I just want to remind you of
the main facts.
The first thing that you have
to do is, of course,
we are going to have to be
doing it several times today.
That is the system we are
trying to solve.
And the first thing you have to
do is find a characteristic
equation which is general form,
although this is not the form
you should use for two-by-two,
is A minus lambda I
equals zero.
And its roots are the
eigenvalues.
And then with each eigenvalue
you then have to calculate its
eigenvector, which you do by
solving the system (A minus
lambda1, let's say,
times I) alpha equals zero
because
the solution is the eigenvector
alpha 1.
And then the final solution
that you make out of the two of
them looks like alpha 1 times e
to the lambda 1t.
Of course you do that for each
eigenvalue.
You get the associated
eigenvector.
And then the general solution
is made up out of a linear
combination of these individual
guys with constant coefficients.
The lecture today is devoted to
the two cases where things do
not go as smoothly as they seem
to in the homework problems you
have been doing up until now.
The first one will take
probably most of the period.
It deals with what happens when
an eigenvalue gets repeated.
But I think since the situation
is a little more complicated
than it is where the case of a
characteristic root gets
repeated in the case of a
second-order equation as we saw
it, you know what to do in that
case, here there are different
possibilities.
And I thought the best thing to
do would be to illustrate them
on an example.
So here is a problem.
It came out of a mild
nightmare, but I won't bore you
with the details.
Anyway, we have this circular
fish tank.
It is a very modern fish tank.
It is divided into three
compartments because one holds
Siamese fighting fish and one
goldfish, and one-- They should
not eat each other.
And it is going to be a simple
temperature problem.
The three actual compartments
have to be kept at different
temperatures because one is for
tropical fish and one is for
arctic fish and one is for
everyday garden variety fish.
But the guy forgets to turn on
the heater so the temperatures
start out what they are supposed
to be, tropical,
icy, and normal.
But as the day wears on,
of course, the three
compartments trade their heat
and sort of tend to all end up
at the same temperature.
So we are going to let (x)i
equal the temperature in tank i.
Now, these are separated from
each other by glass things.
Everything is identical,
each has the same volume,
and the same glass partition
separates them out and no heat
can escape.
This is very well-insulated
with very double-thick
Thermopane glass or something
like that.
You can see in,
but heat cannot get out very
well.
Heat essentially is conducted
from one of these cells to the
other.
And let's assume that the water
in each tank is kept stirred up
because the fish are swimming
around in it.
That should be a pretty decent
way of stirring a fish tank.
The question is how do each of
these, as a function of time,
and I want to know how they
behave over time,
so find these functions.
Well, we are going to find them
in solutions to differential
equations.
And the differential equations
are not hard to set up.
They are very much like the
diffusion equation you had for
homework or the equations we
studied in the beginning of the
term.
Let's do one carefully because
the others go exactly the same
way.
What determines the flow,
the change in temperature?
Well, it is the conductivity
across the barriers.
But there are two barriers
because heat can flow into this
first cell, both from this guy
and it can flow across this
glass pane from the other cell.
We have to take account of both
of those possibilities.
It is like in your homework.
The little diffusion cell that
was in the middle could get
contributions from both sides,
whereas, the two guys on the
end could only get contribution
from one.
But here, nobody is on the end.
It is circular table.
Everyone is dying equally.
Everybody can get input from
the other two cells.
x1 prime is some
constant of conductivity times
the temperature difference
between tank three and tank one.
And then there is another term
which comes from tank two.
So a times tank two minus the
temperature difference,
tank two minus tank one.
Let's write this out.
Remember there will be other
equations, too.
But instead of doing this,
let's do a more careful job
with this first equation.
When I write it out,
remember, the important thing
is you are going to have x1,
x2, x3 down the left,
so they have to occur in the
same order on the right in order
to use these standard eigenvalue
techniques.
The coefficient of x1 is going
to be minus a x1 and then
another minus a x1.
In other words,
it is going to be minus 2 ax1.
And then the x2 term will be
plus a x2.
And the x3 term will be plus a
x3.
Well, you can see now that is
the equation for x1 prime in
terms of the other variables.
But there is symmetry.
There is no difference between
this tank, that tank,
and that tank as far as the
differential equations are
concerned.
And, therefore,
I can get the equations for the
other two tanks by just changing
1 to 2, just switching the
subscripts.
When I finally do it all,
the equations are going to be,
I will write them first out as
a system.
Let's take a equal 1
because I am going to want to
solve them numerically,
and I want you to be able to
concentrate on what is
important, what is new now and
not fuss because I don't want to
have an extra a floating around
everywhere just contributing
nothing but a mild confusion to
the proceedings.
So x1 prime,
I am going to take a equal 1
and simply write it minus 2 x1
plus x2 plus x3.
And so now what would the
equation for x2 prime
be?
Well, here x2 plays the role
that x1 played before.
And the only way to tell that
x1 was the main guy here was it
occurred with a coefficient
negative 2, whereas,
the other guys occurred with
coefficient 1.
That must be what happens here,
too.
Since x2 prime is our
main man, this is minus x2 and
this must be x1 here plus x3.
And finally the last one is no
different, x3 prime is x1 plus
x2.
And now it is the x3 that
should get negative 2 for the
coefficient.
There is a perfectly
reasonable-looking set of
equations.
Just how reasonable they are
depends upon what their
characteristic polynomial turns
out to be.
And all the work in these
problems is trying to find nice
models where you won't have to
use Matlab to calculate the
roots, the eigenvalues,
the roots of the characteristic
polynomial.
So we have to now find the
characteristic polynomial.
The matrix that we are talking
about is the matrix,
well, let's right away write A
minus lambda I
I cannot use the trace and
determinant form for this
equation because it is not a
two-by-two matrix.
It is a three-by-three matrix.
I have to use the original form
for the characteristic equation.
But what is this going to be?
Well, what is A?
A is minus 2.
I am going to leave a little
space here.
1, 1, 1 minus 2,
1.
And finally 1,
1, negative 2.
subtract lambda from
the main diagonal,
minus 2 minus lambda
minus 2 minus lambda,
minus 2 minus lambda.
And now that equals zero is the
characteristic equation.
The term with the most lambdas
in it is the main diagonal.
That is always true,
notice.
Now, each of these I would be
happier writing lambda plus 2,
so there would be a negative
sign, negative sign,
negative sign.
The product of three negative
signs is still a negative sign
because three is an odd number.
So it is minus the principle
term.
The product of these three is
minus lambda plus 2 cubed.
Now, the rest of the terms are
going to be easy.
There is another term 1 times 1
times 1, another term 1 times 1
times 1.
So to that I add 2,
1 and 1 for those two other
terms.
And now I have the three going
in this direction,
but each one of them has to be
prefaced with a minus sign.
What does each one of them come
to?
Well, this is minus 2 minus
lambda when I multiply those
three numbers together.
And so are the other guys.
This is 1 times 1 times minus 2
minus lambda,
the same thing.
There are three of them.
Minus because they are going
this way, minus 3 because there
are three of them,
and what each one of them is is
negative 2 negative lambda.
That is equal to zero,
and that is the characteristic
equation.
Now, it doesn't look very
promising.
On the other hand,
I have selected it for the
lecture.
Simple psychology should tell
you that it is going to come out
okay.
What I am going to do is expand
this.
First imagine changing the
sign.
I hate to have a minus sign in
front of a lambda cubed,
so let's make this plus and we
will make this minus and we will
make this plus.
I will just change all the
signs, which is okay since it is
an equation equals zero.
That doesn't change its roots
any.
And now we are going to expand
it out.
What is this?
Lambda plus 2 cubed.
Lambda cubed plus,
and don't get confused because
it is this 2 that will kill you
when you use the binomial
theorem.
If there is 1 here everybody
knows what to do.
If there is an A there
everybody knows what to do.
It is when that is a number not
1 that everybody makes mistakes,
including me.
The binomial coefficients are
1, 3, 3, 1 because it is a
cubed, I am expanding.
So it is lambda cubed plus 3
times lambda squared times 2.
I won't explain to you what I
am doing.
I will just do it and hope that
you all know what I am doing.
Plus 3 times lambda times 2
squared plus the last term,
which is 2 cubed.
And now we have the other term.
All that is plus because I
changed its sign.
The next thing is negative 2.
And then the last thing is plus
3 times (minus 2 minus lambda).
Let's keep it.
So what is the actual
characteristic equation?
Maybe I can finish it.
I should stay over here instead
of recopying all of it.
Well, there is a lot more work
to do.
Let's see if we can at least
write down the equation.
What is it?
It is lambda cubed.
What is the lambda squared
erm?
It is six and that is all there
is.
How about the lambda term?
Well, we have 12 lambda minus 3
lambda which makes plus 9
lambda.
That looks good but constant
terms have a way of screwing
everything up.
What is the constant term?
It is A minus 2 minus 6.
Zero.
The constant term is zero.
That converts this from a hard
problem to an easy problem.
Now it is a cinch to calculate
the stuff.
Let's go to this board and
continue the work over here.
The equation is lambda cubed
plus 6 lambda squared plus 9
lambda is zero.
It is very easy to calculate
the roots of that.
You factor it.
Lambda is a common factor.
And what is left?
Lambda squared plus 6 lambda
plus 9.
That is the sort of thing you
got all the time when you were
studying critical damping.
It is the square of lambda plus
3. Lambda squared plus 6 lambda
plus 9 equals zero.
So the eigenvalues,
the roots are what?
Well, they are lambda equals
zero from this factor and then
lambda equals minus 3.
But what is new is that the
minus 3 is a double root.
That is a double root.
Now, that, of course,
is what is going to cause the
trouble.
Because, for each one of these,
I am supposed to calculate the
eigenvector and make up the
solution.
But that assumed that I had
three things to get three
different solutions.
Here I have only got two
things.
It is the same trouble we ran
into when there was a repeated
root.
We were studying second or
third order differential
equations and the characteristic
equation had a repeated root.
And I had to go into a song and
dance and stand on my head and
multiply things by t and so on.
And then talk very hard arguing
why that was a good thing to do
to get the answer.
Now, I am not going to do the
same thing here.
Instead, I am going to try to
solve the problem instead.
Let's get two points by at
least doing the easy part of it.
Lambda equals zero.
What am I supposed to do with
lambda equals zero?
I am looking for the alpha that
goes with that.
And I find that eigenvector by
solving this system of
equations.
Let's write out what that
system of equations is.
Well, if lambda is zero,
this isn't there.
It is just the matrix A times
alpha equals zero.
And the matrix A is,
I never even wrote it anywhere.
I never wrote A.
I thought I would get away
without having to do it,
but you never get away with
anything.
It's the principle of life.
That is A.
If I subtract zero from the
main diagonal,
that doesn't do a great deal to
A.
And the resulting system of
equations is those same things,
except you have the a1's there,
too.
There is one.
a1 minus 2 a2 plus a3 equals
zero.
I am just subtracting zero from
the main diagonal so there is
nothing to do.
a2 minus 2 a3 equals zero.
Now I am supposed to solve
those.
Of course we could do it.
Well, how do you know how to
solve a system of three linear
equations?
Well, elimination.
You can always solve by
elimination.
Now we are much more
sophisticated than that.
You all have pocket calculators
so you could use the inverse
matrix, right?
No.
You cannot use the inverse
matrix.
What will happen if you punch
in those coefficients and then
punch in A inverse.
What answer will it give you?
0, 0, 0.
No, I am sorry.
It won't give you any answer.
What will it say?
It will say I cannot calculate
the inverse to that matrix
because the whole purpose of
this exercise was to find a
value of lambda such that this
system of equations is
dependent.
The coefficient determinant is
zero and, therefore,
the coefficient matrix does not
have an inverse matrix.
You cannot use that method.
In other words,
the inverse matrix will never
work in these problems because
the system of equations you will
be trying to solve is always a
non-independent system.
And, therefore,
its determinant is always zero.
And, therefore,
there is no inverse matrix
because the determinant of the
coefficient is zero.
All you can do is use
elimination or physical insight
and common sense.
Now, because I teach
differential equations everybody
assumes, mistakenly,
as I think, that I really know
something about them.
I get now and then graduate
students, not in mathematics,
but some obscure field of
engineering or whatever drift
into my office and say I see you
teach differential equations.
Do you have a minute here?
And before I can say no they
write their differential
equation on the board.
And almost invariably it is
nothing I have ever seen before.
And they so look at me
hopefully and expectantly.
So what do I ask them?
I don't ask them what they have
tried.
What I ask them is where did
this come from?
What field did it come from?
Because each field has its own
little tricks.
It gets the same differential
equations all the time and has
its own little tricks for
solving them.
You should do the same thing
here.
Well, of course we can solve
this.
And by now most of you have
solved it just by inspection,
just by sort of psyching out
the answer.
But a better way is to say
look, suppose we had the
solution, what would the
solution look like?
Well, it would look like (a1,
a2, a3), whatever the values of
those variables were which gave
me the solution to the equation,
times e to the 0t.
But what is this?
e to the 0t is one
for all time.
And, therefore,
this is a constant solution.
What I am asking is to find a
constant solution.
Now, can I, by inspection,
find a constant solution to
this?
If so it must be the one.
Well, there is an obvious
constant solution.
All the cells have the same
temperature.
If that is true then there is
no reason why it should ever
change as time goes on.
The physical problem itself
suggests what the answer must
be.
You don't have to solve
equations.
In other words,
any constant like (1,
1, 1).
Well, could it be (20,
20, 20)?
Yeah, that is a constant
multiple of (1,
1, 1).
That is included.
My basic constant solution,
therefore, is simply (1,
1, 1) times e to the 0t.
You don't have to include e to
the 0t because it is one.
Now, just to check,
is (1, 1, 1) a solution to
these equations?
It certainly is.
1 plus 1 minus 2 is zero in
every case.
The equations are essentially
the same, except they use
different variables.
By inspection or,
if you like,
by elimination,
but not by finding the inverse
matrix you solve those
equations.
And we have our first solution.
Now let's go onto the second
one.
For the second one,
we are going to have to use the
eigenvalue lambda equals
negative 3.
And now what is the system of
equations?
Well, now I have to take this
and I have to subtract negative
3 from the diagonal elements.
Minus 2 minus negative 3 is
plus 1, right?
Got that?
Each of the diagonal elements,
after I subtract minus 3 turns
into plus 1.
And, therefore,
the system becomes,
the system I have to solve is
a1 plus a2 plus a3 equals zero.
And what is the second
equation?
Symmetry is preserved.
All the equations are
essentially the same,
except for the names of the
variables so they all must give
you the same thing after I
subtract minus 3 from the main
diagonal.
Well, that is what we call a
dependent system of equations.
All I have is the same equation
repeated twice,
but I still have to solve it.
Now, what you see is that there
are lots of solutions to this.
Let me write down one of them.
For example,
suppose I made a1 equal to 1
and I made a2 a 0,
then a3 would be negative 1.
So here is a solution.
That is the eigenvector.
And with it,
I can make the solution by
multiplying by e to the negative
3t.
There is a solution.
But that is not the only alpha
I could have chosen.
Suppose I chose this one
instead.
Suppose I kept this 1,
but this time made a3 zero.
Well, in that case,
there would be a2 that had to
be minus 1.
Now, is this essentially
different from that one?
It would still be multiplied by
e to the minus 3t,
but don't be fooled by the e to
the minus 3t.
That is our scalar.
That is not what is essential.
What is essential is the
content of these two vectors.
Is either one a multiple of the
other?
The answer is no.
Therefore, they are
independent.
They are pointing in two
different directions in three
space, these two vectors.
And, therefore,
I have two independent
solutions just by picking two
different vectors that solve
those three equations.
This is also a solution.
If I call this the eigenvector
alpha 1, then I ought to call
this one the alpha 2.
Hey, we can keep on going
through this.
Why not make the first one
zero?
Well, what would happen if I
made the first one 0,
and then 1, and minus 1?
The answer is this one is no
longer independent of those two.
I can get it by taking a
combination of those two.
Do you see what combination I
should take?
This one minus that one.
This guy minus that guy gives
me that guy, isn't that right?
1 minus 1, 0 minus minus 1,
minus 1 minus 0.
This is not a new one.
It looks new,
but it is not.
I can get it by taking a linear
combination of these two.
It is not independent delta.
And that would be true for any
other possible solution you
could get for these equations.
Once you found two solutions,
all the others will be linear
combinations of them.
Well, I cannot use that one.
It is not new.
And the general solution,
therefore, will be a
combination, c1 times that one
plus a constant times this one.
Plus the first one that I found
c3 times (1, 1,
1) e to the 0t,
which I don't have to write in.
That is the general solution to
the system, (x1,x2, x3).
What happens as time goes to
infinity?
Regardless of what the values
of these two C's this term goes
to zero, that term goes to zero
and what I am left with is a
constant solution.
So all of these solutions tend
to be the solution where all the
cells are at the same
temperature.
Well, of course there must be
some vocabulary word in this.
There is.
There are two vocabulary words.
This is a good eigenvalue.
There are also bad eigenvalues.
This is a good repeated
eigenvalue, but good is not the
official word.
An eigenvalue like this,
which is repeated but where you
can find enough eigenvectors,
if lambda is a repeated
eigenvalue, it occurs multiply
in the characteristic polynomial
as a root.
But you can find enough
independent eigenvectors --
Forget the "but."
-- to make up the needed number
of independent solutions.
For example,
if it is repeated once,
that is it occurs doubly then
somehow I have got to get two
solutions out of that as I was
able to here.
If it occurred triply,
I have got to get three
solutions out of it.
I would look for three
independent eigenvectors and
hope I could find them.
That is the good case because
it tells you how to make up as
many solutions as you need.
And this kind of eigenvalue is
called in the literature the
complete eigenvalue.
Now, how about the kind in
which you cannot?
Well, unfortunately,
all my life I have called it
incomplete, which seems to be a
perfectly reasonable thing to
call it.
However, terminology changes
slowly over time.
The notes, because I wrote
them, call it an incomplete
eigenvalue.
But the accepted term nowadays
is defective.
I don't like that.
It violates the "eigenvalues
with disabilities act" or
something.
But I have to give it to you
because that is the word I am
going to try to use from now on,
at least if I remember to use
it.
It would be the word,
for example,
used in the linear algebra
course 18.06 "plug,
plug," defective otherwise.
A defective eigenvalue is one
where you can get one
eigenvector.
If it is double,
for example,
if it a double eigenvalue.
It is defective if you can get
one eigenvector that goes with
it, but you cannot find an
independent one.
The only other ones you can
find are multiples of the first
one.
Then you are really in trouble
because you just don't have
enough solutions that you are
supposed to get out of that,
and you have to do something.
What you do is turn to problem
two on your problem set and
solve it because that tells you
what to do.
And I even give you an example
to work.
Problem two,
that little matrix has a
defective eigenvalue.
It doesn't look defective,
but you cannot tell.
It is defective.
But you, nonetheless,
will be able to find two
solutions because you will be
following instructions.
Now, the only other thing I
should tell you is one of the
most important theorems in
linear algebra,
which is totally beyond the
scope of this course and is
beyond the scope of most
elementary linear algebra
courses as I have taught around
the country but,
of course, not at MIT.
But, nonetheless,
it is the last theorem in the
course.
That means it is liable to use
stuff.
The theorem goes by different
names.
Sometimes it is called the
principle axis theorem.
Sometimes it is called the
spectral theorem.
But, anyway,
what it says is,
if A is a real end-by-end
matrix which is symmetric,
you know what a symmetric
matrix is?
The formal definition is it is
equal to its transpose.
What that means is if you flip
it around the main diagonal it
looks just the same as before.
Somewhere on this board,
right there,
in fact, is a symmetric matrix.
What happened to it?
Right here was the symmetric
matrix.
I erased the one thing which I
had to have.
Minus 2, 1, 1;
1, minus 2, 1;
1, 1 minus 2.
That was our matrix A.
The matrix is symmetric because
if I flip it around the diagonal
it looks the same as it did
before.
Well, not exactly.
The ones are sort of lying on
their side, but you have to take
account of that.
Is that right?
The twos are backward.
Well, you know what I mean.
Put that element there,
this one here,
that one there.
Exchange these two.
Notice the diagonal elements
don't all have to be minus 2 for
that.
No matter what they were,
they are the guys that aren't
moved when you do the flipping.
Therefore, there is no
condition on them.
It is these other guys.
Each guy here has to use the
same guy there.
This one has to be the same as
that one, and so on.
Then it will be real and
symmetric.
If you have a matrix that is
real and symmetric,
like the one we have been
working with,
the theorem is that all its
eigenvalues are complete.
That is a very unobvious
theorem.
All its eigenvalues are
automatically complete.
And it is a remarkable fact
that you can prove that purely
generally with a certain amount
of pure reasoning no calculation
at all.
But it has to be true,
and it is true.
You will find there are whole
branches of applied differential
equations.
You know, equilibrium theory,
all the matrices that you deal
with are always symmetric.
And, therefore,
this repeated eigenvalues is
not something you have to worry
about, finding extra solutions.
Well, I guess that is the end
of the first part of the
lecture.
I have a third of it left.
Let's talk fast.
I would like to,
with the remaining time,
explain to you what to do if
you were to get complex
eigenvalues.
Now, actually,
the answer is follow the same
program.
In other words,
if you solve the characteristic
equation and you get a complex
root, follow the program,
calculate the corresponding
complex eigenvectors.
In other words,
solve the equations.
Everything will be the same
except that the eigenvectors
will turn out to be complex,
that is will have complex
entries.
Don't worry about it.
Then form the solutions.
The solutions are now going to
look once again like alpha times
e to the a plus bi to the t.
This will be complex and that
will be complex,
too.
This will have complex entries.
And then, finally,
take the real and imaginary
parts.
Those will be real and they
will give real and two
solutions.
In other words,
the program is exactly like
what we did for second-order
differential equations.
We used the complex numbers,
got complex solutions.
And then, at the very last
step, we took the real and
imaginary parts to get two real
solutions out of each complex
number.
I would like to give you a
simple example of working that
out.
And it is the system x prime
equals x plus 2y.
And y prime equals minus x
minus y.
Because it is springtime,
it doesn't feel like spring but
it will this weekend as it is
getting warmer.
And since, when I am too tired
to make up problem sets for you
late at night,
I watch reruns of Seinfeld.
I am from New York.
It is just in my bloodstream.
Of course, the most interesting
character on Seinfeld is George.
We are going to consider Susan
who is the girlfriend who got
killed by licking poison
envelopes.
And George carried on their
love affair until Susan was
disposed of by the writers by
this strange death.
And we are going to consider x
is modeling Susan's love for
George.
That is x.
And George's love for Susan
will be y.
Now, I don't mean the absolute
love.
If x and y are zero,
I don't mean that they don't
love each other.
I just mean that that is the
equilibrium value of the love.
Everything else is measured as
departures from that.
So (0, 0) represents the normal
amount of love,
if love is measured.
I don't know what love units
are.
Hearts, I guess.
Six hearts, let's say.
Now, in what sense does this
model it?
This is a normal equation and
this is a neurotic equation.
That is why this is George and
this is Susan who seemed very
normal to me.
Susan is a normal person.
When y is positive that means
that George seems to be loving
her more today than yesterday,
and her natural response is to
be more in love with him.
That is what most people are.
If y is negative,
hey, what's the matter with
George?
He doesn't feel so good.
Maybe there is something wrong
with him.
She gets a little mad at him
and this goes down.
x prime is negative.
And the same way why is this
positive?
Well, again,
it is a psychological thing,
but all the world loves a
lover.
When Susan is in love,
as she feels x is high,
that makes her feel good.
And she loves everything,
in fact.
Not just George.
It is one of those things.
You all know what I am talking
about.
Now, George,
of course, is what makes the
writers happy.
George is neurotic and,
therefore, is exactly the
opposite.
He sees one day that he feels
more in love with Susan than he
was yesterday.
Does this make him happy?
Not at all.
Not at all.
It makes y prime more negative.
Why?
Because all he can think of is,
my God, suppose I am really in
love with this girl?
Suppose I marry her.
Oh, my God, 40 years of seeing
the same person at breakfast all
the time.
I must be crazy.
And so it goes down.
Here is our neurotic model.
The question for differential
equations is,
what do the solutions to that
look like?
In other words,
how does, in fact,
their love affair go?
Now, there is a reason why the
writers picked that model,
as you will see.
It means they were able to get
a year's worth of episodes out
of it.
And why is that so?
Well, let's solve it.
The characteristic equation is
lambda squared.
The matrix that governs this
system is A equals (1,
2; negative 1,
negative 1)
The trace of that matrix,
the sum of the diagonal
elements is zero.
There is the zero lambda here.
The determinant,
which is the constant term,
is negative 1,
minus negative 2,
which is plus 1.
So the characteristic equation,
by calculating the trace and
determinant is lambda squared
plus 1 equals 0.
The eigenvalues are plus and
minus i.
Now, you don't have to pick
both of them because the
negative one lead to essentially
the same solutions but with
negative signs.
Either one will do just as we
solved second order equations.
The system for finding the
eigenvectors,
well, we are going to have to
accept the complex eigenvector.
What is the system going to be?
Well, I take the matrix and I
subtract i.
We will use i.
Subtract i from the main
diagonal.
So the system is (1 minus i)
times a1 plus 2a2 is zero.
And let's, for good measure,
write the other one down,
too.
It is negative a1 plus minus (1
minus i) times a2.
Then what is the solution?
Well, you get the solution the
usual way.
Let's take a1 equal to 1.
Then what is a2?
a2 is 1 minus i divided by 2
from the first equation.
So the complex solution is 1
minus i over 2 times
e to the it.
Now you have to take the real
and imaginary parts of that.
This is the only part which
technically I would not trust
you to do without having someone
show you how to do it.
What do you do?
Well, of course,
you know how to separate the
real and imaginary parts of
that.
It is the first thing is to
separate the vectors.
I don't know how to explain
this.
Just watch.
The real part of it is 1,
one-half.
It should be negative 1,
so minus this plus that because
I didn't put that on the right
side.
It is minus one-half plus i
times (0, one-half).
Anybody want to fight?
1 plus i times 0 minus one-half
plus one-half times i.
You saw how I did that?
Okay.
When you do these problem you
do it the same way,
but don't ask me to explain
what I just did.
Here it is cosine t plus i sine
t.
And so the real part will give
me one solution.
The imaginary part will give me
another.
Since I have a limited amount
of time, let's just calculate
the real part.
What is it?
Well, it is (1,
minus ½) times cosine t,
i squared is negative 1,
so minus (0,
one-half), the negative 1 from
the i squared,
times sine t.
Now, what solution is that?
This is (x, y).
Take the final step.
It doesn't have to look like
that.
x equals cosine t.
Do you see that?
x equals cosine t plus 0 times
sine t.
What is y?
y is minus one-half times
cosine t, minus one-half times
sine t, plus sine t.
Now, you may have the pleasure
of showing eliminating t.
You get a quadratic polynomial
in x and y equals zero.
This is an ellipse.
As t varies,
you can see this repeats its
values at intervals of 2pi,
this gives an ellipse.
And if you want to use a little
computer program,
linear phase,
this is not in the assignment,
but the ellipses look like this
and go around that way.
And that is the model of George
and Susan's love.
x, Susan.
y, George.
They go round and round in this
little love circle,
and it stretches on for 26
episodes.
