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GILBERT STRANG: OK.

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We're still solving systems
of differential equations

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with a matrix A in them.

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And now I want to
create the exponential.

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It's just natural to produce
e to the A, or e to the A t.

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The exponential of a matrix.

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So if we have one
equation, small a,

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then we know the solution
is an e to the A t,

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times the starting value.

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Now we have n equations with
a matrix A and a vector y.

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And the solution should be,
at time t, e to the A t,

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times the starting value.

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It should be a perfect
match with this one, where

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this had a number
in the exponent

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and this has a matrix
in the exponent.

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OK.

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No problem.

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We just use the series
for e to the A t.

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We plug in a matrix
instead of a number.

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So the identity, plus A
t, plus 1/2 A t squared,

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plus 1/6 of A t cubed, forever.

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It's the same.

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It's the exponential series.

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The most important series
in mathematics, I think.

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And it gives us an answer.

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And that answer is a matrix.

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Everything here, every
term, is a matrix.

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OK.

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Now, is that the right answer?

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We check that by putting it
into the differential equation.

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So I want to put that
solution into the equation.

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So I need to take
the derivative.

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The derivative of
this is the derivative

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of-- that's a constant.

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The derivative of that is A.
The derivative of this is 1/2.

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I have an A squared,
and I have a t squared.

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The derivative of t squared
is 2t, so that'll just be a t.

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The 2 and the 2 cancel.

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OK.

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Now I have A cubed here.

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t cubed?

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The derivative of t cubed is 3t
squared, so I have a t squared.

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And the 3 cancels
the 3 and the 6,

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and leaves 1 over 2
factorial, and so on.

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And I look at that.

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And I say it's very
much like the one above.

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Look.

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This series is just
A times this one.

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Multiply the top one by A.
A times I is A. A times A t

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is A squared t.

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Term by term, it
just has a factor A.

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So it's A e to the A
t, is the derivative

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of my matrix exponential.

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It brings down an A.
Just what we want.

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Just what we want.

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So then if I add
a y of 0 in here,

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that's just a constant vector.

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I'll have a y of 0.

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I'll have a y of 0 here.

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When I put this into the
differential equation,

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it works.

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It works.

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Now, is it better than what
we had before, which was using

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eigenvalues and eigenvectors?

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It's better in one way.

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This exponential, this
series, is totally fine

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whether we have n independent
eigenvectors or not.

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We could have
repeated eigenvalues.

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I'll do an example.

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So for with repeated eigenvalues
and missing eigenvectors,

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e to the A t is still
the correct answer.

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Still the correct answer.

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But if we want to use
eigenvalues and eigenvectors

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to compute e to
the A t, because we

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don't want to add up an
infinite series very often,

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then we would want n
independent eigenvectors.

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So what am I saying?

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I'm saying that this e
to the A t-- All right,

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suppose we have n
independent eigenvectors.

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And we know that that
means, in that case,

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a is V times lambda
times V inverse.

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And we can write V
inverse because the matrix

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V has the eigenvectors.

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This is the eigenvector matrix.

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If I have n independent
eigenvectors,

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that matrix is invertible.

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I have that nice formula.

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And now I can see
what is-- e to the A

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t is always identity plus A.

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I'm now going to use
the diagonalization,

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the eigenvectors, and
the eigenvalues for A.

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So I'm doing the
good case now, when

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there are a full set of
independent eigenvectors.

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Then the A t is V
lambda V inverse t.

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That's right, that's I, plus
A t, plus 1/2 A t squared.

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Right?

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So I need A squared.

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So everybody remembers
what A squared is.

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A squared is V lambda V inverse,
times V lambda V inverse.

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And those cancel out to give
V lambda squared V inverse,

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times t squared, and so on.

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You remember this A squared,
so I'll take that away.

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And look at what I've got.

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Look what I've got it.

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Yes.

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Factor V out of the
start, and factor

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V inverse out of the end.

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And in here I have V times V
inverse is I, so that's fine.

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V times V inverse,
I have a lambda t.

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V and a V inverse, so I
have a 1/2 half lambda

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squared t squared.

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And so on, times V inverse.

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This is all just
what we hope for.

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We expect that a V goes out
at the far left at the front.

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This V inverse comes
out at the far right.

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And what do you
see in the middle?

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You see-- so this is now my
formula for e to the A t, is V.

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And what do I have there?

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I have the exponential
series for lambda t.

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So it's e to the
lambda t V inverse.

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And what is e to the lambda t?

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Let's just understand
the matrix exponential.

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When the matrix is diagonal,
the best possible matrix,

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this will be V. What
does my matrix look like?

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V inverse.

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If I'm looking at
this, looking at this.

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Lambda is diagonal.

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All these matrices are
diagonal with lambdas.

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So that'll be e to the lambda
1t down to e to the lambda nt.

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I'm not doing anything
brilliant here.

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I'm just using the
standard diagonalization

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to produce our exponential
from the eigenvector matrix

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and from the eigenvalues.

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So I'm just taking
the exponentials

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of the n different eigenvalues.

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So e to the A t.

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This would lead to e to the A
t y at 0, would be-- y of 0 is

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some combination.

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And then there's an e to the
lambda 1t coming from here.

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And there's an x eigenvector x1,
plus C2 e to the lambda 2t x2,

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so on.

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That's the solution
that we had last time.

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That's the solution that using
eigenvalues and eigenvectors.

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Now.

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Can Can I get
something new here?

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Something new will be,
suppose there are not

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a full set of n
independent eigenvectors.

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e to the A t is still OK.

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But this formula is no good.

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That formula depends
on V and V inverse.

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And suppose we have an example.

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So all that is very nice.

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That's what we expect.

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But we could have a
matrix like this one.

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A equals-- well,
here's an extreme case.

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What are the eigenvalues
of that matrix?

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It's a diagonal matrix.

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The eigenvalues are 0 and 0.

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The eigenvalue of 0 is repeated.

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It's a double eigenvalue.

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And we hope for
two eigenvectors,

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but we don't find them.

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That has only one
line of eigenvectors.

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It only has an x1
equals 1, 0, I think.

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If I multiply that A, times
that x1, gives me 0 times x1.

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That's an eigenvector.

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Well, because the
eigenvalue is 0,

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I'm looking for the null space.

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There is in the null
space, but the null space

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is only one-dimensional.

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Only one eigenvector.

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Missing an eigenvector.

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Still, still, I can
do e to the A t.

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That's still completely correct.

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That series will work.

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So to do this series I
need to know a squared.

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So I'm actually going
to use the series,

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but you'll see that
it cuts off very fast.

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a squared, if you work
that out, it's all 0's.

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So our e to the A t is just
I, plus A t, plus STOP.

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A squared is all 0's.

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A cubed is all 0's.

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So the matrix e to the A
t is identity, a times t.

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a is this, times t is
going to put a t there.

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There you go.

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That's a case of the
matrix exponential,

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which would lead us to the
solution of the equations.

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Of course, it's a pretty
simple exponential.

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But it comes from
pretty simple equations.

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The equations dy dt, that
system of two equations,

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with that matrix in it.

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Our system of equations
is just dy1 dt,

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I have a 1 there so
it would be a y2.

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And dy2 dt is 0
on the second row.

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Well, that's pretty
easy to solve.

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In fact, this tells you how
to solve-- you could naturally

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ask the question, how do we
solve differential equations

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when the matrix doesn't
have n eigenvectors?

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Here's an example.

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This matrix has only
one eigenvector.

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But the equation that we just
solved by, you could say,

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back substitution.

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This gives Y2 equal constant.

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And then that equation, dy1 dt
equal that constant, gives me

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y1 equals t times constant.

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That's what I'm seeing.

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Oh.

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00:13:01,690 --> 00:13:03,490
Yeah.

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00:13:03,490 --> 00:13:06,570
Are you surprised to
see a t show up here?

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Normally I don't see a t
in matrix exponentials.

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But in this repeated
case, that's

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the t that we're
always seeing when

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we have repeated solutions.

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Everybody remembers that when
we have second-order equations,

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and we have the two
exponents are the same.

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So we only get one solution
of that, e to the st.

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00:13:33,260 --> 00:13:35,090
And we have to look
for another one.

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And that other one is?

219
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te to the st. It's
that same t there.

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OK.

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00:13:42,470 --> 00:13:46,170
There is an example
of how a matrix

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00:13:46,170 --> 00:13:54,840
with a missing eigenvector,
the exponential pops a t in.

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The exponential pops a t in.

224
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And if I had two
missing eigenvectors,

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then in the exponential.

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00:14:01,800 --> 00:14:05,910
Shall I just show you an example
with two missing eigenvectors?

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00:14:05,910 --> 00:14:13,490
Let a be-- well, here it
would be 0, 0, 0, 0, 0,

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triple 0, with, let's say.

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There's a matrix with
three 0 eigenvalues,

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but only one eigenvector.

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So it's missing
two eigenvectors.

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00:14:30,260 --> 00:14:33,260
And I would, in the
end, in e to the A t

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00:14:33,260 --> 00:14:40,300
here, I would see
probably 1, 1, 1, t, t,

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and probably I'll see
a 1/2 t squared there.

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A little bit like that.

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But one step worse.

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Because the triple
eigenvalue, well, that's

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not going to happen
very often in reality.

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But we see what it produces.

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It produces a t squared
as well as the t's.

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OK.

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So, the x matrix
exponential gives

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a beautiful, concise, short
formula for the solution.

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And it gives a formula
that's correct,

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even in the case of
missing eigenvectors.

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Thank you.