtracting zero from

265
00:17:22,000 --> 00:17:28,000
the main diagonal so there is
nothing to do.

266
00:17:26,000 --> 00:17:32,000
a2 minus 2 a3 equals zero.
Now I am supposed to solve

267
00:17:33,000 --> 00:17:39,000
those.

268
00:17:40,000 --> 00:17:46,000
Of course we could do it.
Well, how do you know how to

269
00:17:43,000 --> 00:17:49,000
solve a system of three linear
equations?

270
00:17:45,000 --> 00:17:51,000
Well, elimination.
You can always solve by

271
00:17:48,000 --> 00:17:54,000
elimination.
Now we are much more

272
00:17:50,000 --> 00:17:56,000
sophisticated than that.
You all have pocket calculators

273
00:17:54,000 --> 00:18:00,000
so you could use the inverse
matrix, right?

274
00:17:56,000 --> 00:18:02,000
No.
You cannot use the inverse

275
00:17:58,000 --> 00:18:04,000
matrix.
What will happen if you punch

276
00:18:03,000 --> 00:18:09,000
in those coefficients and then
punch in A inverse.

277
00:18:08,000 --> 00:18:14,000
What answer will it give you?
0, 0, 0.

278
00:18:12,000 --> 00:18:18,000
No, I am sorry.
It won't give you any answer.

279
00:18:16,000 --> 00:18:22,000
What will it say?
It will say I cannot calculate

280
00:18:21,000 --> 00:18:27,000
the inverse to that matrix
because the whole purpose of

281
00:18:27,000 --> 00:18:33,000
this exercise was to find a
value of lambda such that this

282
00:18:33,000 --> 00:18:39,000
system of equations is
dependent.

283
00:18:38,000 --> 00:18:44,000
The coefficient determinant is
zero and, therefore,

284
00:18:42,000 --> 00:18:48,000
the coefficient matrix does not
have an inverse matrix.

285
00:18:46,000 --> 00:18:52,000
You cannot use that method.
In other words,

286
00:18:49,000 --> 00:18:55,000
the inverse matrix will never
work in these problems because

287
00:18:54,000 --> 00:19:00,000
the system of equations you will
be trying to solve is always a

288
00:18:59,000 --> 00:19:05,000
non-independent system.
And, therefore,

289
00:19:03,000 --> 00:19:09,000
its determinant is always zero.
And, therefore,

290
00:19:06,000 --> 00:19:12,000
there is no inverse matrix
because the determinant of the

291
00:19:11,000 --> 00:19:17,000
coefficient is zero.
All you can do is use

292
00:19:14,000 --> 00:19:20,000
elimination or physical insight
and common sense.

293
00:19:18,000 --> 00:19:24,000
Now, because I teach
differential equations everybody

294
00:19:23,000 --> 00:19:29,000
assumes, mistakenly,
as I think, that I really know

295
00:19:27,000 --> 00:19:33,000
something about them.
I get now and then graduate

296
00:19:32,000 --> 00:19:38,000
students, not in mathematics,
but some obscure field of

297
00:19:36,000 --> 00:19:42,000
engineering or whatever drift
into my office and say I see you

298
00:19:40,000 --> 00:19:46,000
teach differential equations.
Do you have a minute here?

299
00:19:45,000 --> 00:19:51,000
And before I can say no they
write their differential

300
00:19:49,000 --> 00:19:55,000
equation on the board.
And almost invariably it is

301
00:19:52,000 --> 00:19:58,000
nothing I have ever seen before.
And they so look at me

302
00:19:57,000 --> 00:20:03,000
hopefully and expectantly.
So what do I ask them?

303
00:20:01,000 --> 00:20:07,000
I don't ask them what they have
tried.

304
00:20:05,000 --> 00:20:11,000
What I ask them is where did
this come from?

305
00:20:08,000 --> 00:20:14,000
What field did it come from?
Because each field has its own

306
00:20:14,000 --> 00:20:20,000
little tricks.
It gets the same differential

307
00:20:17,000 --> 00:20:23,000
equations all the time and has
its own little tricks for

308
00:20:22,000 --> 00:20:28,000
solving them.
You should do the same thing

309
00:20:26,000 --> 00:20:32,000
here.
Well, of course we can solve

310
00:20:29,000 --> 00:20:35,000
this.
And by now most of you have

311
00:20:33,000 --> 00:20:39,000
solved it just by inspection,
just by sort of psyching out

312
00:20:37,000 --> 00:20:43,000
the answer.
But a better way is to say

313
00:20:40,000 --> 00:20:46,000
look, suppose we had the
solution, what would the

314
00:20:44,000 --> 00:20:50,000
solution look like?
Well, it would look like (a1,

315
00:20:48,000 --> 00:20:54,000
a2, a3), whatever the values of
those variables were which gave

316
00:20:52,000 --> 00:20:58,000
me the solution to the equation,
times e to the 0t.

317
00:20:57,000 --> 00:21:03,000
But what is this?
e to the 0t is one

318
00:21:02,000 --> 00:21:08,000
for all time.
And, therefore,

319
00:21:04,000 --> 00:21:10,000
this is a constant solution.
What I am asking is to find a

320
00:21:09,000 --> 00:21:15,000
constant solution.
Now, can I, by inspection,

321
00:21:13,000 --> 00:21:19,000
find a constant solution to
this?

322
00:21:15,000 --> 00:21:21,000
If so it must be the one.
Well, there is an obvious

323
00:21:19,000 --> 00:21:25,000
constant solution.
All the cells have the same

324
00:21:23,000 --> 00:21:29,000
temperature.
If that is true then there is

325
00:21:26,000 --> 00:21:32,000
no reason why it should ever
change as time goes on.

326
00:21:32,000 --> 00:21:38,000
The physical problem itself
suggests what the answer must

327
00:21:35,000 --> 00:21:41,000
be.
You don't have to solve

328
00:21:37,000 --> 00:21:43,000
equations.
In other words,

329
00:21:38,000 --> 00:21:44,000
any constant like (1,
1, 1).

330
00:21:40,000 --> 00:21:46,000
Well, could it be (20,
20, 20)?

331
00:21:42,000 --> 00:21:48,000
Yeah, that is a constant
multiple of (1,

332
00:21:44,000 --> 00:21:50,000
1, 1).
That is included.

333
00:21:45,000 --> 00:21:51,000
My basic constant solution,
therefore, is simply (1,

334
00:21:48,000 --> 00:21:54,000
1, 1) times e to the 0t.

335
00:21:51,000 --> 00:21:57,000
You don't have to include e to
the 0t because it is one.

336
00:21:54,000 --> 00:22:00,000
Now, just to check,
is (1, 1, 1) a solution to

337
00:21:57,000 --> 00:22:03,000
these equations?
It certainly is.

338
00:22:01,000 --> 00:22:07,000
1 plus 1 minus 2 is zero in
every case.

339
00:22:04,000 --> 00:22:10,000
The equations are essentially
the same, except they use

340
00:22:08,000 --> 00:22:14,000
different variables.
By inspection or,

341
00:22:11,000 --> 00:22:17,000
if you like,
by elimination,

342
00:22:14,000 --> 00:22:20,000
but not by finding the inverse
matrix you solve those

343
00:22:18,000 --> 00:22:24,000
equations.
And we have our first solution.

344
00:22:21,000 --> 00:22:27,000
Now let's go onto the second
one.

345
00:22:24,000 --> 00:22:30,000
For the second one,
we are going to have to use the

346
00:22:28,000 --> 00:22:34,000
eigenvalue lambda equals
negative 3.

347
00:22:33,000 --> 00:22:39,000
And now what is the system of
equations?

348
00:22:36,000 --> 00:22:42,000
Well, now I have to take this
and I have to subtract negative

349
00:22:41,000 --> 00:22:47,000
3 from the diagonal elements.
Minus 2 minus negative 3 is

350
00:22:46,000 --> 00:22:52,000
plus 1, right?

351
00:22:49,000 --> 00:22:55,000
Got that?
Each of the diagonal elements,

352
00:22:53,000 --> 00:22:59,000
after I subtract minus 3 turns
into plus 1.

353
00:22:58,000 --> 00:23:04,000
And, therefore,
the system becomes,

354
00:23:00,000 --> 00:23:06,000
the system I have to solve is
a1 plus a2 plus a3 equals zero.

355
00:23:05,000 --> 00:23:11,000
And what is the second

356
00:23:09,000 --> 00:23:15,000
equation?
Symmetry is preserved.

357
00:23:11,000 --> 00:23:17,000
All the equations are
essentially the same,

358
00:23:14,000 --> 00:23:20,000
except for the names of the
variables so they all must give

359
00:23:19,000 --> 00:23:25,000
you the same thing after I
subtract minus 3 from the main

360
00:23:24,000 --> 00:23:30,000
diagonal.
Well, that is what we call a

361
00:23:27,000 --> 00:23:33,000
dependent system of equations.
All I have is the same equation

362
00:23:34,000 --> 00:23:40,000
repeated twice,
but I still have to solve it.

363
00:23:38,000 --> 00:23:44,000
Now, what you see is that there
are lots of solutions to this.

364
00:23:44,000 --> 00:23:50,000
Let me write down one of them.
For example,

365
00:23:49,000 --> 00:23:55,000
suppose I made a1 equal to 1
and I made a2 a 0,

366
00:23:54,000 --> 00:24:00,000
then a3 would be negative 1.

367
00:24:00,000 --> 00:24:06,000
So here is a solution.
That is the eigenvector.

368
00:24:03,000 --> 00:24:09,000
And with it,
I can make the solution by

369
00:24:06,000 --> 00:24:12,000
multiplying by e to the negative
3t.

370
00:24:10,000 --> 00:24:16,000
There is a solution.
But that is not the only alpha

371
00:24:14,000 --> 00:24:20,000
I could have chosen.
Suppose I chose this one

372
00:24:18,000 --> 00:24:24,000
instead.
Suppose I kept this 1,

373
00:24:20,000 --> 00:24:26,000
but this time made a3 zero.
Well, in that case,

374
00:24:24,000 --> 00:24:30,000
there would be a2 that had to
be minus 1.

375
00:24:30,000 --> 00:24:36,000
Now, is this essentially
different from that one?

376
00:24:32,000 --> 00:24:38,000
It would still be multiplied by
e to the minus 3t,

377
00:24:36,000 --> 00:24:42,000
but don't be fooled by the e to
the minus 3t.

378
00:24:39,000 --> 00:24:45,000
That is our scalar.
That is not what is essential.

379
00:24:42,000 --> 00:24:48,000
What is essential is the
content of these two vectors.

380
00:24:45,000 --> 00:24:51,000
Is either one a multiple of the
other?

381
00:24:47,000 --> 00:24:53,000
The answer is no.
Therefore, they are

382
00:24:49,000 --> 00:24:55,000
independent.
They are pointing in two

383
00:24:51,000 --> 00:24:57,000
different directions in three
space, these two vectors.

384
00:24:56,000 --> 00:25:02,000
And, therefore,
I have two independent

385
00:24:59,000 --> 00:25:05,000
solutions just by picking two
different vectors that solve

386
00:25:03,000 --> 00:25:09,000
those three equations.
This is also a solution.

387
00:25:07,000 --> 00:25:13,000
If I call this the eigenvector
alpha 1, then I ought to call

388
00:25:12,000 --> 00:25:18,000
this one the alpha 2.
Hey, we can keep on going

389
00:25:16,000 --> 00:25:22,000
through this.
Why not make the first one

390
00:25:19,000 --> 00:25:25,000
zero?
Well, what would happen if I

391
00:25:21,000 --> 00:25:27,000
made the first one 0,
and then 1, and minus 1?

392
00:25:25,000 --> 00:25:31,000
The answer is this one is no
longer independent of those two.

393
00:25:32,000 --> 00:25:38,000
I can get it by taking a
combination of those two.

394
00:25:34,000 --> 00:25:40,000
Do you see what combination I
should take?

395
00:25:42,000 --> 00:25:48,000
This one minus that one.
This guy minus that guy gives

396
00:25:47,000 --> 00:25:53,000
me that guy, isn't that right?
1 minus 1, 0 minus minus 1,

397
00:25:52,000 --> 00:25:58,000
minus 1 minus 0.
This is not a new one.

398
00:25:56,000 --> 00:26:02,000
It looks new,
but it is not.

399
00:26:00,000 --> 00:26:06,000
I can get it by taking a linear
combination of these two.

400
00:26:04,000 --> 00:26:10,000
It is not independent delta.
And that would be true for any

401
00:26:08,000 --> 00:26:14,000
other possible solution you
could get for these equations.

402
00:26:12,000 --> 00:26:18,000
Once you found two solutions,
all the others will be linear

403
00:26:16,000 --> 00:26:22,000
combinations of them.
Well, I cannot use that one.

404
00:26:20,000 --> 00:26:26,000
It is not new.
And the general solution,

405
00:26:23,000 --> 00:26:29,000
therefore, will be a
combination, c1 times that one

406
00:26:26,000 --> 00:26:32,000
plus a constant times this one.
Plus the first one that I found

407
00:26:32,000 --> 00:26:38,000
c3 times (1, 1,
1) e to the 0t,

408
00:26:36,000 --> 00:26:42,000
which I don't have to write in.
That is the general solution to

409
00:26:42,000 --> 00:26:48,000
the system, (x1,x2, x3).

410
00:26:45,000 --> 00:26:51,000
What happens as time goes to
infinity?

411
00:26:48,000 --> 00:26:54,000
Regardless of what the values
of these two C's this term goes

412
00:26:54,000 --> 00:27:00,000
to zero, that term goes to zero
and what I am left with is a

413
00:27:00,000 --> 00:27:06,000
constant solution.
So all of these solutions tend

414
00:27:06,000 --> 00:27:12,000
to be the solution where all the
cells are at the same

415
00:27:11,000 --> 00:27:17,000
temperature.
Well, of course there must be

416
00:27:15,000 --> 00:27:21,000
some vocabulary word in this.
There is.

417
00:27:19,000 --> 00:27:25,000
There are two vocabulary words.
This is a good eigenvalue.

418
00:27:25,000 --> 00:27:31,000
There are also bad eigenvalues.
This is a good repeated

419
00:27:32,000 --> 00:27:38,000
eigenvalue, but good is not the
official word.

420
00:27:37,000 --> 00:27:43,000
An eigenvalue like this,
which is repeated but where you

421
00:27:44,000 --> 00:27:50,000
can find enough eigenvectors,
if lambda is a repeated

422
00:27:50,000 --> 00:27:56,000
eigenvalue, it occurs multiply
in the characteristic polynomial

423
00:27:57,000 --> 00:28:03,000
as a root.
But you can find enough

424
00:28:03,000 --> 00:28:09,000
independent eigenvectors --
Forget the "but."

425
00:28:30,000 --> 00:28:36,000
-- to make up the needed number
of independent solutions.

426
00:28:36,000 --> 00:28:42,000
For example,
if it is repeated once,

427
00:28:40,000 --> 00:28:46,000
that is it occurs doubly then
somehow I have got to get two

428
00:28:46,000 --> 00:28:52,000
solutions out of that as I was
able to here.

429
00:28:51,000 --> 00:28:57,000
If it occurred triply,
I have got to get three

430
00:28:56,000 --> 00:29:02,000
solutions out of it.
I would look for three

431
00:29:02,000 --> 00:29:08,000
independent eigenvectors and
hope I could find them.

432
00:29:06,000 --> 00:29:12,000
That is the good case because
it tells you how to make up as

433
00:29:12,000 --> 00:29:18,000
many solutions as you need.
And this kind of eigenvalue is

434
00:29:17,000 --> 00:29:23,000
called in the literature the
complete eigenvalue.

435
00:29:30,000 --> 00:29:36,000
Now, how about the kind in
which you cannot?

436
00:29:33,000 --> 00:29:39,000
Well, unfortunately,
all my life I have called it

437
00:29:37,000 --> 00:29:43,000
incomplete, which seems to be a
perfectly reasonable thing to

438
00:29:43,000 --> 00:29:49,000
call it.
However, terminology changes

439
00:29:46,000 --> 00:29:52,000
slowly over time.
The notes, because I wrote

440
00:29:50,000 --> 00:29:56,000
them, call it an incomplete
eigenvalue.

441
00:29:53,000 --> 00:29:59,000
But the accepted term nowadays
is defective.

442
00:29:57,000 --> 00:30:03,000
I don't like that.
It violates the "eigenvalues

443
00:30:02,000 --> 00:30:08,000
with disabilities act" or
something.

444
00:30:06,000 --> 00:30:12,000
But I have to give it to you
because that is the word I am

445
00:30:11,000 --> 00:30:17,000
going to try to use from now on,
at least if I remember to use

446
00:30:18,000 --> 00:30:24,000
it.
It would be the word,

447
00:30:20,000 --> 00:30:26,000
for example,
used in the linear algebra

448
00:30:24,000 --> 00:30:30,000
course 18.06 "plug,
plug," defective otherwise.

449
00:30:30,000 --> 00:30:36,000
A defective eigenvalue is one
where you can get one

450
00:30:33,000 --> 00:30:39,000
eigenvector.
If it is double,

451
00:30:34,000 --> 00:30:40,000
for example,
if it a double eigenvalue.

452
00:30:37,000 --> 00:30:43,000
It is defective if you can get
one eigenvector that goes with

453
00:30:41,000 --> 00:30:47,000
it, but you cannot find an
independent one.

454
00:30:43,000 --> 00:30:49,000
The only other ones you can
find are multiples of the first

455
00:30:47,000 --> 00:30:53,000
one.
Then you are really in trouble

456
00:30:49,000 --> 00:30:55,000
because you just don't have
enough solutions that you are

457
00:30:53,000 --> 00:30:59,000
supposed to get out of that,
and you have to do something.

458
00:30:58,000 --> 00:31:04,000
What you do is turn to problem
two on your problem set and

459
00:31:01,000 --> 00:31:07,000
solve it because that tells you
what to do.

460
00:31:04,000 --> 00:31:10,000
And I even give you an example
to work.

461
00:31:06,000 --> 00:31:12,000
Problem two,
that little matrix has a

462
00:31:09,000 --> 00:31:15,000
defective eigenvalue.
It doesn't look defective,

463
00:31:12,000 --> 00:31:18,000
but you cannot tell.
It is defective.

464
00:31:14,000 --> 00:31:20,000
But you, nonetheless,
will be able to find two

465
00:31:17,000 --> 00:31:23,000
solutions because you will be
following instructions.

466
00:31:32,000 --> 00:31:38,000
Now, the only other thing I
should tell you is one of the

467
00:31:35,000 --> 00:31:41,000
most important theorems in
linear algebra,

468
00:31:37,000 --> 00:31:43,000
which is totally beyond the
scope of this course and is

469
00:31:41,000 --> 00:31:47,000
beyond the scope of most
elementary linear algebra

470
00:31:44,000 --> 00:31:50,000
courses as I have taught around
the country but,

471
00:31:47,000 --> 00:31:53,000
of course, not at MIT.
But, nonetheless,

472
00:31:49,000 --> 00:31:55,000
it is the last theorem in the
course.

473
00:31:51,000 --> 00:31:57,000
That means it is liable to use
stuff.

474
00:31:53,000 --> 00:31:59,000
The theorem goes by different
names.

475
00:31:55,000 --> 00:32:01,000
Sometimes it is called the
principle axis theorem.

476
00:32:00,000 --> 00:32:06,000
Sometimes it is called the
spectral theorem.

477
00:32:04,000 --> 00:32:10,000
But, anyway,
what it says is,

478
00:32:06,000 --> 00:32:12,000
if A is a real end-by-end
matrix which is symmetric,

479
00:32:11,000 --> 00:32:17,000
you know what a symmetric
matrix is?

480
00:32:15,000 --> 00:32:21,000
The formal definition is it is
equal to its transpose.

481
00:32:20,000 --> 00:32:26,000
What that means is if you flip
it around the main diagonal it

482
00:32:25,000 --> 00:32:31,000
looks just the same as before.
Somewhere on this board,

483
00:32:32,000 --> 00:32:38,000
right there,
in fact, is a symmetric matrix.

484
00:32:36,000 --> 00:32:42,000
What happened to it?
Right here was the symmetric

485
00:32:40,000 --> 00:32:46,000
matrix.
I erased the one thing which I

486
00:32:44,000 --> 00:32:50,000
had to have.
Minus 2, 1, 1;

487
00:32:46,000 --> 00:32:52,000
1, minus 2, 1;
1, 1 minus 2.

488
00:32:51,000 --> 00:32:57,000
That was our matrix A.
The matrix is symmetric because

489
00:32:57,000 --> 00:33:03,000
if I flip it around the diagonal
it looks the same as it did

490
00:33:02,000 --> 00:33:08,000
before.
Well, not exactly.

491
00:33:06,000 --> 00:33:12,000
The ones are sort of lying on
their side, but you have to take

492
00:33:10,000 --> 00:33:16,000
account of that.
Is that right?

493
00:33:12,000 --> 00:33:18,000
The twos are backward.
Well, you know what I mean.

494
00:33:15,000 --> 00:33:21,000
Put that element there,
this one here,

495
00:33:18,000 --> 00:33:24,000
that one there.
Exchange these two.

496
00:33:20,000 --> 00:33:26,000
Notice the diagonal elements
don't all have to be minus 2 for

497
00:33:24,000 --> 00:33:30,000
that.
No matter what they were,

498
00:33:26,000 --> 00:33:32,000
they are the guys that aren't
moved when you do the flipping.

499
00:33:32,000 --> 00:33:38,000
Therefore, there is no
condition on them.

500
00:33:34,000 --> 00:33:40,000
It is these other guys.
Each guy here has to use the

501
00:33:38,000 --> 00:33:44,000
same guy there.
This one has to be the same as

502
00:33:42,000 --> 00:33:48,000
that one, and so on.
Then it will be real and

503
00:33:45,000 --> 00:33:51,000
symmetric.
If you have a matrix that is

504
00:33:48,000 --> 00:33:54,000
real and symmetric,
like the one we have been

505
00:33:51,000 --> 00:33:57,000
working with,
the theorem is that all its

506
00:33:54,000 --> 00:34:00,000
eigenvalues are complete.
That is a very unobvious

507
00:33:58,000 --> 00:34:04,000
theorem.
All its eigenvalues are

508
00:34:02,000 --> 00:34:08,000
automatically complete.
And it is a remarkable fact

509
00:34:07,000 --> 00:34:13,000
that you can prove that purely
generally with a certain amount

510
00:34:14,000 --> 00:34:20,000
of pure reasoning no calculation
at all.

511
00:34:18,000 --> 00:34:24,000
But it has to be true,
and it is true.

512
00:34:22,000 --> 00:34:28,000
You will find there are whole
branches of applied differential

513
00:34:28,000 --> 00:34:34,000
equations.
You know, equilibrium theory,

514
00:34:33,000 --> 00:34:39,000
all the matrices that you deal
with are always symmetric.

515
00:34:37,000 --> 00:34:43,000
And, therefore,
this repeated eigenvalues is

516
00:34:40,000 --> 00:34:46,000
not something you have to worry
about, finding extra solutions.

517
00:34:45,000 --> 00:34:51,000
Well, I guess that is the end
of the first part of the

518
00:34:49,000 --> 00:34:55,000
lecture.
I have a third of it left.

519
00:34:52,000 --> 00:34:58,000
Let's talk fast.
I would like to,

520
00:34:55,000 --> 00:35:01,000
with the remaining time,
explain to you what to do if

521
00:34:59,000 --> 00:35:05,000
you were to get complex
eigenvalues.

522
00:35:08,000 --> 00:35:14,000
Now, actually,
the answer is follow the same

523
00:35:11,000 --> 00:35:17,000
program.
In other words,

524
00:35:12,000 --> 00:35:18,000
if you solve the characteristic
equation and you get a complex

525
00:35:17,000 --> 00:35:23,000
root, follow the program,
calculate the corresponding

526
00:35:20,000 --> 00:35:26,000
complex eigenvectors.
In other words,

527
00:35:23,000 --> 00:35:29,000
solve the equations.
Everything will be the same

528
00:35:26,000 --> 00:35:32,000
except that the eigenvectors
will turn out to be complex,

529
00:35:30,000 --> 00:35:36,000
that is will have complex
entries.

530
00:35:34,000 --> 00:35:40,000
Don't worry about it.
Then form the solutions.

531
00:35:37,000 --> 00:35:43,000
The solutions are now going to
look once again like alpha times

532
00:35:43,000 --> 00:35:49,000
e to the a plus bi to the t.

533
00:35:47,000 --> 00:35:53,000
This will be complex and that
will be complex,

534
00:35:51,000 --> 00:35:57,000
too.
This will have complex entries.

535
00:35:54,000 --> 00:36:00,000
And then, finally,
take the real and imaginary

536
00:35:58,000 --> 00:36:04,000
parts.
Those will be real and they

537
00:36:02,000 --> 00:36:08,000
will give real and two
solutions.

538
00:36:04,000 --> 00:36:10,000
In other words,
the program is exactly like

539
00:36:08,000 --> 00:36:14,000
what we did for second-order
differential equations.

540
00:36:12,000 --> 00:36:18,000
We used the complex numbers,
got complex solutions.

541
00:36:16,000 --> 00:36:22,000
And then, at the very last
step, we took the real and

542
00:36:20,000 --> 00:36:26,000
imaginary parts to get two real
solutions out of each complex

543
00:36:25,000 --> 00:36:31,000
number.
I would like to give you a

544
00:36:27,000 --> 00:36:33,000
simple example of working that
out.

545
00:36:30,000 --> 00:36:36,000
And it is the system x prime
equals x plus 2y.

546
00:36:35,000 --> 00:36:41,000
And y prime equals minus x

547
00:36:39,000 --> 00:36:45,000
minus y.
Because it is springtime,

548
00:36:45,000 --> 00:36:51,000
it doesn't feel like spring but
it will this weekend as it is

549
00:36:51,000 --> 00:36:57,000
getting warmer.
And since, when I am too tired

550
00:36:56,000 --> 00:37:02,000
to make up problem sets for you
late at night,

551
00:37:01,000 --> 00:37:07,000
I watch reruns of Seinfeld.
I am from New York.

552
00:37:06,000 --> 00:37:12,000
It is just in my bloodstream.
Of course, the most interesting

553
00:37:12,000 --> 00:37:18,000
character on Seinfeld is George.
We are going to consider Susan

554
00:37:19,000 --> 00:37:25,000
who is the girlfriend who got
killed by licking poison

555
00:37:24,000 --> 00:37:30,000
envelopes.
And George carried on their

556
00:37:28,000 --> 00:37:34,000
love affair until Susan was
disposed of by the writers by

557
00:37:33,000 --> 00:37:39,000
this strange death.
And we are going to consider x

558
00:37:39,000 --> 00:37:45,000
is modeling Susan's love for
George.

559
00:37:43,000 --> 00:37:49,000
That is x.
And George's love for Susan

560
00:37:47,000 --> 00:37:53,000
will be y.
Now, I don't mean the absolute

561
00:37:51,000 --> 00:37:57,000
love.
If x and y are zero,

562
00:37:53,000 --> 00:37:59,000
I don't mean that they don't
love each other.

563
00:37:58,000 --> 00:38:04,000
I just mean that that is the
equilibrium value of the love.

564
00:38:05,000 --> 00:38:11,000
Everything else is measured as
departures from that.

565
00:38:10,000 --> 00:38:16,000
So (0, 0) represents the normal
amount of love,

566
00:38:15,000 --> 00:38:21,000
if love is measured.
I don't know what love units

567
00:38:20,000 --> 00:38:26,000
are.
Hearts, I guess.

568
00:38:22,000 --> 00:38:28,000
Six hearts, let's say.
Now, in what sense does this

569
00:38:27,000 --> 00:38:33,000
model it?
This is a normal equation and

570
00:38:32,000 --> 00:38:38,000
this is a neurotic equation.
That is why this is George and

571
00:38:36,000 --> 00:38:42,000
this is Susan who seemed very
normal to me.

572
00:38:39,000 --> 00:38:45,000
Susan is a normal person.
When y is positive that means

573
00:38:43,000 --> 00:38:49,000
that George seems to be loving
her more today than yesterday,

574
00:38:47,000 --> 00:38:53,000
and her natural response is to
be more in love with him.

575
00:38:51,000 --> 00:38:57,000
That is what most people are.
If y is negative,

576
00:38:54,000 --> 00:39:00,000
hey, what's the matter with
George?

577
00:38:57,000 --> 00:39:03,000
He doesn't feel so good.
Maybe there is something wrong

578
00:39:02,000 --> 00:39:08,000
with him.
She gets a little mad at him

579
00:39:06,000 --> 00:39:12,000
and this goes down.
x prime is negative.

580
00:39:10,000 --> 00:39:16,000
And the same way why is this
positive?

581
00:39:13,000 --> 00:39:19,000
Well, again,
it is a psychological thing,

582
00:39:17,000 --> 00:39:23,000
but all the world loves a
lover.

583
00:39:20,000 --> 00:39:26,000
When Susan is in love,
as she feels x is high,

584
00:39:24,000 --> 00:39:30,000
that makes her feel good.
And she loves everything,

585
00:39:28,000 --> 00:39:34,000
in fact.
Not just George.

586
00:39:32,000 --> 00:39:38,000
It is one of those things.
You all know what I am talking

587
00:39:38,000 --> 00:39:44,000
about.
Now, George,

588
00:39:40,000 --> 00:39:46,000
of course, is what makes the
writers happy.

589
00:39:44,000 --> 00:39:50,000
George is neurotic and,
therefore, is exactly the

590
00:39:49,000 --> 00:39:55,000
opposite.
He sees one day that he feels

591
00:39:53,000 --> 00:39:59,000
more in love with Susan than he
was yesterday.

592
00:40:00,000 --> 00:40:06,000
Does this make him happy?
Not at all.

593
00:40:02,000 --> 00:40:08,000
Not at all.
It makes y prime more negative.

594
00:40:05,000 --> 00:40:11,000
Why?
Because all he can think of is,

595
00:40:08,000 --> 00:40:14,000
my God, suppose I am really in
love with this girl?

596
00:40:11,000 --> 00:40:17,000
Suppose I marry her.
Oh, my God, 40 years of seeing

597
00:40:15,000 --> 00:40:21,000
the same person at breakfast all
the time.

598
00:40:18,000 --> 00:40:24,000
I must be crazy.
And so it goes down.

599
00:40:20,000 --> 00:40:26,000
Here is our neurotic model.
The question for differential

600
00:40:24,000 --> 00:40:30,000
equations is,
what do the solutions to that

601
00:40:27,000 --> 00:40:33,000
look like?
In other words,

602
00:40:31,000 --> 00:40:37,000
how does, in fact,
their love affair go?

603
00:40:34,000 --> 00:40:40,000
Now, there is a reason why the
writers picked that model,

604
00:40:39,000 --> 00:40:45,000
as you will see.
It means they were able to get

605
00:40:44,000 --> 00:40:50,000
a year's worth of episodes out
of it.

606
00:40:47,000 --> 00:40:53,000
And why is that so?
Well, let's solve it.

607
00:40:50,000 --> 00:40:56,000
The characteristic equation is
lambda squared.

608
00:40:54,000 --> 00:41:00,000
The matrix that governs this
system is A equals (1,

609
00:40:59,000 --> 00:41:05,000
2; negative 1,
negative 1)

610
00:41:02,000 --> 00:41:08,000
The trace of that matrix,

611
00:41:06,000 --> 00:41:12,000
the sum of the diagonal
elements is zero.

612
00:41:10,000 --> 00:41:16,000
There is the zero lambda here.
The determinant,

613
00:41:13,000 --> 00:41:19,000
which is the constant term,
is negative 1,

614
00:41:16,000 --> 00:41:22,000
minus negative 2,
which is plus 1.

615
00:41:19,000 --> 00:41:25,000
So the characteristic equation,
by calculating the trace and

616
00:41:23,000 --> 00:41:29,000
determinant is lambda squared
plus 1 equals 0.

617
00:41:28,000 --> 00:41:34,000
The eigenvalues are plus and

618
00:41:32,000 --> 00:41:38,000
minus i.
Now, you don't have to pick

619
00:41:35,000 --> 00:41:41,000
both of them because the
negative one lead to essentially

620
00:41:40,000 --> 00:41:46,000
the same solutions but with
negative signs.

621
00:41:44,000 --> 00:41:50,000
Either one will do just as we
solved second order equations.

622
00:41:49,000 --> 00:41:55,000
The system for finding the
eigenvectors,

623
00:41:53,000 --> 00:41:59,000
well, we are going to have to
accept the complex eigenvector.

624
00:42:00,000 --> 00:42:06,000
What is the system going to be?
Well, I take the matrix and I

625
00:42:05,000 --> 00:42:11,000
subtract i.
We will use i.

626
00:42:07,000 --> 00:42:13,000
Subtract i from the main
diagonal.

627
00:42:10,000 --> 00:42:16,000
So the system is (1 minus i)
times a1 plus 2a2 is zero.

628
00:42:15,000 --> 00:42:21,000
And let's, for good measure,

629
00:42:19,000 --> 00:42:25,000
write the other one down,
too.

630
00:42:22,000 --> 00:42:28,000
It is negative a1 plus minus (1
minus i) times a2.

631
00:42:28,000 --> 00:42:34,000
Then what is the solution?

632
00:42:33,000 --> 00:42:39,000
Well, you get the solution the
usual way.

633
00:42:39,000 --> 00:42:45,000
Let's take a1 equal to 1.

634
00:42:44,000 --> 00:42:50,000
Then what is a2?
a2 is 1 minus i divided by 2

635
00:42:50,000 --> 00:42:56,000
from the first equation.

636
00:42:58,000 --> 00:43:04,000
So the complex solution is 1
minus i over 2 times

637
00:43:02,000 --> 00:43:08,000
e to the it.
Now you have to take the real

638
00:43:05,000 --> 00:43:11,000
and imaginary parts of that.
This is the only part which

639
00:43:09,000 --> 00:43:15,000
technically I would not trust
you to do without having someone

640
00:43:13,000 --> 00:43:19,000
show you how to do it.
What do you do?

641
00:43:15,000 --> 00:43:21,000
Well, of course,
you know how to separate the

642
00:43:18,000 --> 00:43:24,000
real and imaginary parts of
that.

643
00:43:20,000 --> 00:43:26,000
It is the first thing is to
separate the vectors.

644
00:43:25,000 --> 00:43:31,000
I don't know how to explain
this.

645
00:43:29,000 --> 00:43:35,000
Just watch.
The real part of it is 1,

646
00:43:33,000 --> 00:43:39,000
one-half.
It should be negative 1,

647
00:43:37,000 --> 00:43:43,000
so minus this plus that because
I didn't put that on the right

648
00:43:45,000 --> 00:43:51,000
side.
It is minus one-half plus i

649
00:43:49,000 --> 00:43:55,000
times (0, one-half).

650
00:43:54,000 --> 00:44:00,000
Anybody want to fight?

651
00:44:00,000 --> 00:44:06,000
1 plus i times 0 minus one-half
plus one-half times i.

652
00:44:06,000 --> 00:44:12,000
You saw how I did that?
Okay.

653
00:44:10,000 --> 00:44:16,000
When you do these problem you
do it the same way,

654
00:44:16,000 --> 00:44:22,000
but don't ask me to explain
what I just did.

655
00:44:21,000 --> 00:44:27,000
Here it is cosine t plus i sine
t.

656
00:44:30,000 --> 00:44:36,000
And so the real part will give
me one solution.

657
00:44:34,000 --> 00:44:40,000
The imaginary part will give me
another.

658
00:44:39,000 --> 00:44:45,000
Since I have a limited amount
of time, let's just calculate

659
00:44:45,000 --> 00:44:51,000
the real part.
What is it?

660
00:44:48,000 --> 00:44:54,000
Well, it is (1,
minus ½) times cosine t,

661
00:44:52,000 --> 00:44:58,000
i squared is negative 1,
so minus (0,

662
00:44:56,000 --> 00:45:02,000
one-half), the negative 1 from
the i squared,

663
00:45:01,000 --> 00:45:07,000
times sine t.

664
00:45:04,000 --> 00:45:10,000
Now, what solution is that?

665
00:45:10,000 --> 00:45:16,000
This is (x, y).
Take the final step.

666
00:45:13,000 --> 00:45:19,000
It doesn't have to look like
that.

667
00:45:16,000 --> 00:45:22,000
x equals cosine t.

668
00:45:19,000 --> 00:45:25,000
Do you see that?
x equals cosine t plus 0 times

669
00:45:24,000 --> 00:45:30,000
sine t.

670
00:45:27,000 --> 00:45:33,000
What is y?
y is minus one-half times

671
00:45:31,000 --> 00:45:37,000
cosine t, minus one-half times
sine t, plus sine t.

672
00:45:37,000 --> 00:45:43,000
Now, you may have the pleasure

673
00:45:42,000 --> 00:45:48,000
of showing eliminating t.
You get a quadratic polynomial

674
00:45:47,000 --> 00:45:53,000
in x and y equals zero.
This is an ellipse.

675
00:45:51,000 --> 00:45:57,000
As t varies,
you can see this repeats its

676
00:45:55,000 --> 00:46:01,000
values at intervals of 2pi,
this gives an ellipse.

677
00:46:01,000 --> 00:46:07,000
And if you want to use a little
computer program,

678
00:46:05,000 --> 00:46:11,000
linear phase,
this is not in the assignment,

679
00:46:09,000 --> 00:46:15,000
but the ellipses look like this
and go around that way.

680
00:46:14,000 --> 00:46:20,000
And that is the model of George
and Susan's love.

681
00:46:19,000 --> 00:46:25,000
x, Susan.
y, George.

682
00:46:21,000 --> 00:46:27,000
They go round and round in this
little love circle,

683
00:46:25,000 --> 00:46:31,000
and it stretches on for 26
episodes.