PROFESSOR: Welcome
to this recitation on matrix exponential. So here, we're given matrix
A with entries 6, 5, 1, 2. And we're asked to compute
the matrix exponential, exponential A*t, and to use
it to solve the initial value problem u prime of
t equals A*u(t), where here u are
basically vectors, with initial condition,
u of 0 equals [4, 1]. So why don't you
pause the video, work through the problem? And I'll be right back. Welcome back. So first, to go ahead and
compute the matrix exponential, we need to identify the
eigenvalues of matrix A and its eigenvectors. So this is a matrix--
I'll just rewrite here-- that we saw before. And its eigenvalues are again,
solution of 6 minus lambda, 5, 1, 2 minus lambda, equals
to 0, which gives us 6..., 2 minus lambda
minus 5 equals to 0. Lambda square minus 8*lambda. Then we have a 12 minus 5. So you can verify that the two
eigenvalues would be 1 and 7. Lambda_1 equals to 1. And lambda_1 equals to 7. So now, we need to seek
the eigenvectors associated to each one of the eigenvalues. So the idea here is
to basically move toward a diagonalization
of the matrix A. So let's seek the eigenvectors. And here, I'm just going
to give them to you, and you can verify
the calculation. And this calculation
was performed in a previous recitation. So the eigenvectors. v_1 associated to the eigenvalue
lambda_1 was, for example, 1, minus 1. And the other one
that we found-- again, this is one form of
the eigenvector-- was 5 and 1. So these are from the notes
of a previous recitation. So you can verify that these
are the two eigenvectors. And from this point, then we
can rewrite this solution, if you recall. I'm just going to
go through the steps toward getting to the definition
of the exponential matrix. So here, if we
didn't know anything about the exponential
matrix, we would be able to write the
solution as c_1 exponential t v_1 plus c_2 exponential 7t v_2,
which basically gives us here, if I write it in this form,
for example, an exponential t, minus exponential t and an
exponential 5t multiplied by the entry of this vector,
an exponential 7t here, multiplying [c_1, c_2]. So this is where the idea of the
matrix exponential comes from. We're basically introducing
the matrix phi of t for which we can write u equals
phi of t multiplied by this [c_1, c_2], general constants. So phi of t would then
be equal to this matrix. But what we want is to be
able to solve an initial value problem for which
e of A of 0 applied to our initial conditions
would give us back our initial condition. So we're seeking for a form for
this exponential matrix that would allow us to do this. So the way that we define the
matrix exponential give us exponential A*t-- now, I
won't go into the proof, but we're just going
to check it together-- multiplied by phi of 0 minus 1. So let's check that if we
use this form of the matrix exponential, we would have e. We will have that at
0 applied to u(0). We have phi(0), phi(0)^(-1)
applied to u(0). This is a matrix
with its inverse, which gives us the identity. And so basically, this
gives us back u of 0. I mean you don't need to
do that when you're asked to find the matrix exponential. But just to remember
where it's coming from, you write down your
system in matrix form. You identify the
matrix phi of t. And then you recall why you want
the matrix exponential to have this form, basically to be
able to solve initial value problems for which the value
u of t is projected to u of 0 when we take t equals 0
for the matrix exponential. So now let's go
back to our problem. So let's compute this
matrix exponential. We have phi of t. So now from this formula, we
know that we need phi of 0. So that give us, basically,
exponential of 0, 5, minus 1, and 1. We need to find its inverse. So recall that the inverse
of a two-by-two matrix is basically just the
determinant, minus b, minus c, and reversing
the diagonal entries. So we can just apply this
to get our phi of 0 minus 1. So here, our
determinant is basically 1 plus 5, which is 1 over 6. And then the entries are
simply 1, 1, minus 5, and 1. So now, we're just left with the
multiplication of two matrices to get our matrix exponential. So our matrix exponential
would give us this one sixth. And we now have to
multiply the entries. So I'm not going to
rewrite everything. I'm just going to
use this space here. So we have exponential
t multiplying 1, plus 5 exponential 7t. Then, we have exponential t
dot minus 5 for this entry. 5 exponential t
multiplying our 1. 7t, thank you. Then for the second
entry, we basically have minus exponential
t 1 exponential 7t 1 minus exponential t minus
5 and exponential 7t 1. So we're done with the
matrix exponential. So now we were asked to solve
for the initial value problem with initial condition 4 and 1. So how do we go about that? Well, recall that
I just reminded you what did we want to use
this matrix exponential for. And what we wanted it for is
to be able to basically project an initial condition into a
solution u of t, t times later. And we constructed this matrix
to be able to basically give us this solution by just
multiplying the matrix by the initial value vector. So basically, to find the
solution of this initial value problem, we simply need
to multiply this matrix by the initial vector
that we were given. And I'm just going to
write it here to not have to rewrite everything. And it was 4 and 1. And this is u of 0. So let me just do
a dash here just so that we can do the computation. And we would end up
with a solution-- I'm going to keep it
in matrix form for now. So we end up with 4 exponential
t minus 5 exponential t, so minus 1 exponential t. And we have a one sixth. Here, 5 exponential 7t, so
we have 20, plus 5, so 25, exponential 7t. Then for the second entry
of the vector solution, we have minus exponential here
minus 4 that we add to a 5, and here, a 7 multiplied
by 4 that we add to a 1. So we have basically
plus 5 exponential 7t. And that basically gives us one
way of writing this solution. And we can split this down,
if we will, into two vectors, plus t; minus 1, 1;
exponential 7t; [25, 5]. And this form is as valid. Yes, thank you. So that ends the
laborious calculations. But basically,
the key point here was just to remember where is
the matrix exponential coming from, basically, from the
eigenvalues and eigenvectors of the original matrix
present in the system, and where is the
definition coming from, why do we define it as
phi of t phi minus 1 of 0, and how to use it then
to give the solution to an initial value problem. So that ends this recitation.
