Here we're given matrix A
which is a three by three matrix.
We're asked to find the eigenvalues of matrix A
and then find a corresponding eigenvector
for each lambda that is a unit vector.
So we'll first find the eigenvalues
by solving this equation here for lambda
where we have the determinate
of the difference of lambda I and A equals zero.
And then we'll find all the eigenvectors of A
corresponding to each lambda
which are the nonzero solutions to this equation here.
Once we find the eigenvectors,
we'll then find a unit eigenvector for each lambda.
So here's the setup of the first equation
to find the eigenvalues.
This is the matrix lambda times I.
IZ three by three identify matrix.
The difference of these two matrices
will give us a determinate shown here.
We notice how for the first row we have lambda minus two,
zero minus negative two is positive two.
Zero minus three is negative three and so on.
So to evaluate this three by three determinate,
we use expansion by minors
of the cofactor method using row one.
The first element in row one
is the quantity lambda minus two.
And then we'll have times the two by two determinate
formed by eliminating row one and column one,
which will give us lambda minus three, two, one,
and lambda minus two.
Then we have minus, the next element in row one is two
times a two by two determinate
formed by eliminating row one column two,
so we have zero, two, zero, lambda minus two
plus the third element in row one is negative three,
then we have times the determinate
formed by eliminating row one column three.
So we'd have zero, lambda minus three, zero, one.
And this must equal zero.
And now the value of each two by two determinate
is equal to this product minus this product.
So we have the quantity lambda minus two times
the quantity lambda minus three times
the quantity lambda minus two, minus two times one,
and then we have minus two times,
here we'll have zero minus zero which is zero.
This will be minus three times zero minus zero,
which is zero.
So we have the quantity lambda minus two times,
let's find this product here.
So we have lambda squared,
minus two lambda, minus three lambda, minus five lambda,
plus six, but then we have minus two,
and of course, this is zero and this is zero.
So this product equals zero.
Continuing on the next slide,
six minus two is four.
So we have the quantity lambda minus two
times the quantity lambda squared minus five lambda
plus four equals zero.
This trinomial factors.
So we'll have two more binomial factors.
The factors of lambda squared are lambda and lambda.
The factor of four that add to negative five,
negative four and negative one.
So we have three solutions.
We have lambda equals two, lambda equals four,
and lambda equals one.
So we have three eigenvalues for the matrix A,
let's call it lambda sub one equals one,
lambda sub two equals two, and lambda sub three equals four.
So this is the first part of our question.
Now we want to the eigenvectors
corresponding to these eigenvalues,
so that we can then find a unit eigenvector for each lambda.
So to find the eigenvectors of A
corresponding to each lambda,
we're going to find the nonzero solutions
to this vector equation here.
Again, I've already set this up.
Here we have the difference of lambda I in matrix A
times vector x which is the eigenvector
equals the zero vector.
The difference of these two matrices
gives us this matrix here
relating the components of vector x
would be x sub one, x sub two, and x sub three,
and here's a zero vector.
So now for lambda sub one equals one,
we substitute one for lambda to form our matrix equation.
So lambda is equal to one.
This first row would be one minus two
that's negative one, two, negative three.
The second row would be zero,
one minus three is negative two, two,
and the third row would be zero, one, negative one
times the vector x
equals the zero vector.
Now let's write this matrix equation as an augmented matrix
and then write the augmented matrix
in reduced row-echelon form.
So the first row would be
negative one, two, negative three, zero.
Second row would be zero, negative two, two, zero.
Third row would be zero, one, negative one, zero.
We've already written several of these
in reduced row-echelon form,
so I'll write this down to save some time.
This is the reduced row-echelon form
of the given augmented matrix.
Notice how we have a row of zeroes which indicates
we have an infinite number of eigenvectors,
but notice how this first row tells us
that x sub one minus two x sub two plus
three x sub three must equal zero.
And the second row indicates that x sub two
minus x sub three must equal zero.
Let's go ahead and solve this first equation for x sub one.
So we'd have x sub one equals two times x sub two
minus three times x sub three.
And the second equation we can write it
as x sub two equals x sub three.
So now to give the eigenvectors corresponding
to lambda equals one, let's parameterize this relationship
by letting x sub three be equal to t.
So if x sub three is equal to t,
the eigenvectors would have to have a z component of t.
And because x sub two equals x sub three,
the y component would also be t.
And so if x sub two and x sub three are both equal to t,
notice that x sub one is equal to
two t minus three t or negative t.
So this would be one way to express the eigenvectors
corresponding to lambda equals one.
We could also write this as t times the vector
with an x component negative one,
a y component of one and a z component of one
where again t can't equal zero,
because the eigenvector must be a nonzero solution.
But the question asked for a unit eigenvector.
So for one eigenvector, if we let t equal one,
we'd have the eigenvector v,
with an x component of negative one,
a y component of one, and a z component of one,
which means a unit eigenvector would be equal to
vector v divided by the magnitude of vector v.
The magnitude of vector v would be equal to
the square root of negative one squared
plus one squared plus one squared
which would be the square root of three.
So one unit eigenvector
would be the vector with an x component
of negative one divided by the square root of three,
a y component of one divided by the square root of three,
and a z component of positive one
divided by the square root of three.
But we could also give the unit vector
in the opposite direction.
So let's call this unit vector sub one,
and let's say unit vector sub two
again could be the opposite unit vector.
So we would just change the sign of each component
so we'd have one divided by the square root of three,
negative one divided by the square root of three,
and negative one divided by the square root of three.
Either of these last two unit vectors
would be a unit eigenvector corresponding
to lambda sub one equals one.
Now, I need to go through the same process
for lambda sub two and lambda sub three.
So for lambda sub two equals two,
we substitute two for lambda in our matrix equation.
So our matrix here would be zero, two, negative three,
then zero, negative one, two, and zero, one, zero.
So the corresponding augmented matrix would be
a three by four matrix.
The first row would be zero, two, negative three, zero.
The second row would be zero, negative one, two, zero.
The third row would be zero, one, zero, zero.
The next step is to write this in reduced row echelon form
which again, to save some time, I've already done.
Notice how this first row indicates
that x sub two minus 3/2,
x sub three equals zero.
And the second row indicates
that x sub three must equal zero.
Let's write this first equation
as x sub two equals 3/2 times x sub three,
and x sub three equals zero.
So the eigenvectors corresponding
to lambda sub two equals two.
It would have to have a z component of zero,
because the z component is x sub three.
Well if x sub three is equal to zero,
notice that x sub two is also equal to zero.
Let's go ahead and let x sub one
or the x component be equal to t.
So this is one way to express the eigenvectors
corresponding to lambda sub two equals two.
We could also write this as t times
the vector with an x component of one
and then y and z component of zero.
But we're asked to find a unit eigenvector
corresponding to lambda.
Notice how if we let t equal one,
we just have the eigenvector one comma zero comma zero,
which is a unit vector.
So unit vector could be the vector
with an x component of one
and then y and z component of zero
or let's call this vector u sub one.
Vector u sub two could be the unit vector
in the opposite direction
which would be negative one, zero, zero.
Either of these two vectors would be acceptable.
And now we need to find a unit eigenvector
for lambda sub three equals four.
So lambda sub there equals four.
So substituting four for lambda into our matrix equation,
we have two, two, negative three.
Then zero, one, two, and zero, one, two.
So again, the corresponding augmented matrix
would be two, two, negative three, zero.
Second row would be zero, one, two, zero.
Third row would be zero, one, two, zero.
The next step would be to write this
in reduced row echelon form,
which again, to save some time, I've already done.
Here's the reduced row echelon form
of the augmented matrix.
So looking at row one and row two.
Row one tells us that x sub one minus 7/2
times x sub three must equal zero.
And the second row tells us x sub two
plus two times x sub three equals zero.
Let's write this first equation
as x sub one equals 7/2 x sub three.
And let's write the second equation
as x sub two equals negative two times x sub three.
Let's parameterize this using t.
So we can say that the eigenvectors
corresponding to lambda equals four
would be the vectors in the form
if x sub three or the z component is equal to t,
then x sub two or the y component would be negative two t,
and the x component or x sub one would be equal to 7/2 t.
Now if we wanted to clear this fraction here,
we could multiply it by two.
Two times this vector
would give us a vector with an x component of seven t,
a y component of negative four t,
and a z component of two t.
We could also write this as t times
the vector with an x component of seven,
a y component of negative four,
and a z component of two,
where, again, t can't equal zero.
So this would be all the eigenvectors
corresponding to lambda equals four.
Remember our goal is to find a unit vector.
So if we let t equal one.
One eigenvector or vector v would have
an x component of seven, a y component of negative four,
and a z component of two.
So if we wanted to find a unit eigenvector,
we could let the unit vector be equal to
vector v divided by the magnitude of vector v,
with a magnitude of vector v will show this one
is equal to the square root of seven squared
plus negative four squared plus two squared,
which is equal to the square root of 49 plus 16 plus four
is equal to 69.
So the magnitude is equal to the square root of 69.
So one unit eigenvector would be the vector
with an x component of seven
divided by the square root of 69,
a y component of negative four
divided by the square root of 69,
and a z component of two divided by the square root of 69.
So this would be one unit eigenvector,
let's call it vector u sub one,
but we could also give the unit eigenvector
in the opposite direction.
So let's say vector u sub two,
which would just be the opposite vector,
so we'd have negative seven
divided by the square root of 69,
positive four divided by the square root of 69
and negative two divided by the square root of 69.
Either of these two unit vectors
would be correct.
I hope you found this helpful.