We're given a matrix A,
which is a three by three matrix,
we're asked to find the
eigenvalues of matrix A
and then find the
corresponding eigenvectors
for each lambda.
So to find the eigenvalues
of the square matrix A,
we need to find the values
of lambda that satisfy
this equation here,
where we have the
determinant of the difference
of lambda I minus A equals zero.
And then to find the eigenvectors of A
corresponding to lambda,
we need to find the
non-zero solutions for this equation here
where we solve this equation for vector X.
Begin, we'll first find the eigenvalues.
So we have the determinant
of the difference of,
here's lambda times I
where I is the three by three
of the itinerary matrix,
minus the given matrix A equals zero.
Performing this matrix subtraction
gives us this determinant here,
where this determinant must equal zero.
So to evaluate this determinant
we'll use expansion by minors,
or the cofactor method using row one.
So starting with the first
element in row one we have lambda
times the determinant
formed by eliminating row one column one,
so I'd have lambda minus four, ten,
zero and lambda minus four.
And we have minus, the next
element in row one is three,
times the two by two determinant,
formed by eliminating row one column two,
so we'd have four, ten,
zero, lambda minus four.
Plus the last element in
row one is negative five,
times the determinant where
the elements are found
by eliminating row one, column three.
So we have four, lambda minus four,
zero, zero, and this
determinant must equal zero.
And now each two by two
determinant is equal to
this product minus this product.
So here we have lambda times
the quantity lambda minus
four, times lambda minus four,
let's write that as
lambda minus four squared
minus 10 times zero that's zero.
And we have minus three times
four times lambda minus four,
minus 10 times zero which is zero.
And here we have minus
five times zero minus zero.
And so this would be zero,
and these products here
contain a common factor
of lambda minus four,
so if we factor out one
factor of lambda minus four
from these two products,
we'd have lambda minus four times,
here we'll have lambda times one factor
of lambda minus four.
And here we just have
minus three times four.
Again we factored one
factor of lambda minus four
from this product, giving us
lambda times lambda minus four.
Then we factored out
lambda minus four from here
leaving us with minus three times four.
Let's continue solving
this on the next slide.
Simplifying inside the brackets,
we'd have lambda squared,
minus four lambda,
minus 12, and this trinomial does factor,
so we have lambda minus four,
we'll have two more binomial factors,
the factors of lambda square
are lambda and lambda.
The factors of negative 12
to that are negative four
and negative six and positive two.
We have three solutions to this equation,
so we have three eigenvalues.
Let's list them in order
from least to greatest.
We'd say lambda one equals negative two,
lambda sub two is equal to positive four,
and lambda sub three is
equal to positive six.
And now you need to find the
corresponding eigenvectors
for each of these eigenvalues.
So again to find the eigenvectors of A
corresponding to lambda,
we need to find the non-zero solutions
to this equation here.
So let's find the eigenvectors
corresponding to lambda
sub one equals negative two.
Again I've already set some of this up.
Here we have the difference
of lambda I minus A,
times vector X equals zero vector.
Subtracting these matrices
gives us this matrix here.
Here are the components of vector X,
and here's the zero vector.
Now for the next step I'll substitute
negative two for lambda,
so we'll have negative
two, three, negative five
for the first row.
Second row will be four, negative six, 10.
The third row'll be
zero, zero, negative six.
And now I'll write the corresponding
augmented matrix
and write it in reduced row echelon form,
this all for X sub one, and
X sub two and X sub three.
So the first row is going
to be negative two, three,
negative five, zero.
The second row four,
negative six, 10, zero.
The third row zero,
zero, negative six, zero.
Let's continue on the next slide.
Let's get a zero in this position here.
Notice that two times negative
two, plus four is zero,
Let's replace row two with two
times row one, plus row two.
Let's get a leading entry of one here
by replacing row three with
negative 1/6 times row three.
The first row stays the same.
Second row is going to be
two times negative two,
plus four that's zero.
Negative two times three
plus negative six is zero.
Two times negative five plus 10 is zero,
and two times zero plus zero is zero.
And the third row is going
to be zero, zero, one, zero.
Now let's get a zero
on this position here,
by replacing row one with
five times row three,
plus row one.
Let's also interchange
row two and row three.
So five times zero plus
negative two is negative two,
five times zero plus three is three,
five times one plus negative five is zero,
five times zero plus zero is zero.
Interchanging row two and row three.
Now for the last step
let's replace row one
with negative one half times row one,
so we have a leading entry of one.
So the first row
would be one, negative
three halves, zero, zero.
Second row zero, zero, one, zero,
third row is all zeros.
So because we have a row
of zeros which is expected,
there's an infinite number of eigenvectors
corresponding to lambda sub
one equals negative two.
But notice how this first row indicates,
x sub one minus three halves
X sub two equals zero.
The second row indicates
that X sub three equals zero.
So we can say X sub one equals
three halves times X sub two,
and X sub three equals zero.
So if you parameterize this with T,
we can say the eigenvectors
corresponding to lambda sub
one equals negative two,
must be in the form where
the Z component must be zero.
If we let X sub two be equal to T,
we could say the X
component would have to be
three halves T.
Now if we don't want
fractions here we could also
perform scalar multiplication
and clear the fractions
by multiplying by 2,
let's go 'head and do that
even though it's not needed.
We could say the
eigenvectors corresponding to
lambda sub one equals negative two,
are in the form of three T
comma, two T, comma zero,
where T can't equal zero because
the eigenvector can't be the zero vector.
Course we could also write this as
vector X equals T times
the vector with an X component of three,
a Y component of two,
and a Z component of zero.
And now I need to do the
same for lambda sub two
and lambda sub three.
So as you can see it, this
is quite a bit of work.
For lambda sub two equals positive four,
we substitute four for lambda here.
So we'd have four, three, negative five,
four, zero, 10, zero, zero, zero.
Again let's write the
corresponding augmented matrix.
So the first row is four,
three, negative five, zero.
Second row is four, zero, 10, zero.
The third row would be
zero, zero, zero, zero.
Let's write this in
reduced row echelon form
to solve for X sub one, X
sub two, and X sub three.
Let's get a zero in this position here,
by replacing row two with
negative one times row one,
plus row two.
So row one and row three stay the same.
So we have negative one times
four, plus four that's zero.
Negative one times three,
plus zero is negative three.
Negative one times five plus 10 is 15.
And negative one times
zero plus zero is zero.
Now let's get a zero in this position here
by replacing row one with
row one plus row two.
And I have a leading entry of one here,
let's replace row two with negative 1/3
times row two.
So, four plus zero is four,
three plus negative three is zero,
negative five plus 15 is 10,
zero plus zero is zero,
1/3 times row two would be
zero, one, negative five, zero.
And for our last step we need
a leading entry of one here,
we'll replace row one
with 1/4 times row one.
Row two and row three stay the same.
So we'd have one, zero, 10
fourths or five halves, zero.
So this first row indicates
X sub one, plus five halves,
X sub three must equal zero.
The second row indicates that X sub two
minus five times X sub
three must equal zero.
So we can also write this as
X sub one equals negative
five halves times X sub three,
and X sub two equals
five times X sub three.
So we can parameterize this relationship
by letting x sub three be equal to T.
So the eigenvectors corresponding to
lambda sub two equals four,
are in the form of
if X sub three is equal to T,
then X sub two would be equal to five T,
and X sub one would be equal
to negative five halves T.
But of course if we do want
to clear the fractions here
we could multiply by two,
and say the eigenvectors
are in the form of
negative five T, comma 10 T, comma two T.
If we want we could also factor out the T
and write this as
the eigenvector's in the form of T times
the vector with an X
component of negative five,
Y component of 10, a Z component of two,
again where T doesn't equal zero,
because the eigenvector
can't be the zero vector.
I now need to find the eigenvector for
lambda sub three equals six.
So for lambda sub three equals six,
in our matrix equation,
this first matrix would be
six, three, negative five.
Second row would be four, two, 10.
Third row would be zero, zero, two.
So the corresponding augmented matrix
would have a first row of six,
three, negative five, zero.
Second row would be four, two, 10, zero.
Third row would be zero, zero, two, zero.
Let's write this in
reduced row echelon form
on the next slide.
Let's get a zero in this position here
by replacing row one
with two times row one,
plus row two.
Notice how I'd have two
times negative five plus 10,
which would give us zero here.
Let's also get a zero in this position.
Notice how negative five times
two plus 10 would be zero.
Let's replace row two with
negative five times row three,
plus row two.
And let's replace row three
with 1/2 times row three,
to get a leading entry of one.
So in the first row we have
two times six plus four,
that's 12 plus four is 16.
And we have two times three
plus two, that's eight.
Two times negative five plus 10 is zero.
And two times zero plus zero is zero.
And then we have negative
five times row three,
plus row two,
so we'd have four, two,
negative five times two,
plus 10 is zero,
negative five times
zero plus zero is zero.
1/2 times row three we'd
have zero, zero, one, zero.
Let's get a zero here
by replacing row two with
negative four times row two,
plus row one, and let's
replace row one with
1/16 times row one.
So the first row would
be one, 1/2, zero, zero.
Second row is going to be
negative four times four,
plus 16 that's zero,
negative four times
two, plus eight is zero,
remaining elements would also be zero.
And the last row is zero, zero, one, zero.
So for our last step, we'll interchange
row two and row three.
So we have one, 1/2, zero, zero.
Zero, zero, one, zero,
and zero, zero, zero, zero.
So the first row indicates that X sub one
plus 1/2 X sub two equals zero.
Second row indicates that
X sub three equals zero,
so we can write this as X sub one equals
negative 1/2 X sub two,
and X sub three must equal zero.
So if your paramertized
relationship using T,
we can say the eigenvectors
corresponding to
lambda sub three equals six
would have to be in the form where
the z component would be zero.
And again if X sub two is equal to T,
then X sub one would be negative 1/2 T.
If we did want to clear the fractions,
we could multiply this by two,
and write this as negative T, two T, zero,
again, where T can't equal zero.
Or we could factor out
the T and write this as
the eigenvectors must be
in the form of T times
the vector with an X
component of negative one,
a Y component of two, and
a Z component of zero.
So the eigenvectors
are any scalar multiple
of this vector here.
We found the three eigenvalues
of the square matrix A.
And the eigenvectors corresponding to
lambda equals negative two,
are the vectors in this form here,
where T doesn't equal zero.
We can also say the eigenspace
corresponding to this lambda
is given by the span of the vector
three comma two comma zero.
The eigenvectors corresponding
to lambda equals four
are these vectors, where
T doesn't equal zero,
and the eigenspace corresponding
to lambda equals four
is given by the span
of the vector with components
negative five comma
10 comma two.
The eigenvectors corresponding
to lambda equals six
are the vectors in this form.
And the eigenspace corresponding
to lambda equals six
is given by the span
of the vector with components
negative one comma two comma zero.
I hope you found this helpful.
