Welcome.
In this section we will learn how to solve
first order linear homogeneous constant coefficient
systems of ODEs using the eigenvalue method.
This introductory lesson will focus on determining
the eigenvalues and corresponding eigenvectors of a matrix.
Supposedly I have a system represented
by the vector equation X prime equals P times X,
where P is a constant matrix, and X and X prime
are vectored valued functions.
We wish to adapt the method
for the single constant coefficient equation
by trying the function E to the lambda T.
However, X is now a vector,
so we try vector X equals vector V
times E to the lambda T.
Times E to the lambda T
where vector V is an arbitrary constant vector.
We plug X and X prime into the differential equation
where X prime is really X prime of T.
To find X prime of T we do have to apply the chain rule,
which on the left, gives us lambda times vector V
times E to lambda T
equals P times vector V times E to lambda T.
Dividing both sides by E to lambda T,
we get the equation below,
lambda times vector V equals P times vector V,
which indicates we're looking for a scaler lambda
and a vector V that satisfies the equation.
To solve this equation,
we need to review some of the near algebra.
Let A be a constant square matrix,
and suppose there is a scalar lambda,
and a non-zero vector V such that matrix A times vector V
equals lambda times vector V.
We call lambda an eigenvalue of matrix A,
and we call vector V a corresponding eigenvector.
As an example, given the two by two matrix
that has an eigenvalue of two,
and a corresponding eigenvector of one zero,
notice the given matrix which we'll call matrix A
times the eigenvector one zero,
is equal to the vector two zero,
which is equal to two times the vector one zero,
where two is lambda the eigenvalue,
and the vector one zero is the corresponding eigenvector.
And now see how we determine eigenvalues
for any square matrix.
We begin with the equation matrix A times vector V
equals lambda times vector V,
and right out in the form shown here on the right,
where we have the difference of A and lambda I
times vector V equals the zero vector.
This equation has a non-zero solution vector V
only if the difference of A an lambda I is not invertible.
If it were invertible,
we could multiply both sides of the equation
by the inverse of the difference of A and lambda I,
which would imply that vector V is equal to the zero vector.
Therefore, matrix A has the eigenvalue lambda
if and only if lambda solves the equation,
the determinant of the difference of A and lambda I
equals zero.
Recall if this difference is zero, then A minus lambda I
doesn't have an inverse and is not invertible.
Consequently, we can find an eigenvalue of matrix A
without finding a corresponding eigenvector
at the same time.
That eigenvector will be found later once we determine,
once we determine lambda the eigenvalue.
And now to determine the eigenvalues
of the given three by three matrix.
Notice this will be the first step
in solving the system represented by
X prime equals the given three by three matrix
times vector X.
To find the eigenvalues,
again we find the determinant
of the difference of A and lambda I,
set it equal to zero, and solve for lambda.
I've already set this up below.
Simplifying inside the parentheses we have the determinant
of the three by three matrix equals zero.
To find the determinant we use expansion by minors.
Using the third row would be easiest
because there are two zero entries in row three.
Notice a determinant would be equal
to the entry in row three column three
which is two minus lambda times the determinant
of the matrix remaining
after eliminating row three and column three.
Notice a determinant of the remaining
two by two matrix is two minus lambda times itself
minus one times one or just minus one shown below.
Simplifying and factoring, we have
negative lambda minus one
times lambda minus two
minus lambda minus three equals zero,
which gives us lambda equals one,
lambda equals two, and lambda equals three
For the eigenvalues.
For an N by N matrix the polynomial we get
by computing the determinant of the difference of A
and lambda I is of degree N,
and hence in general we have N eigenvalues.
However, some may be repeated and some may be complex.
Define an eigenvector corresponding to an eigenvalue lambda.
We write the difference of A and lambda I
times vector V equals a zero vector
and saw for the non-trivial vector V,
which will be an eigenvector.
If Lambda is an eigenvalue,
then there will be at least one free variable
and free distinct eigenvalue lambda.
We can always find an eigenvector.
So going back to the previous three by three matrix
let's find an eigenvector that corresponds
to the eigenvalue lambda equals three.
To do this again, we set up the equation
shown here above.
We have the difference of the given
three by three matrix and lambda times I
times vector V equals a zero vector.
Simplifying inside the parenthesis,
we have the three by three matrix with entries
negative one one one,
one negative one zero,
and zero zero negative one times vector V
equals the zero vector.
And now we need to solve for vector V.
To do this we can write the corresponding augmented matrix
shown here on the right.
And then write the augmented matrix
and reduce row echelon form
which I've already done here below.
Notice of first show indicates V one
minus V two equals zero.
The second row indicates V three equals zero.
The third row is a row of zeros indicating
we have an infinite number of solutions.
Also notice V two is a free variable.
The first equation also indicates that V one equals V two,
and of course V three equals zero.
So if we pick V one equals one,
V one and V two are both one and V three is equal to zero.
Giving us vector V as the vector one one zero
which is a corresponding eigenvector
to the eigenvalue lambda equals three.
And let's go ahead and check this
by checking to make sure the given three by three matrix
times the eigenvector equals the eigenvalue
times the eigenvector.
This gives us the given three by three matrix
times the vector one one zero,
which is equal to the vector three three zero.
Which can be written as three times the vector one one zero.
Notice three is the eigenvalue
and the vector one one zero is a corresponding eigenvector.
The general procedure for solving these types of systems
of ODE's using the eigenvalue method
will be to find the eigenvalues
and the corresponding eigenvectors.
However remember,
there are three possibilities when determining eigenvalues.
We can have distinct real eigenvalues
we can have complex eigenvalues.
We can also have repeated eigenvalues.
The types of eigenvalues we have will affect
how we express the general solution.
I hope you found this introduction helpful.