We're given the matrix A
and asked to find the eigenvalues
and the corresponding eigenvectors.
So looking at our notes below,
an eigenvalue of A is a scalar lambda
such that the determinant of
lambda I minus A equals zero.
So we need to solve this
equation here for lambda
in order to find the eigenvalues.
And then the eigenvectors of A,
corresponding to lambda
are the nonzero solutions
of this equation here
which I'll be solving for
vector x.
Let's first find the eigenvalues.
And I've already set some of this up.
Here we have the determinant of
lambda times the two
by two identity matrix
which will be this matrix here.
Minus the given matrix A
must equal zero.
Performing the matrix subtraction,
notice how we have lambda minus six,
zero minus negative three which is three.
Zero minus negative two which is two.
And lambda minus one.
So, next step is to find the
value of the determinant.
Which is this product,
minus this product, which
again, must equal zero.
So we have
the quantity lambda minus six
times the quantity lambda minus one
minus three times two.
Must equal zero.
Let's go ahead and multiply this out.
We have lambda squared
minus one lambda minus six lambda.
That's minus seven lambda.
Plus six minus six equals zero.
So, simplifying, we have lambda squared
minus seven lambda equals zero.
Solving by factoring, we have lambda
times the quantity lambda
minus seven equals zero.
So notice how we have lambda equals zero.
Or lambda equals seven.
So the eigenvalues are
lambda sub one equals zero.
And lambda sub two equals seven.
Now that we have the eigenvalues,
we'll find the corresponding eigenvectors.
And again, we do this by
solving this equation here
for vector x.
Let's first find
the eigenvalue corresponding
to lambda equals zero.
So, lambda sub one equals zero.
So again, our equation is here.
We have lambda I which
is this matrix here.
Minus the given matrix A which is here.
Times the eigenvector of vector x
equals the zero vector.
And again, performing
the matrix subtraction
gives us this matrix here.
The components of vector x are
x sub one comma x sub two.
And here's the zero vector.
So for our next step we'll
substitute zero for lambda.
So this matrix would be negative six,
three, two, negative one.
Let's go ahead and solve this system.
Using an augmented matrix.
So the same system
as an augmented matrix, would
be a two by three matrix.
Where the first row would be negative six,
three, zero.
The second row would be
two, negative one, zero.
And now I'll write this in
reduced row echelon form.
Notice three times two plus
negative six would be zero.
Let's replace row two
with three time row two,
plus row one.
And to make this element here positive one
let's replace row one with negative 1/6
times row one.
So negative 1/6 times negative six is one.
Negative 1/6 times three is negative 1/2.
And then negative 1/6 times zero is zero.
And now for row two,
we'll have three times two
plus negative six, that's zero.
And then three times negative one
plus three is also zero.
And three times zero plus zero is zero.
Notice how we have a row of zeros.
Which means there's an infinite
number of eigenvectors.
Looking at this first row,
notice how this tells us that x sub one
minus 1/2 times x sub two must equal zero.
So if we solve this for x sub one
we'd have x sub one equals 1/2
times x sub two.
If we wanted to avoid fractions,
we could multiply both sides by two.
Which would give us two times x sub one
equals x sub two.
So the eigenvectors corresponding to
lambda equals zero would be vector x
which equals.
Because we have an
infinite number of vectors,
let's prioritize this with the variable t.
If we let x sub one be equal to t,
notice how x sub two would be two t.
Which we could also express as t times
the vector with an x component of one
and a y component of two.
But, remember, the eigenvector
cannot be the zero vector.
Let's also say that t can't equal zero.
Now, let's find the eigenvectors
corresponding to lambda equals seven.
So we'll go through
the same process again,
with lambda sub two equals seven.
So again, here's the
equation we have to solve
for vector x performing
the matrix subtraction
gives us the matrix here.
And now we substitute seven for lambda.
So seven minus six is one.
So we have one, three, two
and seven minus one is six.
Again let's write this
as an augmented matrix.
And write it in reduced row echelon form.
So the first row would
be one, three, zero.
The second row would be two, six, zero.
Notice that negative two times one
plus two would be zero.
Let's replace row two with
negative two times row one
plus row two.
So row one stays the same.
Row two would be negative two
times one plus two is zero.
Then we'd have negative two times
three plus six that's zero.
And of course, negative two times
zero plus zero is also zero.
Again, we have a row of
zeros which we do expect.
Looking at this first row, notice how
this tells us that x sub one
plus three times x sub
two must equal zero.
Solving for x sub one, we'd have x sub one
equals negative three times x sub two.
Which means the eigenvectors
corresponding to
lambda equals seven would be vector x
which has an x component of, in this case,
let's let x sub two be equal to t.
And therefore, x sub one would be equal to
negative three t.
Now if we want we can write this as
t times the vector with an x component of
negative three and a y component of one.
So any multiple of this vector
would be a corresponding eigenvector,
again where t can't equal zero.
So to summarize what we found.
We found the eigenvalues of the matrix A
are lambda sub one equals zero.
And lambda sub two equals seven.
The eigenvectors corresponding to
lambda sub one equals zero are
the vectors x in this form here,
where t doesn't equal zero.
We can also say the
eigenspace corresponding to
lambda sub one equals zero is given by
the span of this vector.
And the eigenvectors corresponding to
lambda sub two equals
seven are the vectors x
in this form, again
where t can't equal zero.
And the eigenspace corresponding to
lambda sub two equals seven is given
by this span of this vector.
I hope your found this helpful.
