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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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147
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the main diagonal,
minus 2 minus lambda
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minus 2 minus lambda,
minus 2 minus lambda.
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And now that equals zero is the
characteristic equation.
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The term with the most lambdas
in it is the main diagonal.
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That is always true,
notice.
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Now, each of these I would be
happier writing lambda plus 2,
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so there would be a negative
sign, negative sign,
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negative sign.
The product of three negative
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signs is still a negative sign
because three is an odd number.
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So it is minus the principle
term.
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The product of these three is
minus lambda plus 2 cubed.
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Now, the rest of the terms are
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going to be easy.
There is another term 1 times 1
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times 1, another term 1 times 1
times 1.
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So to that I add 2,
1 and 1 for those two other
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terms.
And now I have the three going
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in this direction,
but each one of them has to be
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prefaced with a minus sign.
What does each one of them come
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to?
Well, this is minus 2 minus
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lambda when I multiply those
three numbers together.
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And so are the other guys.
This is 1 times 1 times minus 2
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minus lambda,
the same thing.
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There are three of them.
Minus because they are going
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this way, minus 3 because there
are three of them,
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and what each one of them is is
negative 2 negative lambda.
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That is equal to zero,
and that is the characteristic
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equation.
Now, it doesn't look very
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promising.
On the other hand,
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I have selected it for the
lecture.
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Simple psychology should tell
you that it is going to come out
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okay.
What I am going to do is expand
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this.
First imagine changing the
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sign.
I hate to have a minus sign in
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front of a lambda cubed,
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so let's make this plus and we
will make this minus and we will
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make this plus.
I will just change all the
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signs, which is okay since it is
an equation equals zero.
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That doesn't change its roots
any.
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And now we are going to expand
it out.
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What is this?
Lambda plus 2 cubed.
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Lambda cubed plus,
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and don't get confused because
it is this 2 that will kill you
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when you use the binomial
theorem.
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If there is 1 here everybody
knows what to do.
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If there is an A there
everybody knows what to do.
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It is when that is a number not
1 that everybody makes mistakes,
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including me.
The binomial coefficients are
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1, 3, 3, 1 because it is a
cubed, I am expanding.
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So it is lambda cubed plus 3
times lambda squared times 2.
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I won't explain to you what I
am doing.
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I will just do it and hope that
you all know what I am doing.
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Plus 3 times lambda times 2
squared plus the last term,
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which is 2 cubed.
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And now we have the other term.
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All that is plus because I
changed its sign.
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The next thing is negative 2.
And then the last thing is plus
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3 times (minus 2 minus lambda).
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Let's keep it.
So what is the actual
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characteristic equation?
Maybe I can finish it.
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I should stay over here instead
of recopying all of it.
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208
00:13:27 --> 00:13:33
Well, there is a lot more work
to do.
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00:13:29 --> 00:13:35
Let's see if we can at least
write down the equation.
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What is it?
It is lambda cubed.
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What is the lambda squared
212
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erm?
It is six and that is all there
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is.
How about the lambda term?
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Well, we have 12 lambda minus 3
lambda which makes plus 9
215
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lambda.
That looks good but constant
216
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terms have a way of screwing
everything up.
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What is the constant term?
It is A minus 2 minus 6.
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00:14:00 --> 00:14:06
Zero.
The constant term is zero.
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00:14:02 --> 00:14:08
That converts this from a hard
problem to an easy problem.
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Now it is a cinch to calculate
the stuff.
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Let's go to this board and
continue the work over here.
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223
00:14:24 --> 00:14:30
The equation is lambda cubed
plus 6 lambda squared plus 9
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lambda is zero.
225
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It is very easy to calculate
the roots of that.
226
00:14:34 --> 00:14:40
You factor it.
Lambda is a common factor.
227
00:14:37 --> 00:14:43
And what is left?
Lambda squared plus 6 lambda
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plus 9.
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That is the sort of thing you
got all the time when you were
230
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studying critical damping.
It is the square of lambda plus
231
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3. Lambda squared plus 6 lambda
232
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plus 9 equals zero.
So the eigenvalues,
233
00:14:58 --> 00:15:04
the roots are what?
Well, they are lambda equals
234
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zero from this factor and then
lambda equals minus 3.
235
00:15:07 --> 00:15:13
But what is new is that the
236
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minus 3 is a double root.
That is a double root.
237
00:15:14 --> 00:15:20
Now, that, of course,
is what is going to cause the
238
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trouble.
Because, for each one of these,
239
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I am supposed to calculate the
eigenvector and make up the
240
00:15:26 --> 00:15:32
solution.
But that assumed that I had
241
00:15:29 --> 00:15:35
three things to get three
different solutions.
242
00:15:32 --> 00:15:38
Here I have only got two
things.
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00:15:34 --> 00:15:40
It is the same trouble we ran
into when there was a repeated
244
00:15:38 --> 00:15:44
root.
We were studying second or
245
00:15:40 --> 00:15:46
third order differential
equations and the characteristic
246
00:15:43 --> 00:15:49
equation had a repeated root.
And I had to go into a song and
247
00:15:47 --> 00:15:53
dance and stand on my head and
multiply things by t and so on.
248
00:15:51 --> 00:15:57
And then talk very hard arguing
why that was a good thing to do
249
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to get the answer.
Now, I am not going to do the
250
00:15:57 --> 00:16:03
same thing here.
Instead, I am going to try to
251
00:16:02 --> 00:16:08
solve the problem instead.
Let's get two points by at
252
00:16:06 --> 00:16:12
least doing the easy part of it.
Lambda equals zero.
253
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What am I supposed to do with
254
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lambda equals zero?
I am looking for the alpha that
255
00:16:17 --> 00:16:23
goes with that.
And I find that eigenvector by
256
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solving this system of
equations.
257
00:16:24 --> 00:16:30
Let's write out what that
system of equations is.
258
00:16:29 --> 00:16:35
Well, if lambda is zero,
this isn't there.
259
00:16:32 --> 00:16:38
It is just the matrix A times
alpha equals zero.
260
00:16:38 --> 00:16:44
And the matrix A is,
261
00:16:40 --> 00:16:46
I never even wrote it anywhere.
I never wrote A.
262
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I thought I would get away
without having to do it,
263
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but you never get away with
anything.
264
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It's the principle of life.
That is A.
265
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If I subtract zero from the
main diagonal,
266
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that doesn't do a great deal to
A.
267
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And the resulting system of
equations is those same things,
268
00:17:10 --> 00:17:16
except you have the a1's there,
too.
269
00:17:13 --> 00:17:19
There is one.
a1 minus 2 a2 plus a3 equals
270
00:17:17 --> 00:17:23
zero.
I am just subtracting zero from
271
00:17:22 --> 00:17:28
the main diagonal so there is
nothing to do.
272
00:17:26 --> 00:17:32
a2 minus 2 a3 equals zero.
Now I am supposed to solve
273
00:17:33 --> 00:17:39
those.
274
00:17:35 --> 00:17:41
275
00:17:40 --> 00:17:46
Of course we could do it.
Well, how do you know how to
276
00:17:43 --> 00:17:49
solve a system of three linear
equations?
277
00:17:45 --> 00:17:51
Well, elimination.
You can always solve by
278
00:17:48 --> 00:17:54
elimination.
Now we are much more
279
00:17:50 --> 00:17:56
sophisticated than that.
You all have pocket calculators
280
00:17:54 --> 00:18:00
so you could use the inverse
matrix, right?
281
00:17:56 --> 00:18:02
No.
You cannot use the inverse
282
00:17:58 --> 00:18:04
matrix.
What will happen if you punch
283
00:18:03 --> 00:18:09
in those coefficients and then
punch in A inverse.
284
00:18:08 --> 00:18:14
What answer will it give you?
0, 0, 0.
285
00:18:12 --> 00:18:18
No, I am sorry.
It won't give you any answer.
286
00:18:16 --> 00:18:22
What will it say?
It will say I cannot calculate
287
00:18:21 --> 00:18:27
the inverse to that matrix
because the whole purpose of
288
00:18:27 --> 00:18:33
this exercise was to find a
value of lambda such that this
289
00:18:33 --> 00:18:39
system of equations is
dependent.
290
00:18:38 --> 00:18:44
The coefficient determinant is
zero and, therefore,
291
00:18:42 --> 00:18:48
the coefficient matrix does not
have an inverse matrix.
292
00:18:46 --> 00:18:52
You cannot use that method.
In other words,
293
00:18:49 --> 00:18:55
the inverse matrix will never
work in these problems because
294
00:18:54 --> 00:19:00
the system of equations you will
be trying to solve is always a
295
00:18:59 --> 00:19:05
non-independent system.
And, therefore,
296
00:19:03 --> 00:19:09
its determinant is always zero.
And, therefore,
297
00:19:06 --> 00:19:12
there is no inverse matrix
because the determinant of the
298
00:19:11 --> 00:19:17
coefficient is zero.
All you can do is use
299
00:19:14 --> 00:19:20
elimination or physical insight
and common sense.
300
00:19:18 --> 00:19:24
Now, because I teach
differential equations everybody
301
00:19:23 --> 00:19:29
assumes, mistakenly,
as I think, that I really know
302
00:19:27 --> 00:19:33
something about them.
I get now and then graduate
303
00:19:32 --> 00:19:38
students, not in mathematics,
but some obscure field of
304
00:19:36 --> 00:19:42
engineering or whatever drift
into my office and say I see you
305
00:19:40 --> 00:19:46
teach differential equations.
Do you have a minute here?
306
00:19:45 --> 00:19:51
And before I can say no they
write their differential
307
00:19:49 --> 00:19:55
equation on the board.
And almost invariably it is
308
00:19:52 --> 00:19:58
nothing I have ever seen before.
And they so look at me
309
00:19:57 --> 00:20:03
hopefully and expectantly.
So what do I ask them?
310
00:20:01 --> 00:20:07
I don't ask them what they have
tried.
311
00:20:05 --> 00:20:11
What I ask them is where did
this come from?
312
00:20:08 --> 00:20:14
What field did it come from?
Because each field has its own
313
00:20:14 --> 00:20:20
little tricks.
It gets the same differential
314
00:20:17 --> 00:20:23
equations all the time and has
its own little tricks for
315
00:20:22 --> 00:20:28
solving them.
You should do the same thing
316
00:20:26 --> 00:20:32
here.
Well, of course we can solve
317
00:20:29 --> 00:20:35
this.
And by now most of you have
318
00:20:33 --> 00:20:39
solved it just by inspection,
just by sort of psyching out
319
00:20:37 --> 00:20:43
the answer.
But a better way is to say
320
00:20:40 --> 00:20:46
look, suppose we had the
solution, what would the
321
00:20:44 --> 00:20:50
solution look like?
Well, it would look like (a1,
322
00:20:48 --> 00:20:54
a2, a3), whatever the values of
those variables were which gave
323
00:20:52 --> 00:20:58
me the solution to the equation,
times e to the 0t.
324
00:20:57 --> 00:21:03
But what is this?
e to the 0t is one
325
00:21:02 --> 00:21:08
for all time.
And, therefore,
326
00:21:04 --> 00:21:10
this is a constant solution.
What I am asking is to find a
327
00:21:09 --> 00:21:15
constant solution.
Now, can I, by inspection,
328
00:21:13 --> 00:21:19
find a constant solution to
this?
329
00:21:15 --> 00:21:21
If so it must be the one.
Well, there is an obvious
330
00:21:19 --> 00:21:25
constant solution.
All the cells have the same
331
00:21:23 --> 00:21:29
temperature.
If that is true then there is
332
00:21:26 --> 00:21:32
no reason why it should ever
change as time goes on.
333
00:21:32 --> 00:21:38
The physical problem itself
suggests what the answer must
334
00:21:35 --> 00:21:41
be.
You don't have to solve
335
00:21:37 --> 00:21:43
equations.
In other words,
336
00:21:38 --> 00:21:44
any constant like (1,
1, 1).
337
00:21:40 --> 00:21:46
Well, could it be (20,
20, 20)?
338
00:21:42 --> 00:21:48
Yeah, that is a constant
multiple of (1,
339
00:21:44 --> 00:21:50
1, 1).
That is included.
340
00:21:45 --> 00:21:51
My basic constant solution,
therefore, is simply (1,
341
00:21:48 --> 00:21:54
1, 1) times e to the 0t.
342
00:21:51 --> 00:21:57
You don't have to include e to
the 0t because it is one.
343
00:21:54 --> 00:22:00
Now, just to check,
is (1, 1, 1) a solution to
344
00:21:57 --> 00:22:03
these equations?
It certainly is.
345
00:22:01 --> 00:22:07
1 plus 1 minus 2 is zero in
every case.
346
00:22:04 --> 00:22:10
The equations are essentially
the same, except they use
347
00:22:08 --> 00:22:14
different variables.
By inspection or,
348
00:22:11 --> 00:22:17
if you like,
by elimination,
349
00:22:14 --> 00:22:20
but not by finding the inverse
matrix you solve those
350
00:22:18 --> 00:22:24
equations.
And we have our first solution.
351
00:22:21 --> 00:22:27
Now let's go onto the second
one.
352
00:22:24 --> 00:22:30
For the second one,
we are going to have to use the
353
00:22:28 --> 00:22:34
eigenvalue lambda equals
negative 3.
354
00:22:33 --> 00:22:39
And now what is the system of
equations?
355
00:22:36 --> 00:22:42
Well, now I have to take this
and I have to subtract negative
356
00:22:41 --> 00:22:47
3 from the diagonal elements.
Minus 2 minus negative 3 is
357
00:22:46 --> 00:22:52
plus 1, right?
358
00:22:49 --> 00:22:55
Got that?
Each of the diagonal elements,
359
00:22:53 --> 00:22:59
after I subtract minus 3 turns
into plus 1.
360
00:22:58 --> 00:23:04
And, therefore,
the system becomes,
361
00:23:00 --> 00:23:06
the system I have to solve is
a1 plus a2 plus a3 equals zero.
362
00:23:05 --> 00:23:11
And what is the second
363
00:23:09 --> 00:23:15
equation?
Symmetry is preserved.
364
00:23:11 --> 00:23:17
All the equations are
essentially the same,
365
00:23:14 --> 00:23:20
except for the names of the
variables so they all must give
366
00:23:19 --> 00:23:25
you the same thing after I
subtract minus 3 from the main
367
00:23:24 --> 00:23:30
diagonal.
Well, that is what we call a
368
00:23:27 --> 00:23:33
dependent system of equations.
All I have is the same equation
369
00:23:34 --> 00:23:40
repeated twice,
but I still have to solve it.
370
00:23:38 --> 00:23:44
Now, what you see is that there
are lots of solutions to this.
371
00:23:44 --> 00:23:50
Let me write down one of them.
For example,
372
00:23:49 --> 00:23:55
suppose I made a1 equal to 1
and I made a2 a 0,
373
00:23:54 --> 00:24:00
then a3 would be negative 1.
374
00:24:00 --> 00:24:06
So here is a solution.
That is the eigenvector.
375
00:24:03 --> 00:24:09
And with it,
I can make the solution by
376
00:24:06 --> 00:24:12
multiplying by e to the negative
3t.
377
00:24:10 --> 00:24:16
There is a solution.
But that is not the only alpha
378
00:24:14 --> 00:24:20
I could have chosen.
Suppose I chose this one
379
00:24:18 --> 00:24:24
instead.
Suppose I kept this 1,
380
00:24:20 --> 00:24:26
but this time made a3 zero.
Well, in that case,
381
00:24:24 --> 00:24:30
there would be a2 that had to
be minus 1.
382
00:24:30 --> 00:24:36
Now, is this essentially
different from that one?
383
00:24:32 --> 00:24:38
It would still be multiplied by
e to the minus 3t,
384
00:24:36 --> 00:24:42
but don't be fooled by the e to
the minus 3t.
385
00:24:39 --> 00:24:45
That is our scalar.
That is not what is essential.
386
00:24:42 --> 00:24:48
What is essential is the
content of these two vectors.
387
00:24:45 --> 00:24:51
Is either one a multiple of the
other?
388
00:24:47 --> 00:24:53
The answer is no.
Therefore, they are
389
00:24:49 --> 00:24:55
independent.
They are pointing in two
390
00:24:51 --> 00:24:57
different directions in three
space, these two vectors.
391
00:24:56 --> 00:25:02
And, therefore,
I have two independent
392
00:24:59 --> 00:25:05
solutions just by picking two
different vectors that solve
393
00:25:03 --> 00:25:09
those three equations.
This is also a solution.
394
00:25:07 --> 00:25:13
If I call this the eigenvector
alpha 1, then I ought to call
395
00:25:12 --> 00:25:18
this one the alpha 2.
Hey, we can keep on going
396
00:25:16 --> 00:25:22
through this.
Why not make the first one
397
00:25:19 --> 00:25:25
zero?
Well, what would happen if I
398
00:25:21 --> 00:25:27
made the first one 0,
and then 1, and minus 1?
399
00:25:25 --> 00:25:31
The answer is this one is no
longer independent of those two.
400
00:25:32 --> 00:25:38
I can get it by taking a
combination of those two.
401
00:25:34 --> 00:25:40
Do you see what combination I
should take?
402
00:25:36 --> 00:25:42
403
00:25:42 --> 00:25:48
This one minus that one.
This guy minus that guy gives
404
00:25:47 --> 00:25:53
me that guy, isn't that right?
1 minus 1, 0 minus minus 1,
405
00:25:52 --> 00:25:58
minus 1 minus 0.
This is not a new one.
406
00:25:56 --> 00:26:02
It looks new,
but it is not.
407
00:26:00 --> 00:26:06
I can get it by taking a linear
combination of these two.
408
00:26:04 --> 00:26:10
It is not independent delta.
And that would be true for any
409
00:26:08 --> 00:26:14
other possible solution you
could get for these equations.
410
00:26:12 --> 00:26:18
Once you found two solutions,
all the others will be linear
411
00:26:16 --> 00:26:22
combinations of them.
Well, I cannot use that one.
412
00:26:20 --> 00:26:26
It is not new.
And the general solution,
413
00:26:23 --> 00:26:29
therefore, will be a
combination, c1 times that one
414
00:26:26 --> 00:26:32
plus a constant times this one.
Plus the first one that I found
415
00:26:32 --> 00:26:38
c3 times (1, 1,
1) e to the 0t,
416
00:26:36 --> 00:26:42
which I don't have to write in.
That is the general solution to
417
00:26:42 --> 00:26:48
the system, (x1,x2, x3).
418
00:26:45 --> 00:26:51
What happens as time goes to
infinity?
419
00:26:48 --> 00:26:54
Regardless of what the values
of these two C's this term goes
420
00:26:54 --> 00:27:00
to zero, that term goes to zero
and what I am left with is a
421
00:27:00 --> 00:27:06
constant solution.
So all of these solutions tend
422
00:27:06 --> 00:27:12
to be the solution where all the
cells are at the same
423
00:27:11 --> 00:27:17
temperature.
Well, of course there must be
424
00:27:15 --> 00:27:21
some vocabulary word in this.
There is.
425
00:27:19 --> 00:27:25
There are two vocabulary words.
This is a good eigenvalue.
426
00:27:25 --> 00:27:31
There are also bad eigenvalues.
This is a good repeated
427
00:27:32 --> 00:27:38
eigenvalue, but good is not the
official word.
428
00:27:37 --> 00:27:43
An eigenvalue like this,
which is repeated but where you
429
00:27:44 --> 00:27:50
can find enough eigenvectors,
if lambda is a repeated
430
00:27:50 --> 00:27:56
eigenvalue, it occurs multiply
in the characteristic polynomial
431
00:27:57 --> 00:28:03
as a root.
But you can find enough
432
00:28:03 --> 00:28:09
independent eigenvectors --
Forget the "but."
433
00:28:10 --> 00:28:16
434
00:28:30 --> 00:28:36
-- to make up the needed number
of independent solutions.
435
00:28:36 --> 00:28:42
For example,
if it is repeated once,
436
00:28:40 --> 00:28:46
that is it occurs doubly then
somehow I have got to get two
437
00:28:46 --> 00:28:52
solutions out of that as I was
able to here.
438
00:28:51 --> 00:28:57
If it occurred triply,
I have got to get three
439
00:28:56 --> 00:29:02
solutions out of it.
I would look for three
440
00:29:02 --> 00:29:08
independent eigenvectors and
hope I could find them.
441
00:29:06 --> 00:29:12
That is the good case because
it tells you how to make up as
442
00:29:12 --> 00:29:18
many solutions as you need.
And this kind of eigenvalue is
443
00:29:17 --> 00:29:23
called in the literature the
complete eigenvalue.
444
00:29:23 --> 00:29:29
445
00:29:30 --> 00:29:36
Now, how about the kind in
which you cannot?
446
00:29:33 --> 00:29:39
Well, unfortunately,
all my life I have called it
447
00:29:37 --> 00:29:43
incomplete, which seems to be a
perfectly reasonable thing to
448
00:29:43 --> 00:29:49
call it.
However, terminology changes
449
00:29:46 --> 00:29:52
slowly over time.
The notes, because I wrote
450
00:29:50 --> 00:29:56
them, call it an incomplete
eigenvalue.
451
00:29:53 --> 00:29:59
But the accepted term nowadays
is defective.
452
00:29:57 --> 00:30:03
I don't like that.
It violates the "eigenvalues
453
00:30:02 --> 00:30:08
with disabilities act" or
something.
454
00:30:06 --> 00:30:12
But I have to give it to you
because that is the word I am
455
00:30:11 --> 00:30:17
going to try to use from now on,
at least if I remember to use
456
00:30:18 --> 00:30:24
it.
It would be the word,
457
00:30:20 --> 00:30:26
for example,
used in the linear algebra
458
00:30:24 --> 00:30:30
course 18.06 "plug,
plug," defective otherwise.
459
00:30:30 --> 00:30:36
A defective eigenvalue is one
where you can get one
460
00:30:33 --> 00:30:39
eigenvector.
If it is double,
461
00:30:34 --> 00:30:40
for example,
if it a double eigenvalue.
462
00:30:37 --> 00:30:43
It is defective if you can get
one eigenvector that goes with
463
00:30:41 --> 00:30:47
it, but you cannot find an
independent one.
464
00:30:43 --> 00:30:49
The only other ones you can
find are multiples of the first
465
00:30:47 --> 00:30:53
one.
Then you are really in trouble
466
00:30:49 --> 00:30:55
because you just don't have
enough solutions that you are
467
00:30:53 --> 00:30:59
supposed to get out of that,
and you have to do something.
468
00:30:58 --> 00:31:04
What you do is turn to problem
two on your problem set and
469
00:31:01 --> 00:31:07
solve it because that tells you
what to do.
470
00:31:04 --> 00:31:10
And I even give you an example
to work.
471
00:31:06 --> 00:31:12
Problem two,
that little matrix has a
472
00:31:09 --> 00:31:15
defective eigenvalue.
It doesn't look defective,
473
00:31:12 --> 00:31:18
but you cannot tell.
It is defective.
474
00:31:14 --> 00:31:20
But you, nonetheless,
will be able to find two
475
00:31:17 --> 00:31:23
solutions because you will be
following instructions.
476
00:31:22 --> 00:31:28
477
00:31:32 --> 00:31:38
Now, the only other thing I
should tell you is one of the
478
00:31:35 --> 00:31:41
most important theorems in
linear algebra,
479
00:31:37 --> 00:31:43
which is totally beyond the
scope of this course and is
480
00:31:41 --> 00:31:47
beyond the scope of most
elementary linear algebra
481
00:31:44 --> 00:31:50
courses as I have taught around
the country but,
482
00:31:47 --> 00:31:53
of course, not at MIT.
But, nonetheless,
483
00:31:49 --> 00:31:55
it is the last theorem in the
course.
484
00:31:51 --> 00:31:57
That means it is liable to use
stuff.
485
00:31:53 --> 00:31:59
The theorem goes by different
names.
486
00:31:55 --> 00:32:01
Sometimes it is called the
principle axis theorem.
487
00:32:00 --> 00:32:06
Sometimes it is called the
spectral theorem.
488
00:32:04 --> 00:32:10
But, anyway,
what it says is,
489
00:32:06 --> 00:32:12
if A is a real end-by-end
matrix which is symmetric,
490
00:32:11 --> 00:32:17
you know what a symmetric
matrix is?
491
00:32:15 --> 00:32:21
The formal definition is it is
equal to its transpose.
492
00:32:20 --> 00:32:26
What that means is if you flip
it around the main diagonal it
493
00:32:25 --> 00:32:31
looks just the same as before.
Somewhere on this board,
494
00:32:32 --> 00:32:38
right there,
in fact, is a symmetric matrix.
495
00:32:36 --> 00:32:42
What happened to it?
Right here was the symmetric
496
00:32:40 --> 00:32:46
matrix.
I erased the one thing which I
497
00:32:44 --> 00:32:50
had to have.
Minus 2, 1, 1;
498
00:32:46 --> 00:32:52
1, minus 2, 1;
1, 1 minus 2.
499
00:32:48 --> 00:32:54
500
00:32:51 --> 00:32:57
That was our matrix A.
The matrix is symmetric because
501
00:32:57 --> 00:33:03
if I flip it around the diagonal
it looks the same as it did
502
00:33:02 --> 00:33:08
before.
Well, not exactly.
503
00:33:06 --> 00:33:12
The ones are sort of lying on
their side, but you have to take
504
00:33:10 --> 00:33:16
account of that.
Is that right?
505
00:33:12 --> 00:33:18
The twos are backward.
Well, you know what I mean.
506
00:33:15 --> 00:33:21
Put that element there,
this one here,
507
00:33:18 --> 00:33:24
that one there.
Exchange these two.
508
00:33:20 --> 00:33:26
Notice the diagonal elements
don't all have to be minus 2 for
509
00:33:24 --> 00:33:30
that.
No matter what they were,
510
00:33:26 --> 00:33:32
they are the guys that aren't
moved when you do the flipping.
511
00:33:32 --> 00:33:38
Therefore, there is no
condition on them.
512
00:33:34 --> 00:33:40
It is these other guys.
Each guy here has to use the
513
00:33:38 --> 00:33:44
same guy there.
This one has to be the same as
514
00:33:42 --> 00:33:48
that one, and so on.
Then it will be real and
515
00:33:45 --> 00:33:51
symmetric.
If you have a matrix that is
516
00:33:48 --> 00:33:54
real and symmetric,
like the one we have been
517
00:33:51 --> 00:33:57
working with,
the theorem is that all its
518
00:33:54 --> 00:34:00
eigenvalues are complete.
That is a very unobvious
519
00:33:58 --> 00:34:04
theorem.
All its eigenvalues are
520
00:34:02 --> 00:34:08
automatically complete.
And it is a remarkable fact
521
00:34:07 --> 00:34:13
that you can prove that purely
generally with a certain amount
522
00:34:14 --> 00:34:20
of pure reasoning no calculation
at all.
523
00:34:18 --> 00:34:24
But it has to be true,
and it is true.
524
00:34:22 --> 00:34:28
You will find there are whole
branches of applied differential
525
00:34:28 --> 00:34:34
equations.
You know, equilibrium theory,
526
00:34:33 --> 00:34:39
all the matrices that you deal
with are always symmetric.
527
00:34:37 --> 00:34:43
And, therefore,
this repeated eigenvalues is
528
00:34:40 --> 00:34:46
not something you have to worry
about, finding extra solutions.
529
00:34:45 --> 00:34:51
Well, I guess that is the end
of the first part of the
530
00:34:49 --> 00:34:55
lecture.
I have a third of it left.
531
00:34:52 --> 00:34:58
Let's talk fast.
I would like to,
532
00:34:55 --> 00:35:01
with the remaining time,
explain to you what to do if
533
00:34:59 --> 00:35:05
you were to get complex
eigenvalues.
534
00:35:03 --> 00:35:09
535
00:35:08 --> 00:35:14
Now, actually,
the answer is follow the same
536
00:35:11 --> 00:35:17
program.
In other words,
537
00:35:12 --> 00:35:18
if you solve the characteristic
equation and you get a complex
538
00:35:17 --> 00:35:23
root, follow the program,
calculate the corresponding
539
00:35:20 --> 00:35:26
complex eigenvectors.
In other words,
540
00:35:23 --> 00:35:29
solve the equations.
Everything will be the same
541
00:35:26 --> 00:35:32
except that the eigenvectors
will turn out to be complex,
542
00:35:30 --> 00:35:36
that is will have complex
entries.
543
00:35:34 --> 00:35:40
Don't worry about it.
Then form the solutions.
544
00:35:37 --> 00:35:43
The solutions are now going to
look once again like alpha times
545
00:35:43 --> 00:35:49
e to the a plus bi to the t.
546
00:35:47 --> 00:35:53
This will be complex and that
will be complex,
547
00:35:51 --> 00:35:57
too.
This will have complex entries.
548
00:35:54 --> 00:36:00
And then, finally,
take the real and imaginary
549
00:35:58 --> 00:36:04
parts.
Those will be real and they
550
00:36:02 --> 00:36:08
will give real and two
solutions.
551
00:36:04 --> 00:36:10
In other words,
the program is exactly like
552
00:36:08 --> 00:36:14
what we did for second-order
differential equations.
553
00:36:12 --> 00:36:18
We used the complex numbers,
got complex solutions.
554
00:36:16 --> 00:36:22
And then, at the very last
step, we took the real and
555
00:36:20 --> 00:36:26
imaginary parts to get two real
solutions out of each complex
556
00:36:25 --> 00:36:31
number.
I would like to give you a
557
00:36:27 --> 00:36:33
simple example of working that
out.
558
00:36:30 --> 00:36:36
And it is the system x prime
equals x plus 2y.
559
00:36:35 --> 00:36:41
And y prime equals minus x
560
00:36:39 --> 00:36:45
minus y.
Because it is springtime,
561
00:36:45 --> 00:36:51
it doesn't feel like spring but
it will this weekend as it is
562
00:36:51 --> 00:36:57
getting warmer.
And since, when I am too tired
563
00:36:56 --> 00:37:02
to make up problem sets for you
late at night,
564
00:37:01 --> 00:37:07
I watch reruns of Seinfeld.
I am from New York.
565
00:37:06 --> 00:37:12
It is just in my bloodstream.
Of course, the most interesting
566
00:37:12 --> 00:37:18
character on Seinfeld is George.
We are going to consider Susan
567
00:37:19 --> 00:37:25
who is the girlfriend who got
killed by licking poison
568
00:37:24 --> 00:37:30
envelopes.
And George carried on their
569
00:37:28 --> 00:37:34
love affair until Susan was
disposed of by the writers by
570
00:37:33 --> 00:37:39
this strange death.
And we are going to consider x
571
00:37:39 --> 00:37:45
is modeling Susan's love for
George.
572
00:37:43 --> 00:37:49
That is x.
And George's love for Susan
573
00:37:47 --> 00:37:53
will be y.
Now, I don't mean the absolute
574
00:37:51 --> 00:37:57
love.
If x and y are zero,
575
00:37:53 --> 00:37:59
I don't mean that they don't
love each other.
576
00:37:58 --> 00:38:04
I just mean that that is the
equilibrium value of the love.
577
00:38:05 --> 00:38:11
Everything else is measured as
departures from that.
578
00:38:10 --> 00:38:16
So (0, 0) represents the normal
amount of love,
579
00:38:15 --> 00:38:21
if love is measured.
I don't know what love units
580
00:38:20 --> 00:38:26
are.
Hearts, I guess.
581
00:38:22 --> 00:38:28
Six hearts, let's say.
Now, in what sense does this
582
00:38:27 --> 00:38:33
model it?
This is a normal equation and
583
00:38:32 --> 00:38:38
this is a neurotic equation.
That is why this is George and
584
00:38:36 --> 00:38:42
this is Susan who seemed very
normal to me.
585
00:38:39 --> 00:38:45
Susan is a normal person.
When y is positive that means
586
00:38:43 --> 00:38:49
that George seems to be loving
her more today than yesterday,
587
00:38:47 --> 00:38:53
and her natural response is to
be more in love with him.
588
00:38:51 --> 00:38:57
That is what most people are.
If y is negative,
589
00:38:54 --> 00:39:00
hey, what's the matter with
George?
590
00:38:57 --> 00:39:03
He doesn't feel so good.
Maybe there is something wrong
591
00:39:02 --> 00:39:08
with him.
She gets a little mad at him
592
00:39:06 --> 00:39:12
and this goes down.
x prime is negative.
593
00:39:10 --> 00:39:16
And the same way why is this
positive?
594
00:39:13 --> 00:39:19
Well, again,
it is a psychological thing,
595
00:39:17 --> 00:39:23
but all the world loves a
lover.
596
00:39:20 --> 00:39:26
When Susan is in love,
as she feels x is high,
597
00:39:24 --> 00:39:30
that makes her feel good.
And she loves everything,
598
00:39:28 --> 00:39:34
in fact.
Not just George.
599
00:39:32 --> 00:39:38
It is one of those things.
You all know what I am talking
600
00:39:38 --> 00:39:44
about.
Now, George,
601
00:39:40 --> 00:39:46
of course, is what makes the
writers happy.
602
00:39:44 --> 00:39:50
George is neurotic and,
therefore, is exactly the
603
00:39:49 --> 00:39:55
opposite.
He sees one day that he feels
604
00:39:53 --> 00:39:59
more in love with Susan than he
was yesterday.
605
00:40:00 --> 00:40:06
Does this make him happy?
Not at all.
606
00:40:02 --> 00:40:08
Not at all.
It makes y prime more negative.
607
00:40:05 --> 00:40:11
Why?
Because all he can think of is,
608
00:40:08 --> 00:40:14
my God, suppose I am really in
love with this girl?
609
00:40:11 --> 00:40:17
Suppose I marry her.
Oh, my God, 40 years of seeing
610
00:40:15 --> 00:40:21
the same person at breakfast all
the time.
611
00:40:18 --> 00:40:24
I must be crazy.
And so it goes down.
612
00:40:20 --> 00:40:26
Here is our neurotic model.
The question for differential
613
00:40:24 --> 00:40:30
equations is,
what do the solutions to that
614
00:40:27 --> 00:40:33
look like?
In other words,
615
00:40:31 --> 00:40:37
how does, in fact,
their love affair go?
616
00:40:34 --> 00:40:40
Now, there is a reason why the
writers picked that model,
617
00:40:39 --> 00:40:45
as you will see.
It means they were able to get
618
00:40:44 --> 00:40:50
a year's worth of episodes out
of it.
619
00:40:47 --> 00:40:53
And why is that so?
Well, let's solve it.
620
00:40:50 --> 00:40:56
The characteristic equation is
lambda squared.
621
00:40:54 --> 00:41:00
The matrix that governs this
system is A equals (1,
622
00:40:59 --> 00:41:05
2; negative 1,
negative 1)
623
00:41:02 --> 00:41:08
The trace of that matrix,
624
00:41:06 --> 00:41:12
the sum of the diagonal
elements is zero.
625
00:41:10 --> 00:41:16
There is the zero lambda here.
The determinant,
626
00:41:13 --> 00:41:19
which is the constant term,
is negative 1,
627
00:41:16 --> 00:41:22
minus negative 2,
which is plus 1.
628
00:41:19 --> 00:41:25
So the characteristic equation,
by calculating the trace and
629
00:41:23 --> 00:41:29
determinant is lambda squared
plus 1 equals 0.
630
00:41:28 --> 00:41:34
The eigenvalues are plus and
631
00:41:32 --> 00:41:38
minus i.
Now, you don't have to pick
632
00:41:35 --> 00:41:41
both of them because the
negative one lead to essentially
633
00:41:40 --> 00:41:46
the same solutions but with
negative signs.
634
00:41:44 --> 00:41:50
Either one will do just as we
solved second order equations.
635
00:41:49 --> 00:41:55
The system for finding the
eigenvectors,
636
00:41:53 --> 00:41:59
well, we are going to have to
accept the complex eigenvector.
637
00:42:00 --> 00:42:06
What is the system going to be?
Well, I take the matrix and I
638
00:42:05 --> 00:42:11
subtract i.
We will use i.
639
00:42:07 --> 00:42:13
Subtract i from the main
diagonal.
640
00:42:10 --> 00:42:16
So the system is (1 minus i)
times a1 plus 2a2 is zero.
641
00:42:15 --> 00:42:21
And let's, for good measure,
642
00:42:19 --> 00:42:25
write the other one down,
too.
643
00:42:22 --> 00:42:28
It is negative a1 plus minus (1
minus i) times a2.
644
00:42:28 --> 00:42:34
Then what is the solution?
645
00:42:33 --> 00:42:39
Well, you get the solution the
usual way.
646
00:42:39 --> 00:42:45
Let's take a1 equal to 1.
647
00:42:44 --> 00:42:50
Then what is a2?
a2 is 1 minus i divided by 2
648
00:42:50 --> 00:42:56
from the first equation.
649
00:42:58 --> 00:43:04
So the complex solution is 1
minus i over 2 times
650
00:43:02 --> 00:43:08
e to the it.
Now you have to take the real
651
00:43:05 --> 00:43:11
and imaginary parts of that.
This is the only part which
652
00:43:09 --> 00:43:15
technically I would not trust
you to do without having someone
653
00:43:13 --> 00:43:19
show you how to do it.
What do you do?
654
00:43:15 --> 00:43:21
Well, of course,
you know how to separate the
655
00:43:18 --> 00:43:24
real and imaginary parts of
that.
656
00:43:20 --> 00:43:26
It is the first thing is to
separate the vectors.
657
00:43:25 --> 00:43:31
I don't know how to explain
this.
658
00:43:29 --> 00:43:35
Just watch.
The real part of it is 1,
659
00:43:33 --> 00:43:39
one-half.
It should be negative 1,
660
00:43:37 --> 00:43:43
so minus this plus that because
I didn't put that on the right
661
00:43:45 --> 00:43:51
side.
It is minus one-half plus i
662
00:43:49 --> 00:43:55
times (0, one-half).
663
00:43:54 --> 00:44:00
Anybody want to fight?
664
00:44:00 --> 00:44:06
1 plus i times 0 minus one-half
plus one-half times i.
665
00:44:06 --> 00:44:12
You saw how I did that?
Okay.
666
00:44:10 --> 00:44:16
When you do these problem you
do it the same way,
667
00:44:16 --> 00:44:22
but don't ask me to explain
what I just did.
668
00:44:21 --> 00:44:27
Here it is cosine t plus i sine
t.
669
00:44:30 --> 00:44:36
And so the real part will give
me one solution.
670
00:44:34 --> 00:44:40
The imaginary part will give me
another.
671
00:44:39 --> 00:44:45
Since I have a limited amount
of time, let's just calculate
672
00:44:45 --> 00:44:51
the real part.
What is it?
673
00:44:48 --> 00:44:54
Well, it is (1,
minus ½) times cosine t,
674
00:44:52 --> 00:44:58
i squared is negative 1,
so minus (0,
675
00:44:56 --> 00:45:02
one-half), the negative 1 from
the i squared,
676
00:45:01 --> 00:45:07
times sine t.
677
00:45:04 --> 00:45:10
Now, what solution is that?
678
00:45:10 --> 00:45:16
This is (x, y).
Take the final step.
679
00:45:13 --> 00:45:19
It doesn't have to look like
that.
680
00:45:16 --> 00:45:22
x equals cosine t.
681
00:45:19 --> 00:45:25
Do you see that?
x equals cosine t plus 0 times
682
00:45:24 --> 00:45:30
sine t.
683
00:45:27 --> 00:45:33
What is y?
y is minus one-half times
684
00:45:31 --> 00:45:37
cosine t, minus one-half times
sine t, plus sine t.
685
00:45:37 --> 00:45:43
Now, you may have the pleasure
686
00:45:42 --> 00:45:48
of showing eliminating t.
You get a quadratic polynomial
687
00:45:47 --> 00:45:53
in x and y equals zero.
This is an ellipse.
688
00:45:51 --> 00:45:57
As t varies,
you can see this repeats its
689
00:45:55 --> 00:46:01
values at intervals of 2pi,
this gives an ellipse.
690
00:46:01 --> 00:46:07
And if you want to use a little
computer program,
691
00:46:05 --> 00:46:11
linear phase,
this is not in the assignment,
692
00:46:09 --> 00:46:15
but the ellipses look like this
and go around that way.
693
00:46:14 --> 00:46:20
And that is the model of George
and Susan's love.
694
00:46:19 --> 00:46:25
x, Susan.
y, George.
695
00:46:21 --> 00:46:27
They go round and round in this
little love circle,
696
00:46:25 --> 00:46:31
and it stretches on for 26
episodes.