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We are going to need a few
facts about fundamental

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matrices, and I am worried that
over the weekend this spring

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activities weekend you might
have forgotten them.

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So I will just spend two or
three minutes reviewing the most

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essential things that we are
going to need later in the

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period.
What we are talking about is,

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I will try to color code things
so you will know what they are.

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First of all,
the basic problem is to solve a

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system of equations.
And I am going to make that a

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two-by-two system,
although practically everything

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I say today will also work for
end-by-end systems.

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Your book tries to do it
end-by-end, as usual,

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but I think it is easier to
learn two-by-two first and

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generalize rather than to wade
through the complications of

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end-by-end systems.
So the problem is to solve it.

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And the method I used last time
was to describe something called

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a fundamental matrix.

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A fundamental matrix for the
system or for A,

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whichever you want,
remember what that was.

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That was a two-by-two matrix of
functions of t and whose columns

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were two independent solutions,
x1, x2.

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These were two independent
solutions.

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In other words,
neither was a constant multiple

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of the other.
Now, I spent a fair amount of

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time showing you the two
essential properties that a

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fundamental matrix had.
We are going to need those

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today, so let me remind you the
basic properties of X and the

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properties by which you could
recognize one if you were given

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one.
First of all,

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the easy one,
its determinant shall not be

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zero, is not zero for any t,
for any value of the variable.

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That simply expresses the fact
that its two columns are

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independent, linearly
independent, not a multiple of

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each other.
The other one was more bizarre,

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so I tried to call a little
more attention to it.

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It was that the matrix
satisfies a differential

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equation of its own,
which looks almost the same,

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except it's a matrix
differential equation.

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It is not our column vectors
which are solutions but matrices

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as a whole which are solutions.
In other words,

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if you take that matrix and
differentiate every entry,

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what you get is the same as A
multiplied by that matrix you

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started with.
This, remember,

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expressed the fact,
it was just really formal when

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you analyzed what it was,
but it expressed the fact that

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it says that the columns solved
the system.

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The first thing says the
columns are independent and the

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second says each column
separately is a solution to the

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system.
That is as far,

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more or less.
Then I went in another

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direction and we talked about
variation of parameters.

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I am not going to come back to
variation of parameters today.

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We are going in a different
tack.

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And the tack we are going on is
I want to first talk a little

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more about the fundamental
matrix and then,

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as I said, we will talk about
an entirely different method of

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solving the system,
one which makes no mention of

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eigenvalues or eigenvectors,
if you can believe that.

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But, first, the one confusing
thing about the fundamental

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matrix is that it is not unique.
I have carefully tried to avoid

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talking about the fundamental
matrix because there is no "the"

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fundamental matrix,
there is only "a" fundamental

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matrix.
Why is that?

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Well, because these two columns
can be any two independent

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solutions.
And there are an infinity of

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ways of picking independent
solutions.

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That means there is an infinity
of possible fundamental

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matrices.
Well, that is disgusting,

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but can we repair it a little
bit?

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I mean maybe they are all
derivable from each other in

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some simple way.
And that is,

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of course, what is true.
Now, as a prelude to doing

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that, I would like to show you
what I wanted to show you on

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Friday but, again,
I ran out of time,

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how to write the general
solution --

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-- to the system.
The system I am talking about

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is that pink system.
Well, of course,

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the standard naďve way of doing
it is it's x equals,

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the general solution is an
arbitrary constant times that

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first solution you found,
plus c2, times another

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arbitrary constant,
times the second solution you

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found.
Okay.
Now, how would you abbreviate
that using the fundamental

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matrix?
Well, I did something very

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similar to this on Friday,
except these were called Vs.

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It was part of the variation
parameters method,

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but I promised not to use those
words today so I just said

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nothing.
Okay.

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What is the answer?
It is x equals,

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how do I write this using the
fundamental matrix,

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x1, x2?
Simple.

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It is capital X times the
column vector whose entries are

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c1 and c2.
In other words,

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it is x1, x2 times the column
vector c1, c2,

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isn't it?
Yeah.

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Because if you multiply this
think top row,

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top row, top row c1,
plus top row times c2,

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that exactly gives you the top
row here.

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And the same way the bottom row
here, times this vector,

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gives you the bottom row of
that.

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It is just another way of
writing that,

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but it looks very efficient.
Sometimes efficiency isn't a

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good thing, you have to watch
out for it, but here it is good.

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So, this is the general
solution written out using a

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fundamental matrix.
And you cannot use less symbols

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than that.
There is just no way.

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But that gives us our answer
to, what do all fundamental

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matrices look like?

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Well, they are two columns are
solutions.

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The answer is they look like --
Now, the first column is an

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arbitrary solution.
How do I write an arbitrary

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solution?
There is the general solution.

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I make it a particular one by
giving a particular value to

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that column vector of arbitrary
constants like two,

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three or minus one,
pi or something like that.

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The first guy is a solution,
and I have just shown you I can

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write such a solution like X,
c1 with a column vector,

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a particular column vector of
numbers.

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This is a solution because the
green thing says it is.

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And side by side,
we will write another one.

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And now all I have to do is,
of course, there is supposed to

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be a dependent.
We will worry about that in

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just a moment.
All I have to do is make this

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look better.
Now, I told you last time,

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by the laws of matrix
multiplication,

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if the first column is X c1 and
the second column is X c2,

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using matrix multiplication
that is the same as writing it

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this way.
This square matrix times the

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matrix whose entries are the
first column vector and the

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second column vector.
Now, I am going to call this C.

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It is a square matrix of
constants.

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It is a two-by-two matrix of
constants.

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And so, the final way of
writing it is just what

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corresponds to that,
X times C.

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And so X is a given fundamental
matrix, this one,

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that one, so the most general
fundamental matrix is then the

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one you started with,
and multiply it by an arbitrary

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square matrix of constants,
except you want to be sure that

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the determinant is not zero.
Well, the determinant of this

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guy won't be zero,
so all you have to do is make

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sure that the determinant of C
isn't zero either.

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In other words,
the fundamental matrix is not

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unique, but once you found one
all the other ones are found by

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multiplying it on the right by
an arbitrary square matrix of

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constants, which is nonsingular,
it has determinant nonzero in

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other words.
Well, that was all Friday.

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That's Friday leaking over into
Monday.

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And now we begin the true
Monday.

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Here is the problem.
Once again we have our

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two-by-two system,
or end-by-end if you want to be

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super general.
There is a system.

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What do we have so far by way
of solving it?

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Well, if your kid brother or
sister when you go home said,

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a precocious kid,
okay, tell me how to solve this

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thing, I think the only thing
you will be able to say is well,

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you do this,
you take the matrix and then

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you calculate something called
eigenvalues and eigenvectors.

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Do you know what those are?
I didn't think you did,

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blah, blah, blah,
show how smart I am.

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And you then explain what the
eigenvalues and eigenvectors

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are.
And then you show how out of

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those make up special solutions.
And then you take a combination

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of that.
In other words,

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it is algorithm.
It is something you do,

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a process, a method.
And when it is all done,

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you have the general solution.
Now, that is fine for

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calculating particular problems
with a definite model with

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definite numbers in it where you
want a definite answer.

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And, of course,
a lot of your work in

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engineering and science classes
is that kind of work.

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But the further you get on,
well, when you start reading

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books, for example,
or god forbid start reading

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papers in which people are
telling you, you know,

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they are doing engineering or
they are doing science,

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they don't want a method,
what they want is a formula.

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In other words,
the problem is to fill in the

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blank in the following.
You are writing a paper,

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and you just set up some
elaborate model and A is a

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matrix derived from that model
in some way, represents bacteria

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doing something or bank accounts
doing something,

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I don't know.
And you say,

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as is well-known,
the solution is,

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of course, you only have
letters here,

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no numbers.
This is a general paper.

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The solution is given by the
formula.

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The only trouble is,
we don't have a formula.

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All we have is a method.
Now, people don't like that.

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What I am going to produce for
you this period is a formula,

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and that formula does not
require the calculation of any

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eigenvalues, eigenvectors,
doesn't require any of that.

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It is, therefore,
a very popular way to fill in

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to finish that sentence.
Now the question is where is

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that formula going to come from?
Well, we are,

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for the moment,
clueless.

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If you are clueless the place
to look always is do I know

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anything about this sort of
thing?

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I mean is there some special
case of this problem I can solve

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or that I have solved in the
past?

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And the answer to that is yes.
You haven't solved it for a

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two-by-two matrix but you have
solved it for a one-by-one

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matrix.
A one-by-one matrix also goes

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by the name of a constant.
It is just a thing.

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It's a number.
Just putting brackets around it

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doesn't conceal the fact that it
is just a number.

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Let's look at what the solution
is for a one-by-one matrix,

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a one-by-one case.
If we are looking for a general

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solution for the end-by-end
case, it must work for the

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one-by-one case also.
That is a good reason for us

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starting.
That looks like x,

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doesn't it?
A one-by-one case.

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Well, in that case,
I am trying to solve the

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system.
The system consists of a single

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equation.
That is the way the system

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looks.
How do you solve that?

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Well, you were born knowing how
to solve that.

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Anyway, you certainly didn't
learn it in this course.

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You separate variables,
blah, blah, blah,

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and the solution is x equals,
the basic solution is e to the

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at,
and you multiply that by an

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arbitrary constant.
Now, that is a formula for the

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solution.
And it uses the parameter in

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the equation.
I didn't have to know a special

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number.
I didn't have to put a

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particular number here to use
that.

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Well, the answer is that the
same idea, whatever the answer I

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give here has got to work in
this case, too.

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But let's take a quick look as
to why this works.

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Of course, you separate
variables and use calculus.

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I am going to give you a
slightly different argument that

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has the advantage of
generalizing to the end-by-end

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case.
And the argument goes as

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follows for that.
It uses the definition of the

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exponential function not as the
inverse to the logarithm,

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which is where the fancy
calculus books get it from,

234
00:15:52,000 --> 00:15:58,000
nor as the naďve high school
method, e squared means

235
00:15:56,000 --> 00:16:02,000
you multiply e by itself and e
cubed means you do it three

236
00:16:00,000 --> 00:16:06,000
times and so on.
And e to the one-half means you

237
00:16:05,000 --> 00:16:11,000
do it a half a time or
something.

238
00:16:07,000 --> 00:16:13,000
So, the naďve definition of the
exponential function.

239
00:16:11,000 --> 00:16:17,000
Instead, I will use the
definition of the exponential

240
00:16:14,000 --> 00:16:20,000
function that comes from an
infinite series.

241
00:16:17,000 --> 00:16:23,000
Leaving out the arbitrary
constant that we don't have to

242
00:16:21,000 --> 00:16:27,000
bother with.
e to the a t is the series one

243
00:16:24,000 --> 00:16:30,000
plus at plus a squared t squared
over two factorial.

244
00:16:32,000 --> 00:16:38,000
I will put out one more term
and let's call it quits there.

245
00:16:38,000 --> 00:16:44,000
If I take this then argument
goes let's just differentiate

246
00:16:44,000 --> 00:16:50,000
it.
In other words,

247
00:16:46,000 --> 00:16:52,000
what is the derivative of e to
the at with respect to t?

248
00:16:52,000 --> 00:16:58,000
Well, just differentiating term

249
00:16:57,000 --> 00:17:03,000
by term it is zero plus the
first term is a,

250
00:17:01,000 --> 00:17:07,000
the next term is a squared
times t.

251
00:17:08,000 --> 00:17:14,000
This differentiates to t
squared over two factorial.

252
00:17:13,000 --> 00:17:19,000
And the answer is that this is

253
00:17:17,000 --> 00:17:23,000
equal to a times,
if you factor out the a,

254
00:17:21,000 --> 00:17:27,000
what is left is one plus a t
plus a squared t squared over

255
00:17:26,000 --> 00:17:32,000
two factorial

256
00:17:32,000 --> 00:17:38,000
In other words,
it is simply e to the at.

257
00:17:34,000 --> 00:17:40,000
In other words,

258
00:17:36,000 --> 00:17:42,000
by differentiating the series,
using the series definition of

259
00:17:40,000 --> 00:17:46,000
the exponential and by
differentiating it term by term,

260
00:17:44,000 --> 00:17:50,000
I can immediately see that is
satisfies this differential

261
00:17:48,000 --> 00:17:54,000
equation.
What about the arbitrary

262
00:17:50,000 --> 00:17:56,000
constant?
Well, if you would like,

263
00:17:52,000 --> 00:17:58,000
you can include it here,
but it is easier to observe

264
00:17:56,000 --> 00:18:02,000
that by linearity if e to the a
t solves the equation so does

265
00:18:00,000 --> 00:18:06,000
the constant times it because
the equation is linear.

266
00:18:05,000 --> 00:18:11,000
Now, that is the idea that I am
going to use to solve the system

267
00:18:12,000 --> 00:18:18,000
in general.
What are we doing to say?

268
00:18:16,000 --> 00:18:22,000
Well, what could we say?
The solution to,

269
00:18:21,000 --> 00:18:27,000
well, let's get two solutions
at once by writing a fundamental

270
00:18:28,000 --> 00:18:34,000
matrix.
"A" fundamental matrix,

271
00:18:33,000 --> 00:18:39,000
I don't claim it is "the" one,
for the system x prime equals A

272
00:18:41,000 --> 00:18:47,000
x.
That is what we are trying to

273
00:18:46,000 --> 00:18:52,000
solve.
And we are going to get two

274
00:18:51,000 --> 00:18:57,000
solutions by getting a
fundamental matrix for it.

275
00:18:57,000 --> 00:19:03,000
The answer is e to the a t.

276
00:19:04,000 --> 00:19:10,000
Isn't that what it should be?
I had a little a.

277
00:19:07,000 --> 00:19:13,000
Now we have a matrix.
Okay, just put the matrix up

278
00:19:11,000 --> 00:19:17,000
there.
Now, what on earth?

279
00:19:13,000 --> 00:19:19,000
The first person who must have
thought of this,

280
00:19:16,000 --> 00:19:22,000
it happened about 100 years
ago, what meaning should be

281
00:19:20,000 --> 00:19:26,000
given to e to a matrix power?
Well, clearly the two na•ve

282
00:19:25,000 --> 00:19:31,000
definitions won't work.
The only possible meaning you

283
00:19:30,000 --> 00:19:36,000
could try for is using the
infinite series,

284
00:19:33,000 --> 00:19:39,000
but that does work.
So this is a definition I am

285
00:19:37,000 --> 00:19:43,000
giving you, the exponential
matrix.

286
00:19:40,000 --> 00:19:46,000
Now, notice the A is a
two-by-two matrix multiplying it

287
00:19:44,000 --> 00:19:50,000
by t.
What I have up here is that

288
00:19:46,000 --> 00:19:52,000
it's basically a two-by-two
matrix.

289
00:19:49,000 --> 00:19:55,000
Its entries involve t,
but it's a two-by-two matrix.

290
00:19:53,000 --> 00:19:59,000
Okay.
We are trying to get the analog

291
00:19:56,000 --> 00:20:02,000
of that formula over there.
Well, leave the first term out

292
00:20:02,000 --> 00:20:08,000
just for a moment.
The next term is going to

293
00:20:05,000 --> 00:20:11,000
surely be A times t.
This is a two-by-two matrix,

294
00:20:09,000 --> 00:20:15,000
right?
What should the next term be?

295
00:20:12,000 --> 00:20:18,000
Well, A squared times t squared
over two factorial.

296
00:20:16,000 --> 00:20:22,000
What kind of a guy is that?
Well, if A is a two-by-two

297
00:20:20,000 --> 00:20:26,000
matrix so is A squared.
How about this?

298
00:20:23,000 --> 00:20:29,000
This is just a scalar which
multiplies every entry of A

299
00:20:28,000 --> 00:20:34,000
squared.
And, therefore,

300
00:20:30,000 --> 00:20:36,000
this is still a two-by-two
matrix.

301
00:20:33,000 --> 00:20:39,000
That is a two-by-two matrix.
This is a two-by-two matrix.

302
00:20:36,000 --> 00:20:42,000
No matter how many times you
multiply A by itself it stays a

303
00:20:40,000 --> 00:20:46,000
two-by-two matrix.
It gets more and more

304
00:20:43,000 --> 00:20:49,000
complicated looking but it is
always a two-by-two matrix.

305
00:20:46,000 --> 00:20:52,000
And now I am multiplying every
entry of that by the scalar t

306
00:20:50,000 --> 00:20:56,000
cubed over three factorial.

307
00:20:53,000 --> 00:20:59,000
I am continuing on in that way.
What I get, therefore,

308
00:20:56,000 --> 00:21:02,000
is a sum of two-by-two
matrices.

309
00:21:00,000 --> 00:21:06,000
Well, you can add two-by-two
matrices to each other.

310
00:21:03,000 --> 00:21:09,000
We've never made an infinite
series of them,

311
00:21:06,000 --> 00:21:12,000
we haven't done it,
but others have.

312
00:21:09,000 --> 00:21:15,000
And this is what they wrote.
The only question is,

313
00:21:12,000 --> 00:21:18,000
what should we put in the
beginning?

314
00:21:15,000 --> 00:21:21,000
Over there I have the number
one.

315
00:21:17,000 --> 00:21:23,000
But I, of course,
cannot add the number one to a

316
00:21:20,000 --> 00:21:26,000
two-by-two matrices.
What goes here must be a

317
00:21:23,000 --> 00:21:29,000
two-by-two matrix,
which is the closest thing to

318
00:21:27,000 --> 00:21:33,000
one I can think of.
What should it be?

319
00:21:31,000 --> 00:21:37,000
The I.
Two-by-two I.

320
00:21:32,000 --> 00:21:38,000
Two-by-two identity matrix
looks like the natural candidate

321
00:21:37,000 --> 00:21:43,000
for what to put there.
And, in fact,

322
00:21:40,000 --> 00:21:46,000
it is the right thing to put
there.

323
00:21:42,000 --> 00:21:48,000
Okay.
Now I have a conjecture,

324
00:21:45,000 --> 00:21:51,000
you know, purely formally,
changing only with a keystroke

325
00:21:49,000 --> 00:21:55,000
of the computer,
all the little a's have been

326
00:21:53,000 --> 00:21:59,000
changed to capital A's.
And now all I have to do is

327
00:21:57,000 --> 00:22:03,000
wonder if this is going to work.
Well, what is the basic thing I

328
00:22:03,000 --> 00:22:09,000
have to check to see that it is
the fundamental matrix?

329
00:22:07,000 --> 00:22:13,000
The question is,
I wrote it down all right,

330
00:22:11,000 --> 00:22:17,000
but is this a fundamental
matrix for the system?

331
00:22:14,000 --> 00:22:20,000
Well, I have a way of
recognizing a fundamental matrix

332
00:22:19,000 --> 00:22:25,000
when I see one.
The critical thing is that it

333
00:22:22,000 --> 00:22:28,000
should satisfy this matrix
differential equation.

334
00:22:26,000 --> 00:22:32,000
That is what I should verify.
Does this guy that I have

335
00:22:32,000 --> 00:22:38,000
written down satisfy this
equation?

336
00:22:35,000 --> 00:22:41,000
And the answer is,
number two is,

337
00:22:37,000 --> 00:22:43,000
it satisfies x prime equals Ax.

338
00:22:41,000 --> 00:22:47,000
In other words,
plugging in x equals this e to

339
00:22:45,000 --> 00:22:51,000
the at,
whose definition I just gave

340
00:22:49,000 --> 00:22:55,000
you.
If I substitute that in,

341
00:22:51,000 --> 00:22:57,000
does it satisfy that matrix
differential equation?

342
00:22:56,000 --> 00:23:02,000
The answer is yes.
I am not going to calculate it

343
00:23:00,000 --> 00:23:06,000
out because the calculation is
actually identical to what I did

344
00:23:04,000 --> 00:23:10,000
there.
The only difference is when I

345
00:23:06,000 --> 00:23:12,000
differentiated it term by term,
how do you differentiate

346
00:23:10,000 --> 00:23:16,000
something like this?
Well, you differentiate every

347
00:23:13,000 --> 00:23:19,000
term in it.
But, if you work it out,

348
00:23:15,000 --> 00:23:21,000
this is a constant matrix,
every term of which is

349
00:23:18,000 --> 00:23:24,000
multiplied by t squared over two
factorial.

350
00:23:21,000 --> 00:23:27,000
Well, if you differentiate
every entry of that constant,

351
00:23:25,000 --> 00:23:31,000
of that matrix,
what you are going to get is A

352
00:23:27,000 --> 00:23:33,000
squared times just the
derivative of that part,

353
00:23:30,000 --> 00:23:36,000
which is simply t.
In other words,

354
00:23:34,000 --> 00:23:40,000
the formal calculation looks
absolutely identical to that.

355
00:23:40,000 --> 00:23:46,000
So the answer to this is yes,
by the same calculation as

356
00:23:45,000 --> 00:23:51,000
before, as for the one-by-one
case.

357
00:23:48,000 --> 00:23:54,000
And now the only other thing to
check is that the determinant is

358
00:23:55,000 --> 00:24:01,000
not zero.
In fact, the determinant is not

359
00:23:59,000 --> 00:24:05,000
zero at one point.
That is all you have to check.

360
00:24:04,000 --> 00:24:10,000
What is x of zero?
What is the value of the

361
00:24:08,000 --> 00:24:14,000
determinant of x is e
to the At?

362
00:24:11,000 --> 00:24:17,000
What is the value of this thing
at zero?

363
00:24:14,000 --> 00:24:20,000
Here is my function.
If I plug in t equals zero,

364
00:24:18,000 --> 00:24:24,000
what is it equal to?
I.

365
00:24:20,000 --> 00:24:26,000
What is the determinant of I?
One.

366
00:24:22,000 --> 00:24:28,000
It is certainly not zero.

367
00:24:38,000 --> 00:24:44,000
By writing down this infinite
series, I have my two solutions.

368
00:24:41,000 --> 00:24:47,000
Its columns give me two
solutions to the original

369
00:24:44,000 --> 00:24:50,000
system.
There were no eigenvalues,

370
00:24:47,000 --> 00:24:53,000
no eigenvectors.
I have a formula for the

371
00:24:49,000 --> 00:24:55,000
answer.
What is the formula?

372
00:24:51,000 --> 00:24:57,000
It is e to the At.
And, of course,

373
00:24:54,000 --> 00:25:00,000
anybody reading the paper is
supposed to know what e to the

374
00:24:57,000 --> 00:25:03,000
At is.
It means that.

375
00:25:00,000 --> 00:25:06,000
This is just marvelous.
There must be a fly in the

376
00:25:03,000 --> 00:25:09,000
ointment somewhere.
Only a teeny little fly.

377
00:25:07,000 --> 00:25:13,000
There is a teeny little fly
because it is almost impossible

378
00:25:11,000 --> 00:25:17,000
to calculate that series for all
reasonable times.

379
00:25:15,000 --> 00:25:21,000
However, once in a while it is.
Let me give you an example

380
00:25:20,000 --> 00:25:26,000
where it is possible to
calculate the series and were

381
00:25:24,000 --> 00:25:30,000
you get a nice answer.
Let's work out an example.

382
00:25:36,000 --> 00:25:42,000
By the way, you know,
nowadays, we are not back 50

383
00:25:40,000 --> 00:25:46,000
years, the exponential matrix
has the same status on,

384
00:25:45,000 --> 00:25:51,000
say, a Matlab or Maple or
Mathematica, as the ordinary

385
00:25:51,000 --> 00:25:57,000
exponential function does.
It is just a command you type

386
00:25:56,000 --> 00:26:02,000
in.
You type in your matrix.

387
00:26:00,000 --> 00:26:06,000
And you now say EXP of that
matrix and out comes the answer

388
00:26:04,000 --> 00:26:10,000
to as many decimal places as you
want.

389
00:26:06,000 --> 00:26:12,000
It will be square matrix with
entries carefully written out.

390
00:26:10,000 --> 00:26:16,000
So, in that sense,
the fact that we cannot

391
00:26:13,000 --> 00:26:19,000
calculate it shouldn't bother
us.

392
00:26:15,000 --> 00:26:21,000
There are machines to do the
calculations.

393
00:26:18,000 --> 00:26:24,000
What we are interested in is it
as a theoretical tool.

394
00:26:22,000 --> 00:26:28,000
But, in order to get any
feeling for this at all,

395
00:26:25,000 --> 00:26:31,000
we certainly have to do a few
calculations.

396
00:26:30,000 --> 00:26:36,000
Let's do an easy one.
Let's consider the system x

397
00:26:34,000 --> 00:26:40,000
prime equals y,
y prime equals x.

398
00:26:39,000 --> 00:26:45,000
This is very easily done by
elimination, but that is

399
00:26:43,000 --> 00:26:49,000
forbidden.
First of all,

400
00:26:45,000 --> 00:26:51,000
we write it as a matrix.
It's the system x prime equals

401
00:26:50,000 --> 00:26:56,000
zero, one, one,
zero, x.

402
00:26:53,000 --> 00:26:59,000
Here is my A.

403
00:26:55,000 --> 00:27:01,000
And so e to the At
is going to be --

404
00:27:01,000 --> 00:27:07,000
A is zero, one,
one, zero.

405
00:27:02,000 --> 00:27:08,000
What we want to

406
00:27:05,000 --> 00:27:11,000
calculate is we are going to get
both solutions at once by

407
00:27:08,000 --> 00:27:14,000
calculating it one fell swoop e
to the At.

408
00:27:12,000 --> 00:27:18,000
Okay.
E to the At equals.

409
00:27:13,000 --> 00:27:19,000
I am going to actually write
out these guys.

410
00:27:16,000 --> 00:27:22,000
Well, obviously the hard part,
the part which is normally

411
00:27:20,000 --> 00:27:26,000
going to prevent us from
calculating this series

412
00:27:23,000 --> 00:27:29,000
explicitly, by hand anyway,
because, as I said,

413
00:27:26,000 --> 00:27:32,000
the computer can always do it.
The value, how do we raise a

414
00:27:32,000 --> 00:27:38,000
matrix to a high power?
You just keep multiplying and

415
00:27:37,000 --> 00:27:43,000
multiplying and multiplying.
That looks like a rather

416
00:27:41,000 --> 00:27:47,000
forbidding and unpromising
activity.

417
00:27:44,000 --> 00:27:50,000
Well, here it is easy.
Let's see what happens.

418
00:27:48,000 --> 00:27:54,000
If that is A,
what is A squared?

419
00:27:51,000 --> 00:27:57,000
I am going to have to calculate
that as part of the series.

420
00:27:56,000 --> 00:28:02,000
That is going to be zero,
one, one, zero times zero,

421
00:28:00,000 --> 00:28:06,000
one, one, zero,
which is one,

422
00:28:03,000 --> 00:28:09,000
zero, zero, one.

423
00:28:10,000 --> 00:28:16,000
We got saved.
It is the identity.

424
00:28:13,000 --> 00:28:19,000
Now, from this point on we
don't have to do anymore

425
00:28:17,000 --> 00:28:23,000
calculations,
but I will do them anyway.

426
00:28:21,000 --> 00:28:27,000
What is A cubed?
Don't start from scratch again.

427
00:28:26,000 --> 00:28:32,000
No, no, no.
A cubed is A squared times A.

428
00:28:30,000 --> 00:28:36,000
And A squared is,

429
00:28:34,000 --> 00:28:40,000
in real life,
the identity.

430
00:28:35,000 --> 00:28:41,000
Of course, you would do all
this in your head,

431
00:28:38,000 --> 00:28:44,000
but I am being a good boy and
writing it all out.

432
00:28:41,000 --> 00:28:47,000
This is I, the identity,
times A, which is A.

433
00:28:44,000 --> 00:28:50,000
I will do one more.
What is A to the fourth?

434
00:28:47,000 --> 00:28:53,000
Now, you would be tempted to
say A to the fourth is A

435
00:28:50,000 --> 00:28:56,000
squared, which is I times I,
which is I, but that would be

436
00:28:54,000 --> 00:29:00,000
wrong.
A to the fourth is A cubed

437
00:28:58,000 --> 00:29:04,000
times A,
which is, I have just

438
00:29:02,000 --> 00:29:08,000
calculated is A times A,
right?

439
00:29:05,000 --> 00:29:11,000
And now that is A squared,
which is the identity.

440
00:29:10,000 --> 00:29:16,000
It is clear,
by this argument,

441
00:29:12,000 --> 00:29:18,000
it is going to continue in the
same way each time you add an A

442
00:29:18,000 --> 00:29:24,000
on the right-hand side,
you are going to keep

443
00:29:22,000 --> 00:29:28,000
alternating between the
identity, A, the next one will

444
00:29:27,000 --> 00:29:33,000
be identity, the next will be A.
The end result is that the

445
00:29:34,000 --> 00:29:40,000
first term of the series is
simply the identity;

446
00:29:39,000 --> 00:29:45,000
the next term of the series is
A, but it is multiplied by t.

447
00:29:44,000 --> 00:29:50,000
I will keep the t on the
outside.

448
00:29:47,000 --> 00:29:53,000
Remember, when you multiply a
matrix by a scalar,

449
00:29:52,000 --> 00:29:58,000
that means multiply every entry
by that scalar.

450
00:29:56,000 --> 00:30:02,000
This is the matrix zero,
t, t, zero.

451
00:30:00,000 --> 00:30:06,000
I will do a couple more terms.

452
00:30:05,000 --> 00:30:11,000
The next term would be,
well, A squared we just

453
00:30:08,000 --> 00:30:14,000
calculated as the identity.
That looks like this.

454
00:30:12,000 --> 00:30:18,000
Except now I multiply every
term by t squared over two

455
00:30:16,000 --> 00:30:22,000
factorial.
All right.

456
00:30:19,000 --> 00:30:25,000
I'll go for broke.
The next one will be this times

457
00:30:22,000 --> 00:30:28,000
t cubed over three factorial.

458
00:30:25,000 --> 00:30:31,000
And, fortunately,
I have run out of room.

459
00:30:30,000 --> 00:30:36,000
Okay, let's calculate then.

460
00:30:54,000 --> 00:31:00,000
What is the final answer for e
to At?

461
00:30:57,000 --> 00:31:03,000
I have an infinite series of
two-by-two matrices.

462
00:31:00,000 --> 00:31:06,000
Let's look at the term in the
upper left-hand corner.

463
00:31:03,000 --> 00:31:09,000
It is one plus zero times t
plus one times t squared over

464
00:31:07,000 --> 00:31:13,000
two factorial plus zero
times t.

465
00:31:11,000 --> 00:31:17,000
It is going to be,

466
00:31:12,000 --> 00:31:18,000
in other words,
one plus t squared over two

467
00:31:15,000 --> 00:31:21,000
factorial plus
the next term,

468
00:31:18,000 --> 00:31:24,000
which is not on the board but I
think you can see,

469
00:31:21,000 --> 00:31:27,000
is this.
And it continues on in the same

470
00:31:24,000 --> 00:31:30,000
way.
How about the lower left term?

471
00:31:28,000 --> 00:31:34,000
Well, that is zero plus t plus
zero plus t cubed over three

472
00:31:32,000 --> 00:31:38,000
factorial and so on.

473
00:31:35,000 --> 00:31:41,000
It is t plus t cubed over three
factorial plus t to the fifth

474
00:31:39,000 --> 00:31:45,000
over five factorial.

475
00:31:42,000 --> 00:31:48,000
And the other terms in the

476
00:31:44,000 --> 00:31:50,000
other two corners are just the
same as these.

477
00:31:47,000 --> 00:31:53,000
This one, for example,
is zero plus t plus zero plus t

478
00:31:51,000 --> 00:31:57,000
cubed over three factorial.

479
00:31:55,000 --> 00:32:01,000
And the lower one is one plus
zero plus t squared

480
00:31:59,000 --> 00:32:05,000
and so on.
This is the same as one plus t

481
00:32:04,000 --> 00:32:10,000
squared over two factorial
and so on,

482
00:32:08,000 --> 00:32:14,000
and up here we have t plus t
cubed over three factorial

483
00:32:14,000 --> 00:32:20,000
and so on.
Well, that matrix doesn't look

484
00:32:18,000 --> 00:32:24,000
very square, but it is.
It is infinitely long

485
00:32:22,000 --> 00:32:28,000
physically, but it has one term
here, one term here,

486
00:32:27,000 --> 00:32:33,000
one term here and one term
there.

487
00:32:31,000 --> 00:32:37,000
Now, all we have to do is make
those terms look a little

488
00:32:35,000 --> 00:32:41,000
better.
For here I have to rely on the

489
00:32:38,000 --> 00:32:44,000
culture, which you may or may
not posses.

490
00:32:42,000 --> 00:32:48,000
You would know what these
series were if only they

491
00:32:46,000 --> 00:32:52,000
alternated their signs.
If this were a negative,

492
00:32:50,000 --> 00:32:56,000
negative, negative then the top
would be cosine t and

493
00:32:55,000 --> 00:33:01,000
this would be sine t,
but they don't.

494
00:33:01,000 --> 00:33:07,000
So they are the next best
thing.

495
00:33:04,000 --> 00:33:10,000
They are what?
Hyperbolic.

496
00:33:06,000 --> 00:33:12,000
The topic is not cosine t,
but cosh t.

497
00:33:11,000 --> 00:33:17,000
The bottle is sinh t.

498
00:33:15,000 --> 00:33:21,000
And how do we know this?
Because you remember.

499
00:33:19,000 --> 00:33:25,000
And what if I don't remember?
Well, you know now.

500
00:33:24,000 --> 00:33:30,000
That is why you come to class.

501
00:33:35,000 --> 00:33:41,000
Well, for those of you who
don't, remember,

502
00:33:38,000 --> 00:33:44,000
this is e to the t plus e to
the negative t.

503
00:33:44,000 --> 00:33:50,000
It should be over two,
but I don't have room to put in

504
00:33:48,000 --> 00:33:54,000
the two.
This doesn't mean I will omit

505
00:33:52,000 --> 00:33:58,000
it.
It just means I will put it in

506
00:33:55,000 --> 00:34:01,000
at the end by multiplying every
entry of this matrix by

507
00:34:00,000 --> 00:34:06,000
one-half.
If you have forgotten what cosh

508
00:34:04,000 --> 00:34:10,000
t is, it's e to the t plus e to
the negative t divided by two.

509
00:34:12,000 --> 00:34:18,000
And the similar thing for sinh
t.

510
00:34:14,000 --> 00:34:20,000
There is your first explicit
exponential matrix calculated

511
00:34:19,000 --> 00:34:25,000
according to the definition.
And what we have found is the

512
00:34:24,000 --> 00:34:30,000
solution to the system x prime
equals y,

513
00:34:28,000 --> 00:34:34,000
y prime equals x.
A fundamental matrix.

514
00:34:33,000 --> 00:34:39,000
In other words,
cosh t and sinh t satisfy both

515
00:34:36,000 --> 00:34:42,000
solutions to that system.
Now, there is one thing people

516
00:34:40,000 --> 00:34:46,000
love the exponential matrix in
particular for,

517
00:34:44,000 --> 00:34:50,000
and that is the ease with which
it solves the initial value

518
00:34:48,000 --> 00:34:54,000
problem.
It is exactly what happens when

519
00:34:51,000 --> 00:34:57,000
studying the single system,
the single equation x prime

520
00:34:55,000 --> 00:35:01,000
equals Ax,
but let's do it in general.

521
00:35:00,000 --> 00:35:06,000
Let's do it in general.
What is the initial value

522
00:35:03,000 --> 00:35:09,000
problem?
Well, the initial value problem

523
00:35:07,000 --> 00:35:13,000
is we start with our old system,
but now I want to plug in

524
00:35:11,000 --> 00:35:17,000
initial conditions.
I want the particular solution

525
00:35:15,000 --> 00:35:21,000
which satisfies the initial
condition.

526
00:35:18,000 --> 00:35:24,000
Let's make it zero to avoid
complications,

527
00:35:22,000 --> 00:35:28,000
to avoid a lot of notation.
This is to be some starting

528
00:35:26,000 --> 00:35:32,000
value.
This is a certain constant

529
00:35:29,000 --> 00:35:35,000
vector.
It is to be the value of the

530
00:35:33,000 --> 00:35:39,000
solution at zero.
And the problem is find what x

531
00:35:37,000 --> 00:35:43,000
of t is.
Well, if you are using the

532
00:35:41,000 --> 00:35:47,000
exponential matrix it is a joke.
It is a joke.

533
00:35:45,000 --> 00:35:51,000
Shall I derive it or just do
it?

534
00:35:48,000 --> 00:35:54,000
All right.
The general solution,

535
00:35:51,000 --> 00:35:57,000
let's derive it,
and then I will put up the

536
00:35:55,000 --> 00:36:01,000
final formula in a box so that
you will know it is important.

537
00:36:02,000 --> 00:36:08,000
What is the general solution?
Well, I did that for you at the

538
00:36:06,000 --> 00:36:12,000
beginning of the period.
Once you have a fundamental

539
00:36:09,000 --> 00:36:15,000
matrix, you get the general
solution by multiplying it on

540
00:36:13,000 --> 00:36:19,000
the right by an arbitrary
constant vector.

541
00:36:16,000 --> 00:36:22,000
The general solution is going
to be x equals e to the At.

542
00:36:20,000 --> 00:36:26,000
That is my super fundamental

543
00:36:22,000 --> 00:36:28,000
matrix, found without
eigenvalues and eigenvectors.

544
00:36:27,000 --> 00:36:33,000
And this should be multiplied
by some unknown constant vector

545
00:36:32,000 --> 00:36:38,000
c.
The only question is,

546
00:36:35,000 --> 00:36:41,000
what should the constant vector
c be?

547
00:36:38,000 --> 00:36:44,000
To find c, I will plug in zero.
When t is zero,

548
00:36:42,000 --> 00:36:48,000
here I get x of zero,
here I get e to the A times

549
00:36:47,000 --> 00:36:53,000
zero times c.
Now what is this?

550
00:36:51,000 --> 00:36:57,000
This is the vector of initial
conditions?

551
00:36:55,000 --> 00:37:01,000
What is e to the A times zero?
Plug in t equals zero.

552
00:37:00,000 --> 00:37:06,000
What do you get?
I.

553
00:37:04,000 --> 00:37:10,000
Therefore, c is what?
c is x zero.

554
00:37:11,000 --> 00:37:17,000
It is a total joke.
And the solution is,

555
00:37:17,000 --> 00:37:23,000
the initial value problem is x
equals e to the At

556
00:37:26,000 --> 00:37:32,000
times x zero.
It is just what it would have

557
00:37:32,000 --> 00:37:38,000
been at one variable.
The only difference is that

558
00:37:36,000 --> 00:37:42,000
here we are allowed to put the c
out front.

559
00:37:39,000 --> 00:37:45,000
In other words,
if I asked you to put in the

560
00:37:41,000 --> 00:37:47,000
initial condition,
you would probably write x

561
00:37:44,000 --> 00:37:50,000
equals little x zero times e to
the At.

562
00:37:48,000 --> 00:37:54,000
And you would be tempted to do
the same thing here,

563
00:37:52,000 --> 00:37:58,000
vector x equals vector x zero
times e to the At.

564
00:37:55,000 --> 00:38:01,000
Now, you cannot do that.
And, if you try to Matlab will

565
00:38:00,000 --> 00:38:06,000
hiccup and say illegal
operation.

566
00:38:02,000 --> 00:38:08,000
What is the illegal operation?
Well, x is a column vector.

567
00:38:07,000 --> 00:38:13,000
From the system it is a column
vector.

568
00:38:10,000 --> 00:38:16,000
That means the initial
conditions are also a column

569
00:38:14,000 --> 00:38:20,000
vector.
You cannot multiply a column

570
00:38:17,000 --> 00:38:23,000
vector out front and a square
matrix afterwards.

571
00:38:20,000 --> 00:38:26,000
You cannot.
If you want to multiply a

572
00:38:23,000 --> 00:38:29,000
matrix by a column vector,
it has to come afterwards so

573
00:38:27,000 --> 00:38:33,000
you can do zing,
zing.

574
00:38:31,000 --> 00:38:37,000
There is no zing,
you see.

575
00:38:33,000 --> 00:38:39,000
You cannot put it in front.
It doesn't work.

576
00:38:36,000 --> 00:38:42,000
So it must go behind.
That is the only place you

577
00:38:40,000 --> 00:38:46,000
might get tripped up.
And, as I say,

578
00:38:43,000 --> 00:38:49,000
if you try to type that in
using Matlab,

579
00:38:47,000 --> 00:38:53,000
you will immediately get error
messages that it is illegal,

580
00:38:52,000 --> 00:38:58,000
you cannot do that.
Anyway, we have our solution.

581
00:38:56,000 --> 00:39:02,000
There is our system.
Our initial value problem

582
00:39:00,000 --> 00:39:06,000
anyway is in pink,
and its solution using the

583
00:39:04,000 --> 00:39:10,000
exponential matrix is in green.
Now, the only problem is we

584
00:39:08,000 --> 00:39:14,000
still have to talk a little bit
more about calculating this.

585
00:39:13,000 --> 00:39:19,000
Now, the principle warning with
an exponential matrix is that

586
00:39:17,000 --> 00:39:23,000
once you have gotten by the
simplest things involving the

587
00:39:21,000 --> 00:39:27,000
fact that it solves systems,
it gives you the fundamental

588
00:39:26,000 --> 00:39:32,000
matrix for a system,
then you start flexing your

589
00:39:29,000 --> 00:39:35,000
muscles and say,
oh, well, let's see what else

590
00:39:33,000 --> 00:39:39,000
we can do with this.
For example,

591
00:39:36,000 --> 00:39:42,000
the reason exponentials came
into being in the first place

592
00:39:40,000 --> 00:39:46,000
was because of the exponential
law, right?

593
00:39:43,000 --> 00:39:49,000
I will kill anybody who sends
me emails saying,

594
00:39:46,000 --> 00:39:52,000
what is the exponential law?
The exponential law would say

595
00:39:50,000 --> 00:39:56,000
that e to the A plus B is equal
to e to the A times e to the B.

596
00:39:54,000 --> 00:40:00,000
The law of exponents,

597
00:39:58,000 --> 00:40:04,000
in other words.
It is the thing that makes the

598
00:40:01,000 --> 00:40:07,000
exponential function different
from all other functions that it

599
00:40:05,000 --> 00:40:11,000
satisfies something like that.
Now, first of all,

600
00:40:08,000 --> 00:40:14,000
does this make sense?
That is are the symbols

601
00:40:11,000 --> 00:40:17,000
compatible?
Let's see.

602
00:40:13,000 --> 00:40:19,000
This is a two-by-two matrix,
this is a two-by-two matrix,

603
00:40:16,000 --> 00:40:22,000
so it does make sense to
multiply them,

604
00:40:19,000 --> 00:40:25,000
and the answer will be a
two-by-two matrix.

605
00:40:21,000 --> 00:40:27,000
How about here?
This is a two-by-two matrix,

606
00:40:24,000 --> 00:40:30,000
add this to it.
It is still a two-by-two

607
00:40:27,000 --> 00:40:33,000
matrix.
e to a two-by-two matrix still

608
00:40:29,000 --> 00:40:35,000
comes out to be a two-by-two
matrix.

609
00:40:33,000 --> 00:40:39,000
Both sides are legitimate
two-by-two matrices.

610
00:40:37,000 --> 00:40:43,000
The only question is,
are they equal?

611
00:40:41,000 --> 00:40:47,000
And the answer is not in a
pig's eye.

612
00:40:45,000 --> 00:40:51,000
How could this be?
Well, I didn't make up these

613
00:40:50,000 --> 00:40:56,000
laws.
I just obey them.

614
00:40:52,000 --> 00:40:58,000
I wish I had time to do a
little calculation to show that

615
00:40:58,000 --> 00:41:04,000
it is not true.
It is true in certain special

616
00:41:03,000 --> 00:41:09,000
cases.
It is true in the special case,

617
00:41:06,000 --> 00:41:12,000
and this is pretty much if and
only if, the only case in which

618
00:41:12,000 --> 00:41:18,000
it is true is if A and B are not
arbitrary square matrices but

619
00:41:17,000 --> 00:41:23,000
commute with each other.
You see, if you start writing

620
00:41:22,000 --> 00:41:28,000
out the series to try to check
whether that law is true,

621
00:41:27,000 --> 00:41:33,000
you will get a bunch of terms
here, a bunch of terms here.

622
00:41:34,000 --> 00:41:40,000
And you will find that those
terms are pair-wise equal only

623
00:41:38,000 --> 00:41:44,000
if you are allowed to let the
matrices commute with each

624
00:41:41,000 --> 00:41:47,000
other.
In other words,

625
00:41:43,000 --> 00:41:49,000
if you can turn AB plus BA into
twice AB then

626
00:41:47,000 --> 00:41:53,000
everything will work fine.
But if you cannot do that it

627
00:41:51,000 --> 00:41:57,000
will not.
Now, when do two square

628
00:41:53,000 --> 00:41:59,000
matrices commute with each
other?

629
00:41:56,000 --> 00:42:02,000
The answer is almost never.
It is just a lucky accident if

630
00:42:02,000 --> 00:42:08,000
they do, but there are three
cases of the lucky accident

631
00:42:08,000 --> 00:42:14,000
which you should know.
The three cases,

632
00:42:12,000 --> 00:42:18,000
I feel justified calling it
"the" three cases.

633
00:42:17,000 --> 00:42:23,000
Oh, well, maybe I shouldn't do
that.

634
00:42:21,000 --> 00:42:27,000
The three most significant
examples are,

635
00:42:26,000 --> 00:42:32,000
example number one,
when A is a constant times the

636
00:42:31,000 --> 00:42:37,000
identity matrix.
In other words,

637
00:42:36,000 --> 00:42:42,000
when A is a matrix that looks
like this.

638
00:42:39,000 --> 00:42:45,000
That matrix commutes with every
other square matrix.

639
00:42:43,000 --> 00:42:49,000
If that is A,
then this law is always true

640
00:42:46,000 --> 00:42:52,000
and you are allowed to use this.
Okay, so that is one case.

641
00:42:51,000 --> 00:42:57,000
Another case,
when A is more general,

642
00:42:54,000 --> 00:43:00,000
is when B is equal to negative
A.

643
00:42:59,000 --> 00:43:05,000
I think you can see that that
is going to work because A times

644
00:43:03,000 --> 00:43:09,000
minus A is equal to minus A
times A.

645
00:43:07,000 --> 00:43:13,000
Yeah, they are both equal to A
squared,

646
00:43:11,000 --> 00:43:17,000
except with a negative sign in
front.

647
00:43:14,000 --> 00:43:20,000
And the third case is when B is
equal to the inverse of A

648
00:43:18,000 --> 00:43:24,000
because A A inverse is the same
as A inverse A.

649
00:43:23,000 --> 00:43:29,000
They are both the identity.

650
00:43:26,000 --> 00:43:32,000
Of course, A must have an
inverse.

651
00:43:30,000 --> 00:43:36,000
Okay, let's suppose it does.
Now, of them this is,

652
00:43:34,000 --> 00:43:40,000
I think, the most important one
because it leads to this law.

653
00:43:40,000 --> 00:43:46,000
That is forbidden,
but there is one case of it

654
00:43:44,000 --> 00:43:50,000
which is not forbidden and that
is here.

655
00:43:48,000 --> 00:43:54,000
What will it say?
Well, that will say that e to

656
00:43:52,000 --> 00:43:58,000
the A minus A is equal to e to
the A times e to

657
00:43:58,000 --> 00:44:04,000
the negative A.
This is true,

658
00:44:03,000 --> 00:44:09,000
even though the general law is
false.

659
00:44:05,000 --> 00:44:11,000
That is because A and negative
A commute with each other.

660
00:44:10,000 --> 00:44:16,000
But now what does this say?
What is e to the zero matrix?

661
00:44:14,000 --> 00:44:20,000
In other words,
suppose I take the matrix that

662
00:44:18,000 --> 00:44:24,000
is zero and plug it into the
formula for e?

663
00:44:21,000 --> 00:44:27,000
What do you get?
e to the zero times t is I.

664
00:44:24,000 --> 00:44:30,000
It has to be a two-by-two
matrix if it is going to be

665
00:44:29,000 --> 00:44:35,000
anything.
It is the matrix I.

666
00:44:33,000 --> 00:44:39,000
This side is I.
This side is the exponential

667
00:44:38,000 --> 00:44:44,000
matrix.
And what does that show?

668
00:44:41,000 --> 00:44:47,000
It shows that the inverse
matrix, the e to the A,

669
00:44:47,000 --> 00:44:53,000
is e to the negative A.
That is a very useful fact.

670
00:44:53,000 --> 00:44:59,000
This is the main survivor of
the exponential law.

671
00:45:00,000 --> 00:45:06,000
In general it is false,
but this standard corollary to

672
00:45:05,000 --> 00:45:11,000
the exponential law is true,
is equal to e to the minus A,

673
00:45:10,000 --> 00:45:16,000
just what you would
dream and hope would be true.

674
00:45:16,000 --> 00:45:22,000
Okay.
I have exactly two and a half

675
00:45:19,000 --> 00:45:25,000
minutes left in which to do the
impossible.

676
00:45:23,000 --> 00:45:29,000
All right.
The question is,

677
00:45:25,000 --> 00:45:31,000
how do you calculate e to the
At?

678
00:45:31,000 --> 00:45:37,000
You could use series,
but it rarely works.

679
00:45:34,000 --> 00:45:40,000
It is too hard.
There are a few examples,

680
00:45:38,000 --> 00:45:44,000
and you will have some more for
homework, but in general it is

681
00:45:43,000 --> 00:45:49,000
too hard because it is too hard
to calculate the powers of a

682
00:45:49,000 --> 00:45:55,000
general matrix A.
There is another method,

683
00:45:52,000 --> 00:45:58,000
which is useful only for
matrices which are symmetric,

684
00:45:57,000 --> 00:46:03,000
but like that --
Well, it is more than

685
00:46:01,000 --> 00:46:07,000
symmetric.
These two have to be the same.

686
00:46:04,000 --> 00:46:10,000
But you can handle those,
as you will see from the

687
00:46:07,000 --> 00:46:13,000
homework problems,
by breaking it up this way and

688
00:46:11,000 --> 00:46:17,000
using the exponential law.
This would be zero,

689
00:46:14,000 --> 00:46:20,000
b, b, zero.

690
00:46:16,000 --> 00:46:22,000
See, these two matrices commute
with each other and,

691
00:46:19,000 --> 00:46:25,000
therefore, I could use the
exponential law.

692
00:46:22,000 --> 00:46:28,000
This leaves all other cases.
And here is the way to handle

693
00:46:26,000 --> 00:46:32,000
all other cases.
All other cases.

694
00:46:30,000 --> 00:46:36,000
In other words,
if you cannot calculate the

695
00:46:33,000 --> 00:46:39,000
series, this trick doesn't work,
I have done as follows.

696
00:46:38,000 --> 00:46:44,000
You start with an arbitrary
fundamental matrix,

697
00:46:41,000 --> 00:46:47,000
not the exponential matrix.
You multiply it by its value at

698
00:46:46,000 --> 00:46:52,000
zero, that is a constant matrix,
and you take the inverse of

699
00:46:51,000 --> 00:46:57,000
that constant matrix.
It will have one because,

700
00:46:55,000 --> 00:47:01,000
remember, the fundamental
matrix never has the determinant

701
00:47:00,000 --> 00:47:06,000
zero.
So you can always take its

702
00:47:04,000 --> 00:47:10,000
inverse-ready value of t.
Now, what property does this

703
00:47:09,000 --> 00:47:15,000
have?
It is a fundamental matrix.

704
00:47:12,000 --> 00:47:18,000
How do I know that?
Well, because I found all

705
00:47:16,000 --> 00:47:22,000
fundamental matrices for you.
Take any one,

706
00:47:21,000 --> 00:47:27,000
multiply it by a square matrix
on the right-hand side,

707
00:47:26,000 --> 00:47:32,000
and you get still a fundamental
matrix.

708
00:47:29,000 --> 00:47:35,000
And what is its value at zero?
Well, it is x of zero times x

709
00:47:37,000 --> 00:47:43,000
of zero inverse.
Its value at zero is the

710
00:47:42,000 --> 00:47:48,000
identity.
Now, e to the At has

711
00:47:48,000 --> 00:47:54,000
these same two properties.

712
00:47:56,000 --> 00:48:02,000
Namely, it is a fundamental
matrix and its value at zero is

713
00:48:01,000 --> 00:48:07,000
the identity. Conclusion,
this is e to the At.

714
00:48:05,000 --> 00:48:11,000
And that is the garden variety

715
00:48:08,000 --> 00:48:14,000
method of calculating the
exponential matrix,

716
00:48:11,000 --> 00:48:17,000
if you want to give it
explicitly.

717
00:48:13,000 --> 00:48:19,000
Start with any fundamental
matrix calculated,

718
00:48:16,000 --> 00:48:22,000
you should forgive the
expression using eigenvalues and

719
00:48:20,000 --> 00:48:26,000
eigenvectors and putting the
solutions into the columns.

720
00:48:24,000 --> 00:48:30,000
Evaluate it at zero,
take its inverse and multiply

721
00:48:28,000 --> 00:48:34,000
the two.
And what you end up with has to

722
00:48:32,000 --> 00:48:38,000
be the same as the thing
calculated with that infinite

723
00:48:36,000 --> 00:48:42,000
series.
Okay.
You will get lots of practice
for homework and tomorrow.