PROFESSOR: Welcome
to this recitation.
In this recitation, we're
going to look at linear systems
with complex roots.
So the system we're
examining is the one given
as x dot equals minus 3x minus
2, and y dot equals 5x minus y.
And you're asked
to use the matrix
methods to solve this system.
So why don't you
take a pause here
and try to solve this problem?
And I'll be right back.
Welcome back.
So the first step is to write
this system in matrix form.
So we introduced
a vector [x, y],
the matrix multiplying
column vector [x, y] again.
The coefficients are going to
be minus 3, minus 2, 5, minus 1.
So the first step in
solving this system
is to find the eigenvalues
of the matrix A.
So the eigenvalues of
matrix A are basically
the solutions to this following
determinant equal to 0, minus 3
minus lambda, minus 2,
5, minus 1 minus lambda,
determinant equals to 0.
Here, the lambda are
the unknown eigenvalues.
And to get this
determinant, we're
basically multiplying these
two terms, minus minus 2 dot 5,
which gives us a plus sign.
So here, we're going
to have lambda squared.
3*lambda plus 1*lambda
gives us 4*lambda.
And 3 plus 10 gives us 13.
So this second-order
polynomial in lambda
will give us the two
eigenvalues for this matrix.
So let's examine
the discriminant.
So we have b squared minus 4a*b.
And this gives us minus 36.
So the discriminant is negative.
And that tells us
that we are going
to have two complex roots, which
is the title of the recitation.
And these two complex
roots are going
to be complex conjugate
of each other.
So the formula gives
us plus or minus i of 6
for the root of
the discriminant.
Here, we have minus 4 over 2.
So the two roots are basically
minus 2 plus or minus i*3
or 3i.
So these are our two roots.
So now, let's focus
on one of the roots
to get the eigenvector
associated with the eigenvalue.
So let's focus on the
positive one, for example.
And we could do all
the following again--
AUDIENCE: Minus 2.
PROFESSOR: Minus 2.
Thank you.
We can do all the
following calculation
that I'm going to go do now
for the complex conjugate,
and I will explain at the
end how that basically
not change the result.
So for this eigenvalue,
we need to compute
now the eigenvector.
So to do that, we basically have
to use minus 3 minus-- I'm just
going to write the system here,
let you see what I'm doing.
And we are solving this system.
So where does this
system come from?
It comes from the fact
that we're looking
for an eigenvector, v_plus, that
is defined as a*v_plus equals
lambda_plus v_plus.
And you can then
bring everything
on the left-hand
side, a minus lambda_i
applied to v_plus gives
us the zero vector.
So that's what we have here.
The unknowns are
a_1, a_2, and we're
going to try to solve for this.
So if we plug in now for
the value of lambda_plus
that we have, we have minus 3
plus 2, which gives us minus 1.
And then, we have a
minus 3i and minus 2.
And for the second line,
second entry of this matrix
you have 5, minus
1 minus minus 2.
So we have 2 minus
1, which is 1.
And then, we have minus 3i.
[a 1, a 2] equals to [0, 0].
So here, you can
check for yourself
that these two equations
given by the first line
and the second line
are actually the same.
And so basically,
to get a_1 and a_2,
it is sufficient to just solve,
for example, the first one,
where here, I just wrote
minus 1 minus 3i multiplied
by a_1 minus 2a_2 equals to 0.
And I just brought the
minus 2 on this side.
So here, you can see that
if we pick a_1 equals to 0--
equals to plus 2, which
would be our first entry,
we can then cancel out
these two and just have
a_2 equals to minus 1 minus 3i.
So this would be one
eigenvector associated
with this eigenvalue.
We could have picked other ones.
They're basically
parallel to this one.
So now what?
So what we need to remember
is the meaning of all of this.
Seeking the eigenvalues and
the eigenvectors is basically
equivalent to seeking a
solution in the form exponential
lambda*t with the direction of
the eigenvector associated with
this eigenvalue.
So now that we actually
have this eigenvector
and this eigenvalue, we can
write down the solution.
And I'm just going to write
the solution in x, which
has entries basically x and y.
And one way of writing
it would be just
to basically first start by
writing what we have there.
I'm just going to spell it out.
So we have this
multiplied by 2 minus 3i.
So what do we do with this?
Well, we remember
our earlier formula.
So this is exponential minus
2t plus exponential 3i*t.
So we can split the exponential
3i*t into a cosine and a sine.
And this, we're going to
also be able to split it
into the complex part
and the real part.
And then, we're going
to combine the real part
and the complex part.
So let's do that.
Exponential minus
2t multiplying,
basically, cosine 3t plus i
sine 3t for the entry 2 minus 1
minus 3i.
So we have an i
here and an i here.
So things can be combined
into a real part.
So in the first entry here,
what are we going to have?
We're going to have
cos 3t multiplying 2.
That's going to be
in the real part.
And I'm going to
keep some space.
And another entry here at the
second entry of this vector
is going to give us cosine
3t multiplied by minus 1.
Oops, here, it should be a 3t.
Sorry.
So minus cosine 3t.
Now, where are we going to
have another real part here?
It's going to come from a
multiplication of i sine 3t
by 3i.
So the two i's together
gives a minus 1.
And we end up with
a plus 3 sine 3t.
So we're done for the real part.
Now let's focus on
the imaginary part.
What do we have?
We have an i sine
3t multiplying a 2.
And we have a minus 3i
here multiplying cosine 3t.
So we want to have
a minus 3 cosine 3t,
and finally, this minus 1
multiplying this sine 3t.
So now, we did--
AUDIENCE: [INAUDIBLE].
PROFESSOR: Oh yeah.
Thank you.
2, from this operation.
So now, we did
split our solution
into a real part and
an imaginary part.
So how can we write the
general solution of the system?
Well, we knew that for this
linear system of equations,
if we have a complex
number that is
a solution to the
linear equation,
then its real part
and its imaginary part
are also two
independent solutions.
So we can write the general
solution of the system
as a linear combination
of the real part
and the imaginary part.
And I can just label this u_1 of
t and u_2 of t here and vector.
And we can then write
the general solution
in terms of any constant--
that would be determined
by the initial condition if
we had one-- exponential minus
2t along u_1 plus c_2
exponential minus 2t
along vector u_2, which
are also functions of t,
just the difference
from what we had before.
So here, basically we seek
for the eigenvalues values
of the matrix.
We looked for the
eigenvector associated
with the complex eigenvalue.
We were able to write
the full solution.
And then, because of
the linearity property,
we were able to
just then extract
two linearly independent
solutions, the real part
of the solution we had and the
imaginary part of the solution
we had.
So what I mentioned
earlier was that we
could do this whole calculation
for the other eigenvalue
with a minus.
If you try to do it and
trickle down your minus,
you would see that
basically you would just
end up with minus
signs here basically
in front of the sines.
And what you could do
then is just simply
absorb that minus sign
for the general solution
in c_1 and c_2.
And basically, it gives
you exactly the same form
for the general solution.
So you don't need to redo
it for the second one.
You would still end up with
only two linearly independent
solutions, not four.
OK, so that ends
this recitation.
