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PROFESSOR: Welcome
to this recitation

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on matrix exponential.

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So here, we're given matrix
A with entries 6, 5, 1, 2.

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And we're asked to compute
the matrix exponential,

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exponential A*t, and to use
it to solve the initial value

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problem u prime of
t equals A*u(t),

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where here u are
basically vectors,

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with initial condition,
u of 0 equals [4, 1].

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So why don't you
pause the video, work

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through the problem?

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And I'll be right back.

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Welcome back.

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So first, to go ahead and
compute the matrix exponential,

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we need to identify the
eigenvalues of matrix

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A and its eigenvectors.

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So this is a matrix--
I'll just rewrite here--

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that we saw before.

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And its eigenvalues are again,
solution of 6 minus lambda, 5,

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1, 2 minus lambda, equals
to 0, which gives us 6...,

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2 minus lambda
minus 5 equals to 0.

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Lambda square minus 8*lambda.

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Then we have a 12 minus 5.

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So you can verify that the two
eigenvalues would be 1 and 7.

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Lambda_1 equals to 1.

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And lambda_1 equals to 7.

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So now, we need to seek
the eigenvectors associated

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to each one of the eigenvalues.

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So the idea here is
to basically move

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toward a diagonalization
of the matrix A.

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So let's seek the eigenvectors.

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And here, I'm just going
to give them to you,

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and you can verify
the calculation.

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And this calculation
was performed

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in a previous recitation.

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So the eigenvectors.

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v_1 associated to the eigenvalue
lambda_1 was, for example, 1,

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minus 1.

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And the other one
that we found--

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again, this is one form of
the eigenvector-- was 5 and 1.

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So these are from the notes
of a previous recitation.

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So you can verify that these
are the two eigenvectors.

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And from this point, then we
can rewrite this solution,

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if you recall.

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I'm just going to
go through the steps

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toward getting to the definition
of the exponential matrix.

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So here, if we
didn't know anything

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about the exponential
matrix, we would

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be able to write the
solution as c_1 exponential t

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v_1 plus c_2 exponential 7t v_2,
which basically gives us here,

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if I write it in this form,
for example, an exponential t,

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minus exponential t and an
exponential 5t multiplied

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by the entry of this vector,
an exponential 7t here,

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multiplying [c_1, c_2].

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So this is where the idea of the
matrix exponential comes from.

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We're basically introducing
the matrix phi of t

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for which we can write u equals
phi of t multiplied by this

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[c_1, c_2], general constants.

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So phi of t would then
be equal to this matrix.

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But what we want is to be
able to solve an initial value

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problem for which
e of A of 0 applied

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to our initial conditions
would give us back

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our initial condition.

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So we're seeking for a form for
this exponential matrix that

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would allow us to do this.

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So the way that we define the
matrix exponential give us

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exponential A*t-- now, I
won't go into the proof,

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but we're just going
to check it together--

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multiplied by phi of 0 minus 1.

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So let's check that if we
use this form of the matrix

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exponential, we would have e.

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We will have that at
0 applied to u(0).

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We have phi(0), phi(0)^(-1)
applied to u(0).

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This is a matrix
with its inverse,

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which gives us the identity.

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And so basically, this
gives us back u of 0.

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I mean you don't need to
do that when you're asked

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to find the matrix exponential.

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But just to remember
where it's coming from,

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you write down your
system in matrix form.

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You identify the
matrix phi of t.

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And then you recall why you want
the matrix exponential to have

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this form, basically to be
able to solve initial value

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problems for which the value
u of t is projected to u of 0

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when we take t equals 0
for the matrix exponential.

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So now let's go
back to our problem.

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So let's compute this
matrix exponential.

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We have phi of t.

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So now from this formula, we
know that we need phi of 0.

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So that give us, basically,
exponential of 0,

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5, minus 1, and 1.

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We need to find its inverse.

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So recall that the inverse
of a two-by-two matrix

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is basically just the
determinant, minus b,

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minus c, and reversing
the diagonal entries.

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So we can just apply this
to get our phi of 0 minus 1.

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So here, our
determinant is basically

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1 plus 5, which is 1 over 6.

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And then the entries are
simply 1, 1, minus 5, and 1.

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So now, we're just left with the
multiplication of two matrices

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to get our matrix exponential.

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So our matrix exponential
would give us this one sixth.

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And we now have to
multiply the entries.

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So I'm not going to
rewrite everything.

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I'm just going to
use this space here.

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So we have exponential
t multiplying

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1, plus 5 exponential 7t.

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Then, we have exponential t
dot minus 5 for this entry.

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5 exponential t
multiplying our 1.

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7t, thank you.

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Then for the second
entry, we basically

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have minus exponential
t 1 exponential 7t 1

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minus exponential t minus
5 and exponential 7t 1.

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So we're done with the
matrix exponential.

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So now we were asked to solve
for the initial value problem

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with initial condition 4 and 1.

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So how do we go about that?

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Well, recall that
I just reminded you

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what did we want to use
this matrix exponential for.

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And what we wanted it for is
to be able to basically project

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an initial condition into a
solution u of t, t times later.

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And we constructed this matrix
to be able to basically give us

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this solution by just
multiplying the matrix

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by the initial value vector.

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So basically, to find the
solution of this initial value

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problem, we simply need
to multiply this matrix

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by the initial vector
that we were given.

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And I'm just going to
write it here to not

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have to rewrite everything.

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And it was 4 and 1.

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And this is u of 0.

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So let me just do
a dash here just so

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that we can do the computation.

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And we would end up
with a solution--

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I'm going to keep it
in matrix form for now.

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So we end up with 4 exponential
t minus 5 exponential t,

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so minus 1 exponential t.

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And we have a one sixth.

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Here, 5 exponential 7t, so
we have 20, plus 5, so 25,

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exponential 7t.

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Then for the second entry
of the vector solution,

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we have minus exponential here
minus 4 that we add to a 5,

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and here, a 7 multiplied
by 4 that we add to a 1.

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So we have basically
plus 5 exponential 7t.

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And that basically gives us one
way of writing this solution.

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And we can split this down,
if we will, into two vectors,

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plus t; minus 1, 1;
exponential 7t; [25, 5].

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And this form is as valid.

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Yes, thank you.

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So that ends the
laborious calculations.

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But basically,
the key point here

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was just to remember where is
the matrix exponential coming

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from, basically, from the
eigenvalues and eigenvectors

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of the original matrix
present in the system,

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and where is the
definition coming from,

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why do we define it as
phi of t phi minus 1 of 0,

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and how to use it then
to give the solution

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to an initial value problem.

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So that ends this recitation.