Here, we're given a matrix A
which is a two by two matrix
and we're asked to find
the eigenvalues of A
and then, find a corresponding eigenvector
for each lambda that is a unit vector.
So, to find the eigenvalues
values of a square matrix A,
we want to find the scalars lambda
that satisfy this equation here
where you have the
determinant of the difference
of lambda, I minus A equals zero.
And then, to find the eigenvectors of A
corresponding to each lambda,
we define the solutions
to this equation here.
So, I've already set
up the equation we use
to find the eigenvalues
where we have the determinant of,
again, lambda, I would be lambda times
the two-by-two identity matrix
which is this matrix here.
Then, we have minus matrix A equals zero.
Performing the matrix subtraction,
we have lambda minus one,
zero minus negative 3/2
which is a positive 3/2,
zero minus 1/2 which is negative 1/2,
and lambda minus negative
one which is lambda plus one.
Now, the determinant is equal to
this product minus this product.
So, in this case, we'd have
the quantity lambda minus one
times the quantity
lambda plus one minus 3/2
times negative 1/2 equals zero.
Multiplying this out here,
we have lambda squared
plus lambda minus lambda, that's zero.
And then, minus one
and this becomes plus 3/4 equals zero.
Well, negative one plus
3/4 is negative 1/4.
So, we have lambda squared
minus 1/4 equals zero.
This does factor, but we can
also add 1/4 to both sides
to get lambda squared equals positive 1/4.
And then, we'll take the square root
of both sides of the equation.
But, we are going to
have two solutions here.
We'll have lambda equals
plus or minus 1/2.
The square root of 1/4 is 1/2.
So, let's go ahead and say
lambda sub one equals negative 1/2
and lambda sub two equals positive 1/2.
So, now that we have the
eigenvalues for the given matrix,
we'll find the corresponding eigenvectors
that are unit vectors.
Again, to do this, we'll first
find all the eigenvectors
by solving this equation
here for vector x.
Let's do this on the next slide.
Let's begin by determining
the eigenvectors
corresponding to lambda sub one
equals negative 1/2.
Again, I've already set up
the equation that we use
to find the corresponding eigenvectors
which is the difference
of lambda, I minus A
times vector x equals a zero vector.
So, again, subtracting these two matrices
gives us this matrix here.
The components of vector
x or x sub one, x sub two
and the zero vector is here.
So, now, we substitute
negative 1/2 for lambda.
So, we'd have negative 1/2 minus one,
that's negative 3/2, 3/2.
Second row would be negative 1/2.
And negative 1/2 plus one is positive 1/2
times vector x
equals the zero vector.
Now, let's write the
corresponding augmented matrix
and write that in reduced row echelon form
to solve for x sub one and x sub two.
So, the corresponding augmented matrix
would have a first row of
negative 3/2, positive 3/2, zero
and a second row of negative
1/2, positive 1/2, zero.
And now, let's get a zero
in this position here.
Notice that negative
three times negative 1/2
would be positive 3/2.
If we add that to negative
3/2, you'd get zero.
Let's replace row two with negative three
times row two plus row one.
Let's get a leading entry of one here
by replacing row one with
negative 2/3 times row one.
Replacing the first row with
negative 2/3 times row one,
we'd have one,
negative one, zero
and then, negative three times
negative 1/2 is positive 3/2
plus negative 3/2, that's zero.
Negative three times 1/2 is negative 3/2
plus positive 3/2, that's zero.
Then, we have zero.
As expected, we have a row of zeros
which means you do have an
infinite number of eigenvectors
corresponding to lambda sub
one equals negative 1/2.
But the first row indicates that
x sub one minus x sub two must equal zero.
So, this also tells us that
x sub one must equal x sub two.
So, if we parametrize this
relationship in terms of t,
we can say the eigenvectors
corresponding to
lambda sub one equals
negative 1/2, vector x equals
all the vectors where the x-component is t
and the y-component is also t.
Which you can also express as t times
the vector where the x-component is one
and the y-component is one.
But remember, the eigenvectors
can't be the zero vector.
So, we should say t can't equal zero.
But, this doesn't quite
answer our question.
We do want to find a unit vector
corresponding to lambda
equals negative 1/2.
So, notice how if we let t equal one,
we would just have the
vector one-comma-one.
So, let's say one unit vector
corresponding to lambda
equals negative 1/2
would be equal to vector x
divided by the magnitude of vector x.
Again, where t equals one, the
vector is just one-comma-one
and therefore, the magnitude would just be
the square root of one
squared plus one squared,
the square root of two.
So, one unit vector corresponding to
lambda equals negative 1/2
would be the vector with an x-component of
one divided by square root two
and a y-component of one
divided by square root two.
Now, it's also true we could use
a unit vector in the opposite direction.
So, we could also say the unit vector
with an x-component of negative one
divided by square root two
and a y-component of negative one
divided by square root two.
Let's go ahead and use
this unit vector here.
And now, I need to go
through this process again
for lambda equals positive 1/2.
So, for lambda sub two
equals positive 1/2,
we would have 1/2 minus
one, that's negative 1/2,
3/2, negative 1/2,
and 1/2 plus one is positive 3/2
times vector x
equals the zero vector.
And the corresponding augmented matrix
would have a first row of
negative 1/2, positive 3/2, zero
and a second row of negative
1/2, positive 3/2, and zero.
Notice how these two rows
are actually the same.
So, let's get a row of zeros here
by replacing row two with negative one
times row two plus row one.
Let's get a leading entry of
one here by replacing row one
with negative two times row one.
So, the first row is going to be one
and then, 3/2 times negative
two is negative three
and we have zero.
And then, again, because row
one and row two are the same,
negative one times row two plus row one
is going to give us a row of zeros.
So, this first row indicates
that x sub one minus three
times x sub two must equal zero
or we can also say that x sub one
equals three times x sub two.
Which means, all the
eigenvectors corresponding to
lambda sub two equals
1/2 would be the vector x
where if we let x sub two be equal to t,
we could say the x-component
would have to be in the form of three t
and the y-component would
have to be in the form of t.
If we wanted to, we could
say t times the vector
with an x-component of three
and a y-component of one.
Again, where t can't equal zero
because the eigenvector
can't be the zero vector.
But, again, we're asked
to find a unit vector
corresponding to lambda equals 1/2.
So, if we let t equal one,
we'd have the vector three-comma-one.
So, again, if the unit vector is equal to
vector x divided by its magnitude,
let's say when t equals one,
the x-component would be three
divided by the square root of 10
because the magnitude of
vector x when t is one
would be the square root of
three squared plus one squared
of the square root of 10.
And the y-component would be
one over the square root of 10.
Of course, we could also
give the unit vector
in the opposite direction.
So, we could also say thinking of vector
with the x-component of negative three
divided by the square root 10
and a y-component of negative
one divided square root 10.
But it only says, "Give a
corresponding eigenvector...
"that is a unit vector."
So, we'll go ahead and
give the final answer
in this form here.
So, to summarize what we found,
we found the eigenvectors of matrix A
where lambda equals plus or minus 1/2.
I think I interchanged
the subscripts here,
but I think you get the idea.
Where the eigenvectors
corresponding to lambda equals 1/2
are in the form t times the vector
with an x-component of three
and a y-component of one
where t doesn't equal zero.
We can also say the eigenspace
corresponding to lambda equals 1/2
is given by the span of this vector here.
Where unit vector when t equals one
would be this unit vector here.
And then, for the eigenvectors
corresponding to lambda
equals negative 1/2,
the eigenvectors x would be in this form.
Again, where t doesn't equal zero.
We can say the eigenspace
corresponding to lambda
equals negative 1/2
is given by the span of this vector.
And a unit vector when t
equals one is this vector here.
Remember, for the unit vectors though,
we could also get the unit vectors
in the opposite direction.
I hope you found this helpful.
