Welcome to a lesson
on how to determine the eigenvectors
corresponding to the eigenvalues of a two by two matrix.
In the previous lesson,
we found the eigenvalues of matrix A
to be negative three and positive two.
Once we have the eigenvalues of a given matrix,
we can find the eigenvectors
by solving the eigenvector equation,
which is equivalent to either of the equations shown below.
We will use this first form
where we have the difference of matrix A
and Lambda times the identity matrix
times vector X equals the zero vector,
where X as an eigenvector.
So setting up the equation.
Let's let Lambda sub one equal negative three.
Using our equation,
we have matrix A minus negative three
times two by two identity matrix
times vector X equals the zero vector.
From here, let's write the corresponding matrix equation,
wherefore the coefficient matrix,
the first row is negative five plus three times one,
which is negative two.
And then we have two plus three times zero,
which is two.
In the second row,
we have negative seven plus three times zero,
which is negative seven.
And four plus three times one,
which is seven.
The first role of the corresponding unweighted matrix
is negative two, two, zero.
And the second row is negative seven, seven, zero.
Writing the augmented matrix in reduced row echelon form,
on the far-right,
notice how there's no pivot in column two,
which means X two is a free variable.
The first row indicates that X one minus X two equals zero.
And because X two is a free variable,
we have X two equals X two.
solving the first equation for X one,
we have X one equals X two.
We can parametrize the solution by letting X two equal T.
In this case,
notice both X one and X two are equal to T.
And therefore, we can say,
all the eigenvectors X are the vectors in the form of TT
or T times the vector one, one,
which indicates the eigenvectors
are all the scalar multiples of the vector one, one,
except remember, an eigenvector can't be the zero vector,
and therefore, T can't equal zero.
So because the eigenvectors are in the form of T
times the vector, one, one,
we can say that for Lambda sub one equals negative three,
a basic eigenvector is the vector one, one.
And let's just check
that it does satisfy the eigenvector equation.
Matrix A times vector X equals Lambda times vector X.
Looking at the work below,
we can see the equation is true for the vector one, one,
because the vector negative three, negative three
is equal to negative three times the vector one, one.
If Lambda is an eigenvalue of matrix A,
then the eigenspace of matrix A corresponding to Lambda
is a set of all vectors corresponding to Lambda.
And for the eigenspace, we do include the zero vector.
So the eigenspace of matrix A
corresponding to Lambda equals negative three
are all the scalar multiples of the vector one, one.
Or we can say it's equal to the span
of the set containing the the vector one, one.
We can also state that a basis
for the eigenspace of matrix A
corresponding to Lambda equals negative three
is the vector one, one.
And now, let's consider
the second eigenvalue of positive two.
Again, we find the corresponding eigenvectors
by setting up the equation,
the difference of matrix A
and the product of Lambda in the identity matrix
times vector X equals a zero vector,
which I've already set up here.
And now, let's write the corresponding matrix equation,
wherefore the first row of the coefficient matrix,
we have negative five minus two times one,
which is negative seven.
And then we have two minus two times zero, which is two.
In the second row,
we have negative seven minus two times zero,
which is negative seven.
And then we have four minus two times one,
which is two.
Next, we have the augmented matrix.
And then we have the augmented matrix written in.
We do show echelon form.
Once again, notice how X two is a free variable.
The first row indicates
X one minus 2/7 X two equals zero.
X two is a free variable.
Solving the first equation for X one,
we have X one equals 2/7 X two.
Again, let's parametrize the solution
by letting X two equal S where S as in any real number,
which gives us X one equals 2/7 S
and X two equals S,
which indicates vector X,
which represents all the eigenvectors
corresponding to Lambda equals two
or the vectors in the form of 2/7 S S
or S times the vector 2/7 one.
If we want to clear the fraction of 2/7,
we can let R equal seven S,
which gives us R times the vector 2/7.
But once again,
we don't include the zero vector as an eigenvector.
So whichever form we use,
S and R can't equal zero.
So using the form where we don't have a fraction,
we can say that the eigenvectors
corresponding to Lambda sub two equals two
are all the scalar multiples of the vector 2/7,
except the zero vector.
We can also say for Lambda sub two equals two
a basic eigenvector is the vector 2/7.
Checking to make sure it satisfies our eigenvector equation,
we can say below it is satisfied
because the vector 4/14
equals two times the vector 2/7.
And then finally, for the eigenspace,
we can say the eigenspace of matrix A
corresponding to Lambda equals two
is equal to all the scalar multiples of the vector 2/7
or the span of the set containing the vector 2/7.
And we can also state
that a basis for the eigenspace of matrix A
corresponding to Lambda equals two is the vector 2/7.
I hope you found this helpful.