In this lesson,
we'll consider solving a linear first order system
of ODEs in the form of x prime equals A times x,
using the eigenvalue method,
when matrix A has repeated eigenvalues.
Let's begin by considering the matrix A,
which is the two by two matrix,
with entries three, zero, zero, three.
Let's determine the eeigenvalues.
Recall, to do this, we set up the equation,
the determinant of the difference of A and lambda I,
equals zero, and solve for lambda.
So here we have our setup.
Simplifying, and solving for lambda,
we have lambda equals three.
But notice in this case, the solution of three,
or the root of three, has a multiplicity of two.
Or, we can say the eigenvalue is repeated.
We call the multiplicity of the eigenvalue
in the characteristic equation,
the algebraic multiplicity.
In this case,
there also exists two linearly independent eigenvectors,
which will not always be the case.
Here, two linearly independent eigenvectors
are the vectors one, zero, and zero, one,
corresponding to the eigenvalue of three.
This means the so-called geometric multiplicity
of the eigenvalue, is also two.
Let's review how we found these eigenvectors.
We set up the equation,
the difference of A and lambda I,
times vector V, equals the zero vector,
and then we determine an eigenvector V.
So for lambda equals three, here's the setup, simplifying.
Notice how we have the zero matrix to vector V
equals a zero vector,
which indicates both V1 and V2 are free variables,
which we use
to determine two linearly independent eigenvectors.
For example, we can let V1 equal one, and V2 equal zero.
We can also let V2 equal one, and V1 equal zero,
giving us two linearly independent eigenvectors.
And all the theorems
where we required a matrix to have n distinct eigenvalues,
we only really needed
to have n linearly independent eigenvectors.
For example, X prime times AX, in this case,
has a general solution,
X equals C1 times the eigenvector V1 times e to the 3t,
plus c2 times the eigenvector V2 times e to the 3t.
Again, this only works,
because we were able to find two linearly independent
eigenvectors, which will not always be the case.
Let us restate the theorem about real eigenvalues.
In the following theorem,
we will repeat eigenvalues
according to algebra multiplicity.
So for our matrix A,
we will say that it has eigenvalues three, and three,
or lambda sub one equals three,
and lambda sub two equals three.
Our theorem states,
"Suppose that an n by n matrix P has real eigenvalues,
"not necessarily distinct,
"which are lambda sub one through lambda sub n,
"and there are n linearly independent
"corresponding eigenvectors,
"the vectors V1 through Vn."
Then the general solution to x prime
equals p times x can be written as,
x equals c1 times the eigenvector v1,
times e the power of lambda sub one t,
plus all the way through,
plus c sub n times the eigenvector Vn,
times e the power of lanbda sub nt.
The geometric multiplicity of an eigenvalue
of algebraic multiplicity n,
is equal to the number of corresponding
linearly independent eigenvectors.
The geometric multiplicity
is always less than or equal to the algebraic multiplicity.
The theorem handles the case
when these two multiplicities are equal for all eigenvalues.
If for an eigenvalue, the geometric multiplicity
is equal to the algebraic multiplicity,
then we say the eigenvalue is complete.
In other words, the hypothesis of the theorem
could be stated as,
"If all the eigenvalues of p are complete,
"then there are n linearly independent eigenvectors."
Thus we have the given general solution.
If the geometric multiplicity of the eigenvalue
is two, or greater,
than the set of linearly independent eigenvectors
is not unique up to multiples, as it was before.
For example, for the diagonal matrix A
we considered before,
with entries three, zero, and zero, three,
we could have also picked eigenvectors one, one,
and one, negative one,
or in fact any two pair of linearly independent vectors.
The number
of linearly independent eigenvectors,
corresponding to lambda, is the number of free variables
we obtain when solving a tons v equals lambda times v.
We picked specific values for these free variables,
to obtain eigenvectors.
If we pick different values,
we may get different eigenvectors.
And now let's talk about defective eigenvalues.
If an n by n matrix
has less than n linearly independent eigenvectors,
it is said to be deficient.
Then there is at least one eigenvalue,
with an algebraic multiplicity
that is higher than its geometric multiplicity.
We call the eigenvalue defective,
and the difference between the two multiplicities,
is called the defect.
Let's consider the matrix
with entries three, one, zero, three.
This matrix also has an eigenvalue of three,
with multiplicity two.
Let us try to compute the eigenvectors.
So here we have the setup, simplifying.
Notice in this case, v1 is a free variable,
and v2 must equal zero.
This indicates that every eigenvector
must be in the form of v1, zero.
Any two vectors in the form of v1, zero,
are going to be linearly dependent,
and hence the geometric multiplicity of the eigenvalue
is one.
Therefore the defect is one,
again, because the algebraic multiplicity is two,
and the geometric multiplicity is one,
two minus one is equal to one.
This means we can no longer apply
the eigenvalue method directly to a system of ODEs,
with such a coefficient matrix,
meaning a coefficient matrix in which an eigenvalue
is defective.
Let us continue with the example where matrix A
has entries three, one, zero, three,
and the equation x prime equals Ax.
We found an eigenvalue lambda equals three,
of algebra multiplicity of two, with defect one.
We can find one eigenvector, the vector one, zero,
by letting v1 equal one.
And therefore, we do have one solution,
which is x1 equals the eigenvector one, zero,
times e to the 3t.
But now we are stuck.
We can get no other solutions from standard eigenvectors.
But we need to have two linearly independent solutions,
define the general solution.
Let us try another solution
in the form of x2 equals the sum of v2 and v1 times t,
all times e to the 3t.
Next we differentiate, to find x2 prime.
Notice requires the product rule,
where we have x2 prime equals v1 times e to the 3t,
plus three times the sum of v2, and v1t,
times e to the 3t, which if we distribute,
and gather the e to the 3t, and the t,
e to the 3t terms, we can write x2 prime,
as shown here on the right.
We are assuming that x2 is a solution,
and therefore x2 prime must equal A times x2.
So let's go ahead, and compute A times x2,
which is shown below.
If we distribute and compare the equation on the right,
to the equation above,
by looking at the coefficients of e to the 3t,
and t to the 3t,
notice that 3v2 two plus V1 must equal A times v2,
and 3v1 must equal A times v1.
We can take these two equations,
and write them in the form below,
where we can take the first equation,
and write it in the form
of the difference of A and 3I times v2 equals v1.
And we can write the second equation
in the form of the difference of A and 3I, times v1,
equals the zero vector.
Therefore x2 is a solution,
if these two equations are satisfied.
The second equation is satisfied
if v1 is an eigenvector.
And we found the eigenvector before,
which is the vector one, zero.
So if we can find a vector v2,
that solves the difference of A and 3I, times v2,
equals v1, then we are done,
because we'll have a second independent solution.
And I've set this equation up below.
Again, we have the difference of matrix A, and 3I,
times the vector v2,
which here has components a and b,
equals the eigenvector one, zero.
Simplifying, we get the equation shown here on the right.
Continuing in the next slide,
notice in this case A is a free variable,
and B must equal one.
So if we let A equals zero, we now have a second vector,
vector v2, which is the vector zero, one,
and we can write the general solution.
We recall earlier we found x1
equals the eigenvector v times e the power 3t,
which for the general solution,
gives us x equals c1 times the vector one, zero,
times either the three t,
and then using the vectors v2 and v1,
we now know x2, shown below,
is another independent solution,
which gives us plus c2 times the sum of the vector v2,
which is the vector zero, one,
plus the eigenvector v1,
which is the vector one, zero, times t,
all times e to the 3t,
which if we multiply,
we get the form shown here on the right.
Before we go, let's describe the general algorithm.
Suppose that lambda is an eigenvalue
of multiplicity two, defect one.
We first find an eigenvector v1 of lambda,
that is the vector v1
that solves the difference of A and lambda I
times the vector v1 equals zero.
And then we find a second vector,
the vector v2 such as the difference of A and lambda I,
times the vector v2, equals the eigenvector v1.
This gives us two linearly independent solutions,
in the form shown below,
which we can then use to determine the general solution.
I hope you found this helpful.