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.