We are asked to diagonalize
the following three by three matrix.
The steps to diagonalize an N by N square matrix
are as follows.
Number one, we determine the eigenvalues of the matrix,
which in our case will be lambda sub one
through lambda sub three.
Number two, we construct the diagonal matrix D
using the eigenvalues along the main diagonal.
All other entries are zero.
Number three, we determine N,
linearly independent eigenvectors, which in our case
they'll be the vectors P sub one through P sub three.
Step four, we construct matrix P
using the eigenvectors as the columns.
Number five, we determine the inverse of matrix P.
And then finally, six, we write matrix A at the given matrix
in the form of matrix A equals matrix P times matrix D
times the inverse of matrix P.
So to begin, let's find the eigenvectors
of the given matrix.
To do this, we solve the equation,
the determinant of the difference
of matrix A and the product of lambda
and the identity matrix equal zero.
Notice the matrix resulting from the difference of matrix A
and the product of lambda and the identity matrix
gives us the matrix with entries negative one minus lambda,
negative 19, negative four, zero, negative two
minus Lambda, zero, zero, 15, three minus lambda.
Next, we find the determinant, set at equal
to zero, and solve.
To find the determinant, we will use
the cofactor expansion method.
And I'm going to use row two since row two
has two zero entries.
So notice how the first product and the third product
will be zero because the entries are zero.
And therefore the determinant is equal to the element
in row two, column two times negative one,
raise to the power of two plus two, times the determinant
of the matrix after deleting the row and column
of the entry negative two minus lambda,
which would be row two in column two,
leaving us with the determinant of the matrix
where the entries are negative one minus lambda,
negative four, zero, three minus lambda.
Next, the fourth power of negative one
is equal to positive one.
So we're left with negative two minus lambda
times this two by two determinant, which is equal
to the product of negative one minus lambda
and three minus lambda minus four times zero,
which just gives us the product of negative one
minus lambda and three minus lambda.
So notice how we have the determinant we need
already factored, set equal to zero.
And because the product on the left must be zero,
we know that lambda sub one is equal to negative two,
lambda sub two is equal to negative one,
and lambda sub three is equal to positive three.
So now we have the information we need to form
the diagonal matrix D where the entries
along the main diagonal again are the eigenvalues
of negative two, negative one, three.
All other entries are zero.
And now we need to find the corresponding eigenvectors
that will form the columns of matrix P.
Let's start with lambda sub one equals negative two
to find the corresponding eigenvectors.
We solve the equation.
The difference of matrix A and the product of lambda
and the identity matrix time vector X equals a zero vector,
which I've already set up below.
Notice how I substituted negative two for lambda.
From here, let's go ahead and write the matrix equation.
Let's go ahead and check the coefficient matrix.
The first entry is negative one plus two times one,
which is positive one.
The next two entries don't change.
We have negative 19 and negative four.
In row two, we have zero.
And then the entry in row two, column two
is negative two plus two times one, which is zero.
And the last entry in row two is zero.
And then checking row three, we have zero, 15,
and then three plus two times one, which is equal to five.
I also wrote vector X as the vector X one, X two
and the zero vector as the vector zero, zero.
And now to solve, we write the augmented matrix
and write the augmented matrix and reduce to echelon form,
which you can see I've already done.
Notice how there is no pivot in column three,
which indicates X three is a free variable.
Row one indicates that X one plus 7/3 X three
equals zero or if we solve for X one and we have X one
equals negative 7/3 X three.
The second row indicates X two plus 1/3
X three equals zero.
Solving for X two, we have X two equals
negative 1/3 X three.
And again, X three is a free variable.
We let X three equal X three.
To prioritize the solution, let's let X three equal T.
And therefore all the eigenvectors corresponding to lambda
sub one equals negative two are in the form
of negative 7/3 T, negative 1/3 T, T,
which indicates all the eigenvectors corresponding to lambda
sub one equals negative two are all the scalar multiples
of the vector negative 7/3, negative 1/3, one,
not including the zero vector.
So we can use any eigenvector corresponding to lambda
sub one equals negative two for the first column
of matrix P.
To avoid fractions, let's let T equal three
and use the eigenvector negative seven, negative one, three.
And again, this gives us the first column of matrix P.
And now we need to find an eigenvector
corresponding to lambda sub two equals negative one.
So we go through the same process, solve by setting up
an augmented matrix, and then write the augmented matrix
and reduce to echelon form.
Notice this time, there's no pivot in column one,
which indicates X one is a free variable.
Because X one is a free variable,
we have X one equals X one.
The first row indicates X two equals zero.
And the second row indicates X three equals zero.
To prioritize the solution, let's let X one equal S,
which means all the eigenvectors corresponding to lambda
sub two equals negative one are all the scalar multiples
of the vector one, zero, zero except the zero vector.
So S can't be zero.
Let's use the eigenvector one, zero, zero
for the eigenvector corresponding to lambda sub two
equals negative one, which gives us a second column
of matrix P.
And now we need to find an eigenvector
corresponding to lambda sub three equals three.
So we go through the process one more time,
where we have the original equation, the matrix equation,
followed by the augmented matrix, and then
the augmented matrix written in reduce to echelon form.
Notice we don't have a pivot in column three,
which indicates X three is a free variable.
The first row indicates X one plus X three equals zero.
Solving for X one, we have X one equals negative X three.
The second row indicates X two equals zero.
And again, X three is a free variable.
We have X three equals X three.
Again, let's prioritize the solution
by letting R equal X three.
All the eigenvectors corresponding to lambda sub three
equals three are scalar multiples of the vector
negative one, zero, one, excluding the zero vector.
Let's use the vector negative one, zero, one
as the eigenvector corresponding to lambda sub three
equals three, which gives us a third column of matrix P.
Notice how we do have three
linearly independent eigenvectors.
It's also important to recognize that this matrix P
is not unique but what is important is that the columns
in matrix D and matrix P correspond,
meaning if we have the eigenvector
of negative two in column one of matrix D,
we must have the corresponding eigenvector
in column one of matrix P.
And if we have the eigenvector of negative one in column two
of matrix D, column two must contain
the corresponding eigenvector in matrix P.
And so on for column three.
Now that we have matrix P, the last step is to determine
the inverse of matrix P.
There are a few ways to find the inverse
of a three by three matrix.
I'm going to go ahead and use an augmented matrix
where we have matrix P on the left
and the three by three identity matrix on the right.
And then we perform raw operations
so that we have a three by three
identity matrix on the left.
And then on the right we will have the inverse of matrix P.
So I've already performed all the raw operations.
Notice, now we have the three by three
identity matrix on the left.
And, on the right, columns four, five, and six
form the inverse of matrix P.
So now that we have matrix D, matrix P,
and the inverse of matrix P, we have all the information
we need to diagonalize the given three by three matrix.
Again, the given matrix is equal to matrix P
times matrix D times the inverse of matrix P,
which is the diagonalization of the given matrix.
I hope you found this helpful.