The last time I spent solving a system of equations dealing with
the chilling of this hardboiled egg being put in an ice bath.
We called T1 the temperature of the yoke and T2 the temperature
of the white. What I am going to do is
revisit that same system of equations, but basically the
topic for today is to learn to solve that system of equations
by a completely different method.
It is the method that is normally used in practice.
Elimination is used mostly by people who have forgotten how to
do it any other way. Now, in order to make it a
little more general, I am not going to use the
dependent variables T1 and T2 because they suggest temperature
a little too closely. Let's change them to neutral
variables. I will use x equals T1,
and for T2 I will just use y.
I am not going to re-derive anything.
I am not going to resolve anything.
I am not going to repeat anything of what I did last
time, except to write down to remind you what the system was
in terms of these variables, the system we derived using the
particular conductivity constants, two and three,
respectively. The system was this one,
minus 2x plus 2y. And the y prime was
2x minus 5y. And so we solved this by
elimination. We got a single second-order
equation with constant coefficients,
which we solved in the usual way.
From that I derived what the x was, from that we derived what
the y was, and then I put them all together.
I will just remind you what the final solution was when written
out in terms of arbitrary constants.
It was c1 times e to the negative t plus c2 e to the
negative 6t, and y was c1 over 2 e
to the negative t minus 2c2 e to the negative 6t.
That was the solution we got.
And then I went on to put in initial conditions,
but we are not going to explore that aspect of it today.
We will in a week or so. This was the general solution
because it had two arbitrary constants in it.
What I want to do now is revisit this and do it by a
different method, which makes heavy use of
matrices. That is a prerequisite for this
course, so I am assuming that you reviewed a little bit about
matrices. And it is in your book.
Your book puts in a nice little review section.
Two-by-two and three-by-three will be good enough for 18.03
mostly because I don't want you to calculate all night on bigger
matrices, bigger systems. So nothing serious,
matrix multiplication, solving systems of linear
equations, end-by-end systems. I will remind you at the
appropriate places today of what it is you need to remember.
The very first thing we are going to do is,
let's see. I haven't figured out the color
coding for this lecture yet, but let's make this system in
green and the solution can be in purple.
Invisible purple, but I have a lot of it.
Let's abbreviate, first of all,
the system using matrices. I am going to make a column
vector out of (x, y).
Then you differentiate a column vector by differentiating each
component. I can write the left-hand side
of the system as (x, y) prime.
How about the right-hand side?
Well, I say I can just write the matrix of coefficients to
negative 2, 2, 2, negative 5 times x,y.
And I say that this matrix equation says exactly the same
thing as that green equation and, therefore,
it is legitimate to put it up in green, too.
The top here is x prime. What is the top here?
After I multiply these two I get a column vector.
And what is its top entry? It is negative 2x plus 2y.
There it is.
And the bottom entry the same way is 2x minus 5y,
just as it is down there. Now, what I want to do is,
well, maybe I should translate the solution.
What does the solution look like?
We got that, too. How am I going to write this as a matrix equation? Actually, if I told you to use matrices, use vectors,
the point at which you might be most hesitant is this one right here, the very next step. Because how you should write it
is extremely well-concealed in this notation. But the point is, this is a column vector and I
am adding together two column vectors. And what is in each one of the column vectors?
Think of these two things as a column vector.
Pull out all the scalars from them that you can.
Well, you see that c1 is a common factor of both entries
and so is e to the negative t, that function.
Now, if I pull both of those out of the vector,
what is left of the vector? Well, you cannot even see it.
What is left is a 1 up here and a one-half there.
So I am going to write that in the following form.
I will put out the c1, it's the common factor in both,
and put that out front. Then I will put in the guts of
the vector, even though you cannot see it,
the column vector 1, one-half.
And then I will put the other
scalar function in back. The only reason for putting one
of these in front and one in back is visual so to make it
easy to read. There is no other reason.
You could put the c1 here, you could put it here,
you could put the e negative t in front if you want
to, but people will fire you. Don't do that.
Write it the standard way because that is the way that it
is easiest to read. The constants out front,
the functions behind, and the column vector of
numbers in the middle. And so the other one will be
written how? Well, here, that one is a
little more transparent. c2, 1, 2 and the other thing is
e to the negative 6t.
There is our solution. That is going to need a lot of
purple, but I have it.
And now I want to talk about how the new method of solving
the equation. It is based just on the same
idea as the way we solve second-order equations.
Yes, question.
Oh, here. Sorry. This should be negative two. Thanks very much.
What I am going to use is a trial solution.
Remember when we had a second-order equation with
constant coefficients the very first thing I did was I said we
are going to try a solution of the form e to the rt.
Why that? Well, because Oiler thought of
it and it has been known for or 300 years that that is the
thing you should do. Well, this has not been known
nearly as long because matrices were only invented around
or so, and people did not really use them to solve systems of
differential equations until the middle of the last century,
1950-1960. If you look at books written in
1950, they won't even talk about systems of differential
equations, or talk very little anyway and they won't solve them
using matrices. This is only 50 years old.
I mean, my God, in mathematics that is very up
to date, particularly elementary mathematics.
Anyway, the method of solving is going to use as a trial
solution. Now, if you were left to your
own devices you might say, well, let's try x equals some
constant times e to the lambda1 t and y
equals some other constant times e to the lambda2 t.
Now, if you try that,
it is a sensible thing to try, but it will turn out not to
work. And that is the reason I have
written out this particular solution, so we can see what
solutions look like. The essential point is here is
the basic solution I am trying to find.
Here is another one. Their form is a column vector
of constants. But they both use the same
exponential factor, which is the point.
In other words, I should not use here,
in my trial solution, two different lambdas,
I should use the same lambda. And so the way to write the
trial solution is (x, y) equals two unknown numbers,
that or that or whatever, times e to a single unknown
exponent factor. Let's call it lambda t.
It is called lambda. It is called r.
It is called m. I have never seen it called
anything but one of those three things.
I am using lambda. Your book uses lambda.
It is a common choice. Let's stick with it.
Now what is the next step? Well, we plug into the system.
Substitute into the system. What are we going to get?
Well, let's do it. First of all,
I have to differentiate. The left-hand side asks me to
differentiate this. How do I differentiate this?
Column vector times a function. Well, the column vector acts as
a constant. And I differentiate that.
That is lambda e to the lambda t.
So the (x, y) prime is (a1, a2) times e to the lambda t
times lambda.
Now, it is ugly to put the
lambda afterwards because it is a number so you should put it in
front, again, to make things easier to read.
But this lambda comes from differentiating e to the lambda
t and using the chain rule.
This much is the left-hand side.
That is the derivative (x, y) prime.
I differentiate the x and I differentiated the y.
How about the right-hand side. Well, the right-hand side is
negative 2, 2, 2, negative 5 times what?
Well, times (x,
y), which is (a1, a2) e to the lambda t.
Now, the same thing that
happened a month or a month and a half ago happens now.
The whole point of making that substitution is that the e to
the lambda t, the function part of it drops
out completely. And one is left with what?
An algebraic equation to be solved for lambda a1 and a2.
In other words, by means of that substitution,
and it basically uses the fact that the coefficients are
constant, what you have done is reduced the problem of calculus,
of solving differential equations, to solving algebraic
equations. In some sense that is the only
method there is, unless you do numerical stuff.
You reduce the calculus to algebra.
The Laplace transform is exactly the same thing.
All the work is algebra. You turn the original
differential equation into an algebraic equation for Y of s,
you solve it, and then you use more algebra
to find out what the original little y of t was.
It is not different here. So let's solve this system of
equations. Now, the whole problem with
solving this system, first of all,
what is the system? Let's write it out explicitly.
Well, it is really two equations, isn't it?
The first one says lambda a1 is equal to negative 2 a1
plus 2 a2. That is the first one.
The other one says lambda a2 is equal to 2 a1 minus 5 a2.
Now, purely, if you want to classify that,
that is two equations and three variables, three unknowns.
The a1, a2, and lambda are all unknown.
And, unfortunately, if you want to classify them
correctly, they are nonlinear equations because they are made
nonlinear by the fact that you have multiplied two of the
variables. Well, if you sit down and try
to hack away at solving those without a plan,
you are not going to get anywhere.
It is going to be a mess. Also, two equations and three
unknowns is indeterminate. You can solve three equations
and three unknowns and get a definite answer,
but two equations and three unknowns usually have an
infinity of solutions. Well, at this point it is the
only idea that is required. Well, this was a little idea,
but I assume one would think of that.
And the idea that is required here is, I think,
not so unnatural, it is not to view these a1,
a2, and lambda as equal. Not all variables are created
equal. Some are more equal than
others. a1 and a2 are definitely equal
to each other, and let's relegate lambda to
the background. In other words,
I am going to think of lambda as just a parameter.
I am going to demote it from the status of variable to
parameter. If I demoted it further it
would just be an unknown constant.
That is as bad as you can be. I am going to focus my
attention on the a1, a2 and sort of view the lambda
as a nuisance. Now, as soon as I do that,
I see that these equations are linear if I just look at them as
equations in a1 and a2. And moreover,
they are not just linear, they are homogenous.
Because if I think of lambda just as a parameter,
I should rewrite the equations this way.
I am going to subtract this and move the left-hand side to the
right side, and it is going to look like (minus 2 minus lambda)
times a1 plus 2 a2 is equal to zero.
And the same way for the other
one. It is going to be 2a1 plus,
what is the coefficient, (minus 5 minus lambda) a2
equals zero.
That is a pair of simultaneous linear equations for determining
a1 and a2, and the coefficients involved are parameter lambda.
Now, what is the point of doing that?
Well, now the point is whatever you learned about linear
equations, you should have learned the most fundamental
theorem of linear equations. The main theorem is that you
have a square system of homogeneous equations,
this is a two-by-two system so it is square,
it always has the trivial solution, of course,
a1, a2 equals zero. Now, we don't want that trivial
solution because if a1 and a2 are zero, then so are x and y
zero. Now that is a solution.
Unfortunately, it is of no interest.
If the solution were x, y zero, it corresponds to the
fact that this is an ice bath. The yoke is at zero,
the white is at zero and it stays that way for all time
until the ice melts. So that is the solution we
don't want. We don't want the trivial
solution. Well, when does it have a
nontrivial solution? Nontrivial means non-zero,
in other words. If and only if this determinant
is zero.
In other words, by using that theorem on linear
equations, what we find is there is a condition that lambda must
satisfy, an equation in lambda in order that we would be able
to find non-zero values for a1 and a2.
Let's write it out. I will recopy it over here.
What was it? Negative 2 minus lambda,
two, here it was 2 and minus 5 minus lambda.
All right. You have to expand the
determinant. In other words,
we are trying to find out for what values of lambda is this
determinant zero. Those will be the good values
which lead to nontrivial solutions for the a's.
This is the equation lambda plus 2.
See, this is minus that and minus that, the product of the
two minus ones is plus one. So it is lambda plus 2 times
lambda plus 5,
which is the product of the two diagonal elements,
minus the product of the two anti-diagonal elements,
which is 4, is equal to zero. And if I write that out,
what is that, that is the equation lambda
squared plus 7 lambda,
5 lambda plus 2 lambda, and then the constant term is
10 minus 4 which is 6. How many of you have long
enough memories, two-day memories that you
remember that equation? When I did the method of
elimination, it led to exactly the same equation except it had
r's in it instead of lambda. And this equation,
therefore, is given the same name and another color.
Let's make it salmon.
And it is called the characteristic equation for this
method. All right.
Now I am going to use now the word from last time.
You factor this. From the factorization we get
its root easily enough. The roots are lambda equals
negative 1 and lambda equals negative 6
by factoring the equation.
Now what I am supposed to do? You have to keep the different
parts of the method together. Now I have found the only
values of lambda for which I will be able to find nonzero
values for the a1 and a2. For each of those values of
lambda, I now have to find the corresponding a1 and a2.
Let's do them one at a time. Let's take first lambda equals
negative one. My problem is now to find a1
and a2. Where am I going to find them
from? Well, from that system of
equations over there. I will recopy it over here.
What is the system? The hardest part of this is
dealing with multiple minus signs, but you had experience
with that in determinants so you know all about that.
In other words, there is the system of
equations over there. Let's recopy them here.
Minus 2, minus minus 1 makes minus 1.
What's the other coefficient? It is just plain old 2.
Good. There is my first equation.
And when I substitute lambda equals negative one
for the second equation, what do you get?
2 a1 plus negative 5 minus negative 1 makes negative 4.
There is my system that will find me a1 and a2.
What is the first thing you notice about it?
You immediately notice that this system is fake because this
second equation is twice the first one.
Something is wrong. No, something is right.
If that did not happen, if the second equation were not
a constant multiple of the first one then the only solution of
the system would be a1 equals zero, a2 equals zero because the
determinant of the coefficients would not be zero.
The whole function of this exercise was to find the value
of lambda, negative 1, for which the system would be
redundant and, therefore, would have a
nontrivial solution. Do you get that?
In other words, calculate the system out,
just as I have done here, you have an automatic check on
the method. If one equation is not a
constant multiple of the other you made a mistake.
You don't have the right value of lambda or you substituted
into the system wrong, which is frankly a more common
error. Go back, recheck first the
substitution, and if convinced that is right
then recheck where you got lambda from.
But here everything is going fine so we can now find out what
the value of a1 and a2 are. You don't have to go through a
big song and dance for this since most of the time you will
have two-by-two equations and now and then three-by-three.
For two-by-two all you do is, since we really have the same
equation twice, to get a solution I can assign
one of the variables any value and then simply solve for the
other. The natural thing to do is to
make a2 equal one, then I won't need fractions and
then a1 will be a2. So the solution is (2,
1). I am only trying to find one
solution. Any constant multiple of this
would also be a solution, as long as it wasn't zero,
zero which is the trivial one. And, therefore,
this is a solution to this system of algebraic equations.
And the solution to the whole system of differential equations
is, this is only the (a1, a2) part.
I have to add to it, as a factor,
lambda is negative, therefore, e to the minus t.
There is our purple thing.
See how I got it? Starting with the trial
solution, I first found out through this procedure what the
lambda's have to be. Then I took the lambda and
found what the corresponding a1 and a2 that went with it and
then made up my solution out of that.
Now, quickly I will do the same thing for lambda
equals negative 6. Each one of these must be
treated separately. They are separate problems and
you are looking for separate solutions.
Lambda equals negative 6. What do I do?
How do my equations look now? Well, the first one is minus 2
minus negative 6 makes plus 4. It is 4a1 plus 2a2 equals zero.
Then I hold my breath while I
calculate the second one to see if it comes out to be a constant
multiple. I get 2a1 plus negative 5 minus
negative 6, which makes plus 1. And, indeed,
one is a constant multiple of the other.
I really only have on equation there.
I will just write down immediately now what the
solution is to the system. Well, the (a1,
a2) will be what? Now, it is more natural to make
a1 equal 1 and then solve to get an integer for a2.
If a1 is 1, then a2 is negative 2.
And I should multiply that by e to the negative 6t
because negative 6 is the corresponding value.
There is my other one. And now there is a
superposition principle, which if I get a chance will
prove for you at the end of the hour.
If not, you will have to do it yourself for homework.
Since this is a linear system of equations,
once you have two separate solutions, neither a constant
multiple of the other, you can multiply each one of
these by a constant and it will still be a solution.
You can add them together and that will still be a solution,
and that gives the general solution.
The general solution is the sum of these two,
an arbitrary constant. I am going to change the name
since I don't want to confuse it with the c1 I used before,
times the first solution which is (2, 1) e to the negative t
plus c2, another arbitrary constant, times 1 negative 2 e
to the minus 6t.
Now you notice that is exactly
the same solution I got before. The only difference is that I
have renamed the arbitrary constants.
The relationship between them, c1 over 2,
I am now calling c1 tilda, and c2 I am calling c2 tilda.
If you have an arbitrary constant, it doesn't matter
whether you divide it by two. It is still just an arbitrary a
constant. It covers all values,
in other words. Well, I think you will agree
that is a different procedure, yet it has only one
coincidence. It is like elimination goes
this way and comes to the answer.
And this method goes a completely different route and
comes to the answer, except it is not quite like
that. They walk like this and then
they come within viewing distance of each other to check
that both are using the same characteristic equation,
and then they again go their separate ways and end up with
the same answer.
There is something special of these values.
You cannot get away from those two values of lambda.
Somehow they are really intrinsically connected.
Occurs the exponential coefficient, and they are
intrinsically connected with the problem of the egg that we
started with. Now what I would like to do is
very quickly sketch how this method looks when I remove all
the numbers from it. In some sense,
it becomes a little clearer what is going on.
And that will give me a chance to introduce the terminology
that you need when you talk about it.
Well, you have notes. Let me try to write it down in
general.
I will first write it out two-by-two.
I am just going to sketch. The system looks like (x,
y) equals, I will still put it up in colors.
Except now, instead of using twos and fives,
I will use (a, b; c, d).
The trial solution will look how?
The trial is going to be (a1, a2).
That I don't have to change the name of.
I am going to substitute in, and what the result of
substitution is going to be lambda (a1, a2).
I am going to skip a step and pretend that the e to the lambda
t's have already been canceled out.
Is equal to (a, b; c, d) times (a1,
a2). What does that correspond to?
That corresponds to the system as I wrote it here.
And then we wrote it out in terms of two equations.
And what was the resulting thing that we ended up with?
Well, you write it out, you move the lambda to the
other side. And then the homogeneous system
is we will look in general how? Well, we could write it out.
It is going to look like a minus lambda,
b, c, d minus lambda.
That is just how it looks there
and the general calculation is the same.
Times (a1, a2) is equal to zero.
This is solvable nontrivially. In other words,
it has a nontrivial solution if an only if the determinant of
coefficients is zero.
Let's now write that out, calculate out once and for all
what that determinant is. I will write it out here.
It is a minus lambda times d minus lambda,
the product of the diagonal elements, minus the
anti-diagonal minus bc is equal to zero.
And let's calculate that out.
It is lambda squared minus a lambda minus d lambda plus ad,
the constant term from here, negative bc from there,
plus ad minus bc,
where have I seen that before? This equation is the general
form using letters of what we calculated using the specific
numbers before. Again, I will code it the same
way with that color salmon. Now, most of the calculations
will be for two-by-two systems. I advise you,
in the strongest possible terms, to remember this
equation. You could write down this
equation immediately for the matrix.
You don't have to go through all this stuff.
For God's sakes, don't say let the trial
solution be blah, blah, blah.
You don't want to do that. I don't want you to repeat the
derivation of this every time you go through a particular
problem. It is just like in solving
second order equations. You have a second order
equation. You immediately write down its
characteristic equation, then you factor it,
you find its roots and you construct the solution.
It takes a minute. The same thing,
this takes a minute, too.
What is the constant term? Ad minus bc,
what is that? Matrix is (a,
b; c, d). Ad minus bc is its determinant.
This is the determinant of that matrix.
I didn't give the matrix a name, did I?
I will now give the matrix a name A.
What is this? Well, you are not supposed to
know that until now. I will tell you.
This is called the trace of A. Put that down in your little
books. The abbreviation is trace A,
and the word is trace. The trace of a square matrix is
the sum of the d elements down its main diagonal.
If it were a three-by-three there would be three terms in
whatever you are up to. Here it is a plus b,
the sum of the diagonal elements.
You can immediately write down this characteristic equation.
Let's give it a name. This is a characteristic
equation of what? Of the matrix,
now. Not of the system,
of the matrix. You have a two-by-two matrix.
You could immediately write down its characteristic
equation. Watch out for this sign,
minus. That is a very common error to
leave out the minus sign because that is the way the formula
comes out. Its roots.
If it is a quadratic equation it will have roots;
lambda1, lambda2 for the moment let's assume are real and
distinct.
For the enrichment of your vocabulary, those are called the
eigenvalues.
They are something which belonged to the matrix A.
They are two secret numbers. You can calculate from the
coefficients a, b, and c, and d,
but they are not in the coefficients.
You cannot look at a matrix and see what its eigenvalues are.
You have to calculate something.
But they are the most important numbers in the matrix.
They are hidden, but they are the things that
control how this system behaves. Those are called the
eigenvalues. Now, there are various purists,
there are a fair number of them in the world who do not like
this word because it begins German and ends English.
Eigenvalues were first introduced by a German
mathematician, you know, around the time
matrices came into being in or so.
A little while after eigenvalues came into being,
too. And since all this happened in
Germany they were named eigenvalues in German,
which begins eigen and ends value.
But people who do not like that call them the characteristic
values. Unfortunately,
it is two words and takes a lot more space to write out.
An older generation even calls them something different,
which you are not so likely to see nowadays,
but you will in slightly older books.
You can also call them the proper values.
Characteristic is not a translation of eigen,
but proper is, but it means it in a funny
sense which has almost disappeared nowadays.
It means proper in the sense of belong to.
The only example I can think of is the word property.
Property is something that belongs to you.
That is the use of the word proper.
It is something that belongs to the matrix.
The matrix has its proper values.
It does not mean proper in the sense of fitting and proper or I
hope you will behave properly when we go to Aunt Agatha's or
something like that. But, as I say,
by far the most popular thing, slowly the word eigenvalue is
pretty much taking over the literature.
Just because it's just one word, that is a tremendous
advantage. Okay.
What now is still to be done? Well, there are those vectors
to be found. So the very last step would be
to solve the system to find the vectors a1 and a2.
For each (lambda)i, find the associated vector.
The vector, we will call it (alpha)i.
That is the a1 and a2. Of course it's going to be
indexed. You have to put another
subscript on it because there are two of them.
And a1 and a2 is stretched a little too far.
By solving the system, and the system will be the
system which I will write this way, (a minus lambda,
b, c, d minus lambda).
It is just that system that was
over there, but I will recopy it, (a1, a2) equals zero,
zero. And these are called the
eigenvectors. Each of these is called the
eigenvector associated with or belonging to,
again, in that sense of property.
Eigenvector, let's say belonging to,
I see that a little more frequently, belonging to lambda
i. So we have the eigenvalues,
the eigenvectors and, of course, the people who call
them characteristic values also call these guys characteristic
vectors. I don't think I have ever seen
proper vectors, but that is because I am not
old enough. I think that is what they used
to be called a long time ago, but not anymore.
And then, finally, the general solution will be,
by the superposition principle, (x, y) equals the arbitrary
constant times the first eigenvector times the eigenvalue
times the e to the corresponding eigenvalue.
And then the same thing for the second one, (a1,
a2), but now the second index will be 2 to indicate that it
goes with the eigenvalue e to the lambda 2t.
I have done that twice. And now in the remaining five
minutes I will do it a third time because it is possible to
write this in still a more condensed form.
And the advantage of the more condensed form is A,
it takes only that much space to write, and B,
it applies to systems, not just the two-by-two
systems, but to end-by-end systems.
The method is exactly the same. Let's write it out as it would
apply to end-by-end systems. The vector I started with is
(x, y) and so on, but I will simply abbreviate
this, as is done in 18.02, by x with an arrow over it.
The matrix A I will abbreviate with A, as I did before with
capital A. And then the system looks like
x prime is equal to --
x prime is what? Ax.
That is all there is to it.
There is our green system. Now notice in this form I did
not even tell you whether this a two-by-two matrix or an
end-by-end. And in this condensed form it
will look the same no matter how many equations you have.
Your book deals from the beginning with end-by-end
systems. That is, in my view,
one of its weaknesses because I don't think most students start
with two-by-two. Fortunately,
the book double-talks. The theory is end-by-end,
but all the examples are two-by-two.
So just read the examples. Read the notes instead,
which just do two-by-two to start out with.
The trial solution is x equals what?
An unknown vector alpha times e to the lambda t.
Alpha is what we called a1 and
a2 before. Plug this into there and cancel
the e to the lambda t's.
What do you get? Well, this is lambda alpha e to
the lambda t equals A alpha e to the lambda t.
These two cancel. And the system to be solved,
A alpha equals lambda alpha.
And now the question is how do you solve that system?
Well, you can tell if a book is written by a scoundrel or not by
how they go -- A book, which is in my opinion
completely scoundrel, simply says you subtract one
from the other, and without further ado writes
A minus lambda, and they tuck a little I in
there and write alpha equals zero.
Why is the I put in there? Well, this is what you would
like to write. What is wrong with this
equation? This is not a valid matrix
equation because that is a square end-by-end matrix,
a square two-by-two matrix if you like.
This is a scalar. You cannot subtract the scalar
from a matrix. It is not an operation.
To subtract matrices they have to be the same size,
the same shape. What is done is you make this a
two-by-two matrix. This is a two-by-two matrix
with lambdas down the main diagonal and I elsewhere.
And the justification is that lambda alpha is the same thing
as the lambda I times alpha because I is an identity matrix.
Now, in fact, jumping from here to here is
not something that would occur to anybody.
The way it should occur to you to do this is you do this,
you write that, you realize it doesn't work,
and then you say to yourself I don't understand what these
matrices are all about. I think I'd better write it all
out. And then you would write it all
out and you would write that equation on the left-hand board
there. Oh, now I see what it should
look like. I should subtract lambda from
the main diagonal. That is the way it will come
out. And then say,
hey, the way to save lambda from the main diagonal is put it
in an identity matrix. That will do it for me.
In other words, there is a little detour that
goes from here to here. And one of the ways I judge
books is by how well they explain the passage from this to
that. If they don't explain it at all
and just write it down, they have never talked to
students. They have just written books.
Where did we get finally here? The characteristic equation
from that, I had forgotten what color.
That is in salmon. The characteristic equation,
then, is going to be the thing which says that the determinant
of that is zero. That is the circumstances under
which it is solvable. In general, this is the way the
characteristic equation looks. And its roots,
once again, are the eigenvalues.
And from then you calculate the corresponding eigenvectors.
Okay. Go home and practice.
In recitation you will practice on both two-by-two and
three-by-three cases, and we will talk more next
time.