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.