We are given the three-by-three matrix A
with entries one, three, three, one, one, zero,
and negative one, one, two.
We're asked, what are the eigenvalues?
What are the defects of the eigenvalues?
And we're asked to find the general solution
to x prime equals A times x.
Let's begin by determining the eigenvalues.
We do this by setting up the equation of the determinant
of the difference of matrix A and lambda I equals zero,
and then we solve for lambda.
So here we have our setup,
where we have, again,
the determinant of matrix A minus lambda
times the three-by-three identity matrix equals zero.
Simplifying inside the parentheses,
we now evaluate the three-by-three determinant
using row two, because row two has a zero entry.
The determinant is equal to the entry
in row two, column one, which is one,
times the determinant of the matrix
after deleting row two and column one
and then minus the entry in row two, column two,
which is one minus lambda,
times the determinant of the two-by-two matrix
after deleting row two and column two.
We can stop here because we'd have plus
the entry in row two, column three, which is zero,
times the determinant of the matrix
after deleting row two and column three,
which would all result in zero.
So now we evaluate the two-by-two determinants,
simply and factor,
which leaves us with the square of lambda minus one
times lambda minus two equals zero,
indicating the eigenvalues are one and two.
However, notice we have two factors of lambda minus one,
which means the eigenvalue of one
has an algebraic multiplicity of two.
To keep track of this,
it might be helpful to write lambda sub one
and lambda sub two are equal to one
and lambda sub three equals two.
But I think the important part to recognize here
is the eigenvalue of one
has an algebraic multiplicity of two.
For part b, we're asked to find the defects
of the eigenvalues.
Eigenvalues have defects if the algebraic multiplicity
is more than the geometric multiplicity.
And the geometric multiplicity
corresponds to how many linearly independent eigenvectors
we can find for an eigenvalue.
So in this case, the only possible eigenvalue
that may have a defect would be the eigenvalue of one,
because it has an algebraic multiplicity of two.
So if we're not able to determine
two linearly independent eigenvectors
for the eigenvalue of one,
the eigenvalue of one will be defective.
So for the next step,
we will see if we can find
two linearly independent eigenvectors
for the eigenvalue of one.
To do this, we set up the equation
the difference of A and lambda I
times vector v equals the zero vector,
and then we solve for vector v.
So here we have our setup.
We have matrix A minus one times
the three-by-three identity matrix times vector v
equals the zero vector.
Simplifying, we now need to solve this system on the right.
Let's use an augmented matrix
and write it and reduce to echelon form.
Notice v3 is a free variable.
The first row indicates v1 equals zero.
The second row indicates v2 equals negative v3.
And again, v3 is a free variable.
So if we let v3 equal negative one,
we have the eigenvector v1,
which is the vector zero, one, negative one.
From here, we're not going to be able to determine
a second linearly independent eigenvector
because, if we select a different v3,
it'll be a scalar multiple of the eigenvector v1,
and the two vectors will not be linearly independent.
So again, we cannot find another eigenvector
using this method.
As a result, the eigenvalue of one has one defect.
Lambda equals one has algebraic multiplicity two
and geometric multiplicity of one,
because we only found one eigenvector.
And two minus one is equal to one,
giving us a defect of one.
When it comes to finding a general solution
to x prime equals A times x,
we do need a second linearly independent solution
for lambda equals one.
To do this, we'll have to follow the procedure below,
where we'll have to solve the equation
the difference of A and lambda I times the vector v2
equals the eigenvector v1.
We'll solve this for vector v2,
which will give us a second linearly independent solution
for lambda equals one in the form shown below.
Let's work on this next.
Because we only found one eigenvector
for the eigenvalue of one,
which has algebraic multiplicity of two,
we now need to set up and solve the equation
the difference of A and lambda sub one I
times v2 equals the eigenvector v1.
So I have the setup here.
Again, notice the right side is not the zero vector.
It's the eigenvector v1.
Simplifying, we now need to solve this system on the right,
which, again, we'll do using an augmented matrix
and then write the augmented matrix
and reduce to echelon form.
We have v1 equals one, v2 equals negative v3,
and v3 equals v3.
So if we let v3 equal zero,
we have the vector v2 is the vector one, zero, zero,
which we can now use to determine
a second linearly independent solution
using the eigenvalue of lambda equals one.
So this is information that we now have,
which also means,
one independent solution to x prime equals A times x
is x1 equals the eigenvector v1 times e
to the power of one t.
But we also now know a second independent solution
is x2 equals the sum of the vector v2
and the eigenvector v1 times t,
all times e to the power of one t.
We still need to find a third linearly independent solution
for the eigenvalue of two
before we can determine a general solution
to x prime equals A times x.
Let's do that now.
Using lambda sub three equals two,
we set up our equation.
Simplifying, we now need to solve the system on the right,
which again we'll do using an augmented matrix
and then write it and reduce to echelon form.
Row one indicates that v1 equals negative 3/2 v3.
The second row indicates v2 equals negative 3/2 v3.
And v3 is a free variable.
So if we let v3 equal negative two, for example,
we have the eigenvector three, three, negative two,
which we'll call the eigenvector v3.
So this indicates a third linearly independent solution
to the system indicated by x prime equals A times x
is x3 equals the eigenvector v3
times e to the power of two t.
And now we can put all these pieces together
to write the general solution for this system.
We have x of t equals c1
times the eigenvector zero, one, negative one
times e to the t plus c2
times the vector v2, which is the vector one, zero, zero,
plus the eigenvector v1,
which is the vector zero, one, negative one, times t,
and this sum is multiplied by e to the t,
and then finally, we have plus c3
times the eigenvector v3
times e to the power of two t.
This is our general solution.
I hope you found this helpful.