WEBVTT
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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
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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,
00:12:22.000 --> 00:12:28.000
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
00:13:11.000 --> 00:13:17.000
characteristic equation?
Maybe I can finish it.
00:13:15.000 --> 00:13:21.000
I should stay over here instead
of recopying all of it.
00:13:27.000 --> 00:13:33.000
Well, there is a lot more work
to do.
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Let's see if we can at least
write down the equation.
00:13:33.000 --> 00:13:39.000
What is it?
It is lambda cubed.
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What is the lambda squared
00:13:38.000 --> 00:13:44.000
erm?
It is six and that is all there
00:13:42.000 --> 00:13:48.000
is.
How about the lambda term?
00:13:44.000 --> 00:13:50.000
Well, we have 12 lambda minus 3
lambda which makes plus 9
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lambda.
That looks good but constant
00:13:51.000 --> 00:13:57.000
terms have a way of screwing
everything up.
00:13:54.000 --> 00:14:00.000
What is the constant term?
It is A minus 2 minus 6.
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Zero.
The constant term is zero.
00:14:02.000 --> 00:14:08.000
That converts this from a hard
problem to an easy problem.
00:14:07.000 --> 00:14:13.000
Now it is a cinch to calculate
the stuff.
00:14:11.000 --> 00:14:17.000
Let's go to this board and
continue the work over here.
00:14:24.000 --> 00:14:30.000
The equation is lambda cubed
plus 6 lambda squared plus 9
00:14:27.000 --> 00:14:33.000
lambda is zero.
00:14:31.000 --> 00:14:37.000
It is very easy to calculate
the roots of that.
00:14:34.000 --> 00:14:40.000
You factor it.
Lambda is a common factor.
00:14:37.000 --> 00:14:43.000
And what is left?
Lambda squared plus 6 lambda
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plus 9.
00:14:43.000 --> 00:14:49.000
That is the sort of thing you
got all the time when you were
00:14:47.000 --> 00:14:53.000
studying critical damping.
It is the square of lambda plus
00:14:51.000 --> 00:14:57.000
3. Lambda squared plus 6 lambda
00:14:54.000 --> 00:15:00.000
plus 9 equals zero.
So the eigenvalues,
00:14:58.000 --> 00:15:04.000
the roots are what?
Well, they are lambda equals
00:15:02.000 --> 00:15:08.000
zero from this factor and then
lambda equals minus 3.
00:15:07.000 --> 00:15:13.000
But what is new is that the
00:15:10.000 --> 00:15:16.000
minus 3 is a double root.
That is a double root.
00:15:14.000 --> 00:15:20.000
Now, that, of course,
is what is going to cause the
00:15:18.000 --> 00:15:24.000
trouble.
Because, for each one of these,
00:15:21.000 --> 00:15:27.000
I am supposed to calculate the
eigenvector and make up the
00:15:26.000 --> 00:15:32.000
solution.
But that assumed that I had
00:15:29.000 --> 00:15:35.000
three things to get three
different solutions.
00:15:32.000 --> 00:15:38.000
Here I have only got two
things.
00:15:34.000 --> 00:15:40.000
It is the same trouble we ran
into when there was a repeated
00:15:38.000 --> 00:15:44.000
root.
We were studying second or
00:15:40.000 --> 00:15:46.000
third order differential
equations and the characteristic
00:15:43.000 --> 00:15:49.000
equation had a repeated root.
And I had to go into a song and
00:15:47.000 --> 00:15:53.000
dance and stand on my head and
multiply things by t and so on.
00:15:51.000 --> 00:15:57.000
And then talk very hard arguing
why that was a good thing to do
00:15:54.000 --> 00:16:00.000
to get the answer.
Now, I am not going to do the
00:15:57.000 --> 00:16:03.000
same thing here.
Instead, I am going to try to
00:16:02.000 --> 00:16:08.000
solve the problem instead.
Let's get two points by at
00:16:06.000 --> 00:16:12.000
least doing the easy part of it.
Lambda equals zero.
00:16:11.000 --> 00:16:17.000
What am I supposed to do with
00:16:13.000 --> 00:16:19.000
lambda equals zero?
I am looking for the alpha that
00:16:17.000 --> 00:16:23.000
goes with that.
And I find that eigenvector by
00:16:21.000 --> 00:16:27.000
solving this system of
equations.
00:16:24.000 --> 00:16:30.000
Let's write out what that
system of equations is.
00:16:29.000 --> 00:16:35.000
Well, if lambda is zero,
this isn't there.
00:16:32.000 --> 00:16:38.000
It is just the matrix A times
alpha equals zero.
00:16:38.000 --> 00:16:44.000
And the matrix A is,
00:16:40.000 --> 00:16:46.000
I never even wrote it anywhere.
I never wrote A.
00:16:45.000 --> 00:16:51.000
I thought I would get away
without having to do it,
00:16:49.000 --> 00:16:55.000
but you never get away with
anything.
00:16:53.000 --> 00:16:59.000
It's the principle of life.
That is A.
00:16:58.000 --> 00:17:04.000
If I subtract zero from the
main diagonal,
00:17:01.000 --> 00:17:07.000
that doesn't do a great deal to
A.
00:17:04.000 --> 00:17:10.000
And the resulting system of
equations is those same things,
00:17:10.000 --> 00:17:16.000
except you have the a1's there,
too.
00:17:13.000 --> 00:17:19.000
There is one.
a1 minus 2 a2 plus a3 equals
00:17:17.000 --> 00:17:23.000
zero.
I am just subtracting zero from
00:17:22.000 --> 00:17:28.000
the main diagonal so there is
nothing to do.
00:17:26.000 --> 00:17:32.000
a2 minus 2 a3 equals zero.
Now I am supposed to solve
00:17:33.000 --> 00:17:39.000
those.
00:17:40.000 --> 00:17:46.000
Of course we could do it.
Well, how do you know how to
00:17:43.000 --> 00:17:49.000
solve a system of three linear
equations?
00:17:45.000 --> 00:17:51.000
Well, elimination.
You can always solve by
00:17:48.000 --> 00:17:54.000
elimination.
Now we are much more
00:17:50.000 --> 00:17:56.000
sophisticated than that.
You all have pocket calculators
00:17:54.000 --> 00:18:00.000
so you could use the inverse
matrix, right?
00:17:56.000 --> 00:18:02.000
No.
You cannot use the inverse
00:17:58.000 --> 00:18:04.000
matrix.
What will happen if you punch
00:18:03.000 --> 00:18:09.000
in those coefficients and then
punch in A inverse.
00:18:08.000 --> 00:18:14.000
What answer will it give you?
0, 0, 0.
00:18:12.000 --> 00:18:18.000
No, I am sorry.
It won't give you any answer.
00:18:16.000 --> 00:18:22.000
What will it say?
It will say I cannot calculate
00:18:21.000 --> 00:18:27.000
the inverse to that matrix
because the whole purpose of
00:18:27.000 --> 00:18:33.000
this exercise was to find a
value of lambda such that this
00:18:33.000 --> 00:18:39.000
system of equations is
dependent.
00:18:38.000 --> 00:18:44.000
The coefficient determinant is
zero and, therefore,
00:18:42.000 --> 00:18:48.000
the coefficient matrix does not
have an inverse matrix.
00:18:46.000 --> 00:18:52.000
You cannot use that method.
In other words,
00:18:49.000 --> 00:18:55.000
the inverse matrix will never
work in these problems because
00:18:54.000 --> 00:19:00.000
the system of equations you will
be trying to solve is always a
00:18:59.000 --> 00:19:05.000
non-independent system.
And, therefore,
00:19:03.000 --> 00:19:09.000
its determinant is always zero.
And, therefore,
00:19:06.000 --> 00:19:12.000
there is no inverse matrix
because the determinant of the
00:19:11.000 --> 00:19:17.000
coefficient is zero.
All you can do is use
00:19:14.000 --> 00:19:20.000
elimination or physical insight
and common sense.
00:19:18.000 --> 00:19:24.000
Now, because I teach
differential equations everybody
00:19:23.000 --> 00:19:29.000
assumes, mistakenly,
as I think, that I really know
00:19:27.000 --> 00:19:33.000
something about them.
I get now and then graduate
00:19:32.000 --> 00:19:38.000
students, not in mathematics,
but some obscure field of
00:19:36.000 --> 00:19:42.000
engineering or whatever drift
into my office and say I see you
00:19:40.000 --> 00:19:46.000
teach differential equations.
Do you have a minute here?
00:19:45.000 --> 00:19:51.000
And before I can say no they
write their differential
00:19:49.000 --> 00:19:55.000
equation on the board.
And almost invariably it is
00:19:52.000 --> 00:19:58.000
nothing I have ever seen before.
And they so look at me
00:19:57.000 --> 00:20:03.000
hopefully and expectantly.
So what do I ask them?
00:20:01.000 --> 00:20:07.000
I don't ask them what they have
tried.
00:20:05.000 --> 00:20:11.000
What I ask them is where did
this come from?
00:20:08.000 --> 00:20:14.000
What field did it come from?
Because each field has its own
00:20:14.000 --> 00:20:20.000
little tricks.
It gets the same differential
00:20:17.000 --> 00:20:23.000
equations all the time and has
its own little tricks for
00:20:22.000 --> 00:20:28.000
solving them.
You should do the same thing
00:20:26.000 --> 00:20:32.000
here.
Well, of course we can solve
00:20:29.000 --> 00:20:35.000
this.
And by now most of you have
00:20:33.000 --> 00:20:39.000
solved it just by inspection,
just by sort of psyching out
00:20:37.000 --> 00:20:43.000
the answer.
But a better way is to say
00:20:40.000 --> 00:20:46.000
look, suppose we had the
solution, what would the
00:20:44.000 --> 00:20:50.000
solution look like?
Well, it would look like (a1,
00:20:48.000 --> 00:20:54.000
a2, a3), whatever the values of
those variables were which gave
00:20:52.000 --> 00:20:58.000
me the solution to the equation,
times e to the 0t.
00:20:57.000 --> 00:21:03.000
But what is this?
e to the 0t is one
00:21:02.000 --> 00:21:08.000
for all time.
And, therefore,
00:21:04.000 --> 00:21:10.000
this is a constant solution.
What I am asking is to find a
00:21:09.000 --> 00:21:15.000
constant solution.
Now, can I, by inspection,
00:21:13.000 --> 00:21:19.000
find a constant solution to
this?
00:21:15.000 --> 00:21:21.000
If so it must be the one.
Well, there is an obvious
00:21:19.000 --> 00:21:25.000
constant solution.
All the cells have the same
00:21:23.000 --> 00:21:29.000
temperature.
If that is true then there is
00:21:26.000 --> 00:21:32.000
no reason why it should ever
change as time goes on.
00:21:32.000 --> 00:21:38.000
The physical problem itself
suggests what the answer must
00:21:35.000 --> 00:21:41.000
be.
You don't have to solve
00:21:37.000 --> 00:21:43.000
equations.
In other words,
00:21:38.000 --> 00:21:44.000
any constant like (1,
1, 1).
00:21:40.000 --> 00:21:46.000
Well, could it be (20,
20, 20)?
00:21:42.000 --> 00:21:48.000
Yeah, that is a constant
multiple of (1,
00:21:44.000 --> 00:21:50.000
1, 1).
That is included.
00:21:45.000 --> 00:21:51.000
My basic constant solution,
therefore, is simply (1,
00:21:48.000 --> 00:21:54.000
1, 1) times e to the 0t.
00:21:51.000 --> 00:21:57.000
You don't have to include e to
the 0t because it is one.
00:21:54.000 --> 00:22:00.000
Now, just to check,
is (1, 1, 1) a solution to
00:21:57.000 --> 00:22:03.000
these equations?
It certainly is.
00:22:01.000 --> 00:22:07.000
1 plus 1 minus 2 is zero in
every case.
00:22:04.000 --> 00:22:10.000
The equations are essentially
the same, except they use
00:22:08.000 --> 00:22:14.000
different variables.
By inspection or,
00:22:11.000 --> 00:22:17.000
if you like,
by elimination,
00:22:14.000 --> 00:22:20.000
but not by finding the inverse
matrix you solve those
00:22:18.000 --> 00:22:24.000
equations.
And we have our first solution.
00:22:21.000 --> 00:22:27.000
Now let's go onto the second
one.
00:22:24.000 --> 00:22:30.000
For the second one,
we are going to have to use the
00:22:28.000 --> 00:22:34.000
eigenvalue lambda equals
negative 3.
00:22:33.000 --> 00:22:39.000
And now what is the system of
equations?
00:22:36.000 --> 00:22:42.000
Well, now I have to take this
and I have to subtract negative
00:22:41.000 --> 00:22:47.000
3 from the diagonal elements.
Minus 2 minus negative 3 is
00:22:46.000 --> 00:22:52.000
plus 1, right?
00:22:49.000 --> 00:22:55.000
Got that?
Each of the diagonal elements,
00:22:53.000 --> 00:22:59.000
after I subtract minus 3 turns
into plus 1.
00:22:58.000 --> 00:23:04.000
And, therefore,
the system becomes,
00:23:00.000 --> 00:23:06.000
the system I have to solve is
a1 plus a2 plus a3 equals zero.
00:23:05.000 --> 00:23:11.000
And what is the second
00:23:09.000 --> 00:23:15.000
equation?
Symmetry is preserved.
00:23:11.000 --> 00:23:17.000
All the equations are
essentially the same,
00:23:14.000 --> 00:23:20.000
except for the names of the
variables so they all must give
00:23:19.000 --> 00:23:25.000
you the same thing after I
subtract minus 3 from the main
00:23:24.000 --> 00:23:30.000
diagonal.
Well, that is what we call a
00:23:27.000 --> 00:23:33.000
dependent system of equations.
All I have is the same equation
00:23:34.000 --> 00:23:40.000
repeated twice,
but I still have to solve it.
00:23:38.000 --> 00:23:44.000
Now, what you see is that there
are lots of solutions to this.
00:23:44.000 --> 00:23:50.000
Let me write down one of them.
For example,
00:23:49.000 --> 00:23:55.000
suppose I made a1 equal to 1
and I made a2 a 0,
00:23:54.000 --> 00:24:00.000
then a3 would be negative 1.
00:24:00.000 --> 00:24:06.000
So here is a solution.
That is the eigenvector.
00:24:03.000 --> 00:24:09.000
And with it,
I can make the solution by
00:24:06.000 --> 00:24:12.000
multiplying by e to the negative
3t.
00:24:10.000 --> 00:24:16.000
There is a solution.
But that is not the only alpha
00:24:14.000 --> 00:24:20.000
I could have chosen.
Suppose I chose this one
00:24:18.000 --> 00:24:24.000
instead.
Suppose I kept this 1,
00:24:20.000 --> 00:24:26.000
but this time made a3 zero.
Well, in that case,
00:24:24.000 --> 00:24:30.000
there would be a2 that had to
be minus 1.
00:24:30.000 --> 00:24:36.000
Now, is this essentially
different from that one?
00:24:32.000 --> 00:24:38.000
It would still be multiplied by
e to the minus 3t,
00:24:36.000 --> 00:24:42.000
but don't be fooled by the e to
the minus 3t.
00:24:39.000 --> 00:24:45.000
That is our scalar.
That is not what is essential.
00:24:42.000 --> 00:24:48.000
What is essential is the
content of these two vectors.
00:24:45.000 --> 00:24:51.000
Is either one a multiple of the
other?
00:24:47.000 --> 00:24:53.000
The answer is no.
Therefore, they are
00:24:49.000 --> 00:24:55.000
independent.
They are pointing in two
00:24:51.000 --> 00:24:57.000
different directions in three
space, these two vectors.
00:24:56.000 --> 00:25:02.000
And, therefore,
I have two independent
00:24:59.000 --> 00:25:05.000
solutions just by picking two
different vectors that solve
00:25:03.000 --> 00:25:09.000
those three equations.
This is also a solution.
00:25:07.000 --> 00:25:13.000
If I call this the eigenvector
alpha 1, then I ought to call
00:25:12.000 --> 00:25:18.000
this one the alpha 2.
Hey, we can keep on going
00:25:16.000 --> 00:25:22.000
through this.
Why not make the first one
00:25:19.000 --> 00:25:25.000
zero?
Well, what would happen if I
00:25:21.000 --> 00:25:27.000
made the first one 0,
and then 1, and minus 1?
00:25:25.000 --> 00:25:31.000
The answer is this one is no
longer independent of those two.
00:25:32.000 --> 00:25:38.000
I can get it by taking a
combination of those two.
00:25:34.000 --> 00:25:40.000
Do you see what combination I
should take?
00:25:42.000 --> 00:25:48.000
This one minus that one.
This guy minus that guy gives
00:25:47.000 --> 00:25:53.000
me that guy, isn't that right?
1 minus 1, 0 minus minus 1,
00:25:52.000 --> 00:25:58.000
minus 1 minus 0.
This is not a new one.
00:25:56.000 --> 00:26:02.000
It looks new,
but it is not.
00:26:00.000 --> 00:26:06.000
I can get it by taking a linear
combination of these two.
00:26:04.000 --> 00:26:10.000
It is not independent delta.
And that would be true for any
00:26:08.000 --> 00:26:14.000
other possible solution you
could get for these equations.
00:26:12.000 --> 00:26:18.000
Once you found two solutions,
all the others will be linear
00:26:16.000 --> 00:26:22.000
combinations of them.
Well, I cannot use that one.
00:26:20.000 --> 00:26:26.000
It is not new.
And the general solution,
00:26:23.000 --> 00:26:29.000
therefore, will be a
combination, c1 times that one
00:26:26.000 --> 00:26:32.000
plus a constant times this one.
Plus the first one that I found
00:26:32.000 --> 00:26:38.000
c3 times (1, 1,
1) e to the 0t,
00:26:36.000 --> 00:26:42.000
which I don't have to write in.
That is the general solution to
00:26:42.000 --> 00:26:48.000
the system, (x1,x2, x3).
00:26:45.000 --> 00:26:51.000
What happens as time goes to
infinity?
00:26:48.000 --> 00:26:54.000
Regardless of what the values
of these two C's this term goes
00:26:54.000 --> 00:27:00.000
to zero, that term goes to zero
and what I am left with is a
00:27:00.000 --> 00:27:06.000
constant solution.
So all of these solutions tend
00:27:06.000 --> 00:27:12.000
to be the solution where all the
cells are at the same
00:27:11.000 --> 00:27:17.000
temperature.
Well, of course there must be
00:27:15.000 --> 00:27:21.000
some vocabulary word in this.
There is.
00:27:19.000 --> 00:27:25.000
There are two vocabulary words.
This is a good eigenvalue.
00:27:25.000 --> 00:27:31.000
There are also bad eigenvalues.
This is a good repeated
00:27:32.000 --> 00:27:38.000
eigenvalue, but good is not the
official word.
00:27:37.000 --> 00:27:43.000
An eigenvalue like this,
which is repeated but where you
00:27:44.000 --> 00:27:50.000
can find enough eigenvectors,
if lambda is a repeated
00:27:50.000 --> 00:27:56.000
eigenvalue, it occurs multiply
in the characteristic polynomial
00:27:57.000 --> 00:28:03.000
as a root.
But you can find enough
00:28:03.000 --> 00:28:09.000
independent eigenvectors --
Forget the "but."
00:28:30.000 --> 00:28:36.000
-- to make up the needed number
of independent solutions.
00:28:36.000 --> 00:28:42.000
For example,
if it is repeated once,
00:28:40.000 --> 00:28:46.000
that is it occurs doubly then
somehow I have got to get two
00:28:46.000 --> 00:28:52.000
solutions out of that as I was
able to here.
00:28:51.000 --> 00:28:57.000
If it occurred triply,
I have got to get three
00:28:56.000 --> 00:29:02.000
solutions out of it.
I would look for three
00:29:02.000 --> 00:29:08.000
independent eigenvectors and
hope I could find them.
00:29:06.000 --> 00:29:12.000
That is the good case because
it tells you how to make up as
00:29:12.000 --> 00:29:18.000
many solutions as you need.
And this kind of eigenvalue is
00:29:17.000 --> 00:29:23.000
called in the literature the
complete eigenvalue.
00:29:30.000 --> 00:29:36.000
Now, how about the kind in
which you cannot?
00:29:33.000 --> 00:29:39.000
Well, unfortunately,
all my life I have called it
00:29:37.000 --> 00:29:43.000
incomplete, which seems to be a
perfectly reasonable thing to
00:29:43.000 --> 00:29:49.000
call it.
However, terminology changes
00:29:46.000 --> 00:29:52.000
slowly over time.
The notes, because I wrote
00:29:50.000 --> 00:29:56.000
them, call it an incomplete
eigenvalue.
00:29:53.000 --> 00:29:59.000
But the accepted term nowadays
is defective.
00:29:57.000 --> 00:30:03.000
I don't like that.
It violates the "eigenvalues
00:30:02.000 --> 00:30:08.000
with disabilities act" or
something.
00:30:06.000 --> 00:30:12.000
But I have to give it to you
because that is the word I am
00:30:11.000 --> 00:30:17.000
going to try to use from now on,
at least if I remember to use
00:30:18.000 --> 00:30:24.000
it.
It would be the word,
00:30:20.000 --> 00:30:26.000
for example,
used in the linear algebra
00:30:24.000 --> 00:30:30.000
course 18.06 "plug,
plug," defective otherwise.
00:30:30.000 --> 00:30:36.000
A defective eigenvalue is one
where you can get one
00:30:33.000 --> 00:30:39.000
eigenvector.
If it is double,
00:30:34.000 --> 00:30:40.000
for example,
if it a double eigenvalue.
00:30:37.000 --> 00:30:43.000
It is defective if you can get
one eigenvector that goes with
00:30:41.000 --> 00:30:47.000
it, but you cannot find an
independent one.
00:30:43.000 --> 00:30:49.000
The only other ones you can
find are multiples of the first
00:30:47.000 --> 00:30:53.000
one.
Then you are really in trouble
00:30:49.000 --> 00:30:55.000
because you just don't have
enough solutions that you are
00:30:53.000 --> 00:30:59.000
supposed to get out of that,
and you have to do something.
00:30:58.000 --> 00:31:04.000
What you do is turn to problem
two on your problem set and
00:31:01.000 --> 00:31:07.000
solve it because that tells you
what to do.
00:31:04.000 --> 00:31:10.000
And I even give you an example
to work.
00:31:06.000 --> 00:31:12.000
Problem two,
that little matrix has a
00:31:09.000 --> 00:31:15.000
defective eigenvalue.
It doesn't look defective,
00:31:12.000 --> 00:31:18.000
but you cannot tell.
It is defective.
00:31:14.000 --> 00:31:20.000
But you, nonetheless,
will be able to find two
00:31:17.000 --> 00:31:23.000
solutions because you will be
following instructions.
00:31:32.000 --> 00:31:38.000
Now, the only other thing I
should tell you is one of the
00:31:35.000 --> 00:31:41.000
most important theorems in
linear algebra,
00:31:37.000 --> 00:31:43.000
which is totally beyond the
scope of this course and is
00:31:41.000 --> 00:31:47.000
beyond the scope of most
elementary linear algebra
00:31:44.000 --> 00:31:50.000
courses as I have taught around
the country but,
00:31:47.000 --> 00:31:53.000
of course, not at MIT.
But, nonetheless,
00:31:49.000 --> 00:31:55.000
it is the last theorem in the
course.
00:31:51.000 --> 00:31:57.000
That means it is liable to use
stuff.
00:31:53.000 --> 00:31:59.000
The theorem goes by different
names.
00:31:55.000 --> 00:32:01.000
Sometimes it is called the
principle axis theorem.
00:32:00.000 --> 00:32:06.000
Sometimes it is called the
spectral theorem.
00:32:04.000 --> 00:32:10.000
But, anyway,
what it says is,
00:32:06.000 --> 00:32:12.000
if A is a real end-by-end
matrix which is symmetric,
00:32:11.000 --> 00:32:17.000
you know what a symmetric
matrix is?
00:32:15.000 --> 00:32:21.000
The formal definition is it is
equal to its transpose.
00:32:20.000 --> 00:32:26.000
What that means is if you flip
it around the main diagonal it
00:32:25.000 --> 00:32:31.000
looks just the same as before.
Somewhere on this board,
00:32:32.000 --> 00:32:38.000
right there,
in fact, is a symmetric matrix.
00:32:36.000 --> 00:32:42.000
What happened to it?
Right here was the symmetric
00:32:40.000 --> 00:32:46.000
matrix.
I erased the one thing which I
00:32:44.000 --> 00:32:50.000
had to have.
Minus 2, 1, 1;
00:32:46.000 --> 00:32:52.000
1, minus 2, 1;
1, 1 minus 2.
00:32:51.000 --> 00:32:57.000
That was our matrix A.
The matrix is symmetric because
00:32:57.000 --> 00:33:03.000
if I flip it around the diagonal
it looks the same as it did
00:33:02.000 --> 00:33:08.000
before.
Well, not exactly.
00:33:06.000 --> 00:33:12.000
The ones are sort of lying on
their side, but you have to take
00:33:10.000 --> 00:33:16.000
account of that.
Is that right?
00:33:12.000 --> 00:33:18.000
The twos are backward.
Well, you know what I mean.
00:33:15.000 --> 00:33:21.000
Put that element there,
this one here,
00:33:18.000 --> 00:33:24.000
that one there.
Exchange these two.
00:33:20.000 --> 00:33:26.000
Notice the diagonal elements
don't all have to be minus 2 for
00:33:24.000 --> 00:33:30.000
that.
No matter what they were,
00:33:26.000 --> 00:33:32.000
they are the guys that aren't
moved when you do the flipping.
00:33:32.000 --> 00:33:38.000
Therefore, there is no
condition on them.
00:33:34.000 --> 00:33:40.000
It is these other guys.
Each guy here has to use the
00:33:38.000 --> 00:33:44.000
same guy there.
This one has to be the same as
00:33:42.000 --> 00:33:48.000
that one, and so on.
Then it will be real and
00:33:45.000 --> 00:33:51.000
symmetric.
If you have a matrix that is
00:33:48.000 --> 00:33:54.000
real and symmetric,
like the one we have been
00:33:51.000 --> 00:33:57.000
working with,
the theorem is that all its
00:33:54.000 --> 00:34:00.000
eigenvalues are complete.
That is a very unobvious
00:33:58.000 --> 00:34:04.000
theorem.
All its eigenvalues are
00:34:02.000 --> 00:34:08.000
automatically complete.
And it is a remarkable fact
00:34:07.000 --> 00:34:13.000
that you can prove that purely
generally with a certain amount
00:34:14.000 --> 00:34:20.000
of pure reasoning no calculation
at all.
00:34:18.000 --> 00:34:24.000
But it has to be true,
and it is true.
00:34:22.000 --> 00:34:28.000
You will find there are whole
branches of applied differential
00:34:28.000 --> 00:34:34.000
equations.
You know, equilibrium theory,
00:34:33.000 --> 00:34:39.000
all the matrices that you deal
with are always symmetric.
00:34:37.000 --> 00:34:43.000
And, therefore,
this repeated eigenvalues is
00:34:40.000 --> 00:34:46.000
not something you have to worry
about, finding extra solutions.
00:34:45.000 --> 00:34:51.000
Well, I guess that is the end
of the first part of the
00:34:49.000 --> 00:34:55.000
lecture.
I have a third of it left.
00:34:52.000 --> 00:34:58.000
Let's talk fast.
I would like to,
00:34:55.000 --> 00:35:01.000
with the remaining time,
explain to you what to do if
00:34:59.000 --> 00:35:05.000
you were to get complex
eigenvalues.
00:35:08.000 --> 00:35:14.000
Now, actually,
the answer is follow the same
00:35:11.000 --> 00:35:17.000
program.
In other words,
00:35:12.000 --> 00:35:18.000
if you solve the characteristic
equation and you get a complex
00:35:17.000 --> 00:35:23.000
root, follow the program,
calculate the corresponding
00:35:20.000 --> 00:35:26.000
complex eigenvectors.
In other words,
00:35:23.000 --> 00:35:29.000
solve the equations.
Everything will be the same
00:35:26.000 --> 00:35:32.000
except that the eigenvectors
will turn out to be complex,
00:35:30.000 --> 00:35:36.000
that is will have complex
entries.
00:35:34.000 --> 00:35:40.000
Don't worry about it.
Then form the solutions.
00:35:37.000 --> 00:35:43.000
The solutions are now going to
look once again like alpha times
00:35:43.000 --> 00:35:49.000
e to the a plus bi to the t.
00:35:47.000 --> 00:35:53.000
This will be complex and that
will be complex,
00:35:51.000 --> 00:35:57.000
too.
This will have complex entries.
00:35:54.000 --> 00:36:00.000
And then, finally,
take the real and imaginary
00:35:58.000 --> 00:36:04.000
parts.
Those will be real and they
00:36:02.000 --> 00:36:08.000
will give real and two
solutions.
00:36:04.000 --> 00:36:10.000
In other words,
the program is exactly like
00:36:08.000 --> 00:36:14.000
what we did for second-order
differential equations.
00:36:12.000 --> 00:36:18.000
We used the complex numbers,
got complex solutions.
00:36:16.000 --> 00:36:22.000
And then, at the very last
step, we took the real and
00:36:20.000 --> 00:36:26.000
imaginary parts to get two real
solutions out of each complex
00:36:25.000 --> 00:36:31.000
number.
I would like to give you a
00:36:27.000 --> 00:36:33.000
simple example of working that
out.
00:36:30.000 --> 00:36:36.000
And it is the system x prime
equals x plus 2y.
00:36:35.000 --> 00:36:41.000
And y prime equals minus x
00:36:39.000 --> 00:36:45.000
minus y.
Because it is springtime,
00:36:45.000 --> 00:36:51.000
it doesn't feel like spring but
it will this weekend as it is
00:36:51.000 --> 00:36:57.000
getting warmer.
And since, when I am too tired
00:36:56.000 --> 00:37:02.000
to make up problem sets for you
late at night,
00:37:01.000 --> 00:37:07.000
I watch reruns of Seinfeld.
I am from New York.
00:37:06.000 --> 00:37:12.000
It is just in my bloodstream.
Of course, the most interesting
00:37:12.000 --> 00:37:18.000
character on Seinfeld is George.
We are going to consider Susan
00:37:19.000 --> 00:37:25.000
who is the girlfriend who got
killed by licking poison
00:37:24.000 --> 00:37:30.000
envelopes.
And George carried on their
00:37:28.000 --> 00:37:34.000
love affair until Susan was
disposed of by the writers by
00:37:33.000 --> 00:37:39.000
this strange death.
And we are going to consider x
00:37:39.000 --> 00:37:45.000
is modeling Susan's love for
George.
00:37:43.000 --> 00:37:49.000
That is x.
And George's love for Susan
00:37:47.000 --> 00:37:53.000
will be y.
Now, I don't mean the absolute
00:37:51.000 --> 00:37:57.000
love.
If x and y are zero,
00:37:53.000 --> 00:37:59.000
I don't mean that they don't
love each other.
00:37:58.000 --> 00:38:04.000
I just mean that that is the
equilibrium value of the love.
00:38:05.000 --> 00:38:11.000
Everything else is measured as
departures from that.
00:38:10.000 --> 00:38:16.000
So (0, 0) represents the normal
amount of love,
00:38:15.000 --> 00:38:21.000
if love is measured.
I don't know what love units
00:38:20.000 --> 00:38:26.000
are.
Hearts, I guess.
00:38:22.000 --> 00:38:28.000
Six hearts, let's say.
Now, in what sense does this
00:38:27.000 --> 00:38:33.000
model it?
This is a normal equation and
00:38:32.000 --> 00:38:38.000
this is a neurotic equation.
That is why this is George and
00:38:36.000 --> 00:38:42.000
this is Susan who seemed very
normal to me.
00:38:39.000 --> 00:38:45.000
Susan is a normal person.
When y is positive that means
00:38:43.000 --> 00:38:49.000
that George seems to be loving
her more today than yesterday,
00:38:47.000 --> 00:38:53.000
and her natural response is to
be more in love with him.
00:38:51.000 --> 00:38:57.000
That is what most people are.
If y is negative,
00:38:54.000 --> 00:39:00.000
hey, what's the matter with
George?
00:38:57.000 --> 00:39:03.000
He doesn't feel so good.
Maybe there is something wrong
00:39:02.000 --> 00:39:08.000
with him.
She gets a little mad at him
00:39:06.000 --> 00:39:12.000
and this goes down.
x prime is negative.
00:39:10.000 --> 00:39:16.000
And the same way why is this
positive?
00:39:13.000 --> 00:39:19.000
Well, again,
it is a psychological thing,
00:39:17.000 --> 00:39:23.000
but all the world loves a
lover.
00:39:20.000 --> 00:39:26.000
When Susan is in love,
as she feels x is high,
00:39:24.000 --> 00:39:30.000
that makes her feel good.
And she loves everything,
00:39:28.000 --> 00:39:34.000
in fact.
Not just George.
00:39:32.000 --> 00:39:38.000
It is one of those things.
You all know what I am talking
00:39:38.000 --> 00:39:44.000
about.
Now, George,
00:39:40.000 --> 00:39:46.000
of course, is what makes the
writers happy.
00:39:44.000 --> 00:39:50.000
George is neurotic and,
therefore, is exactly the
00:39:49.000 --> 00:39:55.000
opposite.
He sees one day that he feels
00:39:53.000 --> 00:39:59.000
more in love with Susan than he
was yesterday.
00:40:00.000 --> 00:40:06.000
Does this make him happy?
Not at all.
00:40:02.000 --> 00:40:08.000
Not at all.
It makes y prime more negative.
00:40:05.000 --> 00:40:11.000
Why?
Because all he can think of is,
00:40:08.000 --> 00:40:14.000
my God, suppose I am really in
love with this girl?
00:40:11.000 --> 00:40:17.000
Suppose I marry her.
Oh, my God, 40 years of seeing
00:40:15.000 --> 00:40:21.000
the same person at breakfast all
the time.
00:40:18.000 --> 00:40:24.000
I must be crazy.
And so it goes down.
00:40:20.000 --> 00:40:26.000
Here is our neurotic model.
The question for differential
00:40:24.000 --> 00:40:30.000
equations is,
what do the solutions to that
00:40:27.000 --> 00:40:33.000
look like?
In other words,
00:40:31.000 --> 00:40:37.000
how does, in fact,
their love affair go?
00:40:34.000 --> 00:40:40.000
Now, there is a reason why the
writers picked that model,
00:40:39.000 --> 00:40:45.000
as you will see.
It means they were able to get
00:40:44.000 --> 00:40:50.000
a year's worth of episodes out
of it.
00:40:47.000 --> 00:40:53.000
And why is that so?
Well, let's solve it.
00:40:50.000 --> 00:40:56.000
The characteristic equation is
lambda squared.
00:40:54.000 --> 00:41:00.000
The matrix that governs this
system is A equals (1,
00:40:59.000 --> 00:41:05.000
2; negative 1,
negative 1)
00:41:02.000 --> 00:41:08.000
The trace of that matrix,
00:41:06.000 --> 00:41:12.000
the sum of the diagonal
elements is zero.
00:41:10.000 --> 00:41:16.000
There is the zero lambda here.
The determinant,
00:41:13.000 --> 00:41:19.000
which is the constant term,
is negative 1,
00:41:16.000 --> 00:41:22.000
minus negative 2,
which is plus 1.
00:41:19.000 --> 00:41:25.000
So the characteristic equation,
by calculating the trace and
00:41:23.000 --> 00:41:29.000
determinant is lambda squared
plus 1 equals 0.
00:41:28.000 --> 00:41:34.000
The eigenvalues are plus and
00:41:32.000 --> 00:41:38.000
minus i.
Now, you don't have to pick
00:41:35.000 --> 00:41:41.000
both of them because the
negative one lead to essentially
00:41:40.000 --> 00:41:46.000
the same solutions but with
negative signs.
00:41:44.000 --> 00:41:50.000
Either one will do just as we
solved second order equations.
00:41:49.000 --> 00:41:55.000
The system for finding the
eigenvectors,
00:41:53.000 --> 00:41:59.000
well, we are going to have to
accept the complex eigenvector.
00:42:00.000 --> 00:42:06.000
What is the system going to be?
Well, I take the matrix and I
00:42:05.000 --> 00:42:11.000
subtract i.
We will use i.
00:42:07.000 --> 00:42:13.000
Subtract i from the main
diagonal.
00:42:10.000 --> 00:42:16.000
So the system is (1 minus i)
times a1 plus 2a2 is zero.
00:42:15.000 --> 00:42:21.000
And let's, for good measure,
00:42:19.000 --> 00:42:25.000
write the other one down,
too.
00:42:22.000 --> 00:42:28.000
It is negative a1 plus minus (1
minus i) times a2.
00:42:28.000 --> 00:42:34.000
Then what is the solution?
00:42:33.000 --> 00:42:39.000
Well, you get the solution the
usual way.
00:42:39.000 --> 00:42:45.000
Let's take a1 equal to 1.
00:42:44.000 --> 00:42:50.000
Then what is a2?
a2 is 1 minus i divided by 2
00:42:50.000 --> 00:42:56.000
from the first equation.
00:42:58.000 --> 00:43:04.000
So the complex solution is 1
minus i over 2 times
00:43:02.000 --> 00:43:08.000
e to the it.
Now you have to take the real
00:43:05.000 --> 00:43:11.000
and imaginary parts of that.
This is the only part which
00:43:09.000 --> 00:43:15.000
technically I would not trust
you to do without having someone
00:43:13.000 --> 00:43:19.000
show you how to do it.
What do you do?
00:43:15.000 --> 00:43:21.000
Well, of course,
you know how to separate the
00:43:18.000 --> 00:43:24.000
real and imaginary parts of
that.
00:43:20.000 --> 00:43:26.000
It is the first thing is to
separate the vectors.
00:43:25.000 --> 00:43:31.000
I don't know how to explain
this.
00:43:29.000 --> 00:43:35.000
Just watch.
The real part of it is 1,
00:43:33.000 --> 00:43:39.000
one-half.
It should be negative 1,
00:43:37.000 --> 00:43:43.000
so minus this plus that because
I didn't put that on the right
00:43:45.000 --> 00:43:51.000
side.
It is minus one-half plus i
00:43:49.000 --> 00:43:55.000
times (0, one-half).
00:43:54.000 --> 00:44:00.000
Anybody want to fight?
00:44:00.000 --> 00:44:06.000
1 plus i times 0 minus one-half
plus one-half times i.
00:44:06.000 --> 00:44:12.000
You saw how I did that?
Okay.
00:44:10.000 --> 00:44:16.000
When you do these problem you
do it the same way,
00:44:16.000 --> 00:44:22.000
but don't ask me to explain
what I just did.
00:44:21.000 --> 00:44:27.000
Here it is cosine t plus i sine
t.
00:44:30.000 --> 00:44:36.000
And so the real part will give
me one solution.
00:44:34.000 --> 00:44:40.000
The imaginary part will give me
another.
00:44:39.000 --> 00:44:45.000
Since I have a limited amount
of time, let's just calculate
00:44:45.000 --> 00:44:51.000
the real part.
What is it?
00:44:48.000 --> 00:44:54.000
Well, it is (1,
minus ½) times cosine t,
00:44:52.000 --> 00:44:58.000
i squared is negative 1,
so minus (0,
00:44:56.000 --> 00:45:02.000
one-half), the negative 1 from
the i squared,
00:45:01.000 --> 00:45:07.000
times sine t.
00:45:04.000 --> 00:45:10.000
Now, what solution is that?
00:45:10.000 --> 00:45:16.000
This is (x, y).
Take the final step.
00:45:13.000 --> 00:45:19.000
It doesn't have to look like
that.
00:45:16.000 --> 00:45:22.000
x equals cosine t.
00:45:19.000 --> 00:45:25.000
Do you see that?
x equals cosine t plus 0 times
00:45:24.000 --> 00:45:30.000
sine t.
00:45:27.000 --> 00:45:33.000
What is y?
y is minus one-half times
00:45:31.000 --> 00:45:37.000
cosine t, minus one-half times
sine t, plus sine t.
00:45:37.000 --> 00:45:43.000
Now, you may have the pleasure
00:45:42.000 --> 00:45:48.000
of showing eliminating t.
You get a quadratic polynomial
00:45:47.000 --> 00:45:53.000
in x and y equals zero.
This is an ellipse.
00:45:51.000 --> 00:45:57.000
As t varies,
you can see this repeats its
00:45:55.000 --> 00:46:01.000
values at intervals of 2pi,
this gives an ellipse.
00:46:01.000 --> 00:46:07.000
And if you want to use a little
computer program,
00:46:05.000 --> 00:46:11.000
linear phase,
this is not in the assignment,
00:46:09.000 --> 00:46:15.000
but the ellipses look like this
and go around that way.
00:46:14.000 --> 00:46:20.000
And that is the model of George
and Susan's love.
00:46:19.000 --> 00:46:25.000
x, Susan.
y, George.
00:46:21.000 --> 00:46:27.000
They go round and round in this
little love circle,
00:46:25.000 --> 00:46:31.000
and it stretches on for 26
episodes.