We are asked to diagonalize
the following two by two matrix.
An N by N matrix A is diagonalizable
when there exists an invertible matrix P
and diagonal matrix D such that matrix A
is equal to matrix P times matrix D
times the inverse of matrix P.
To diagonalize a square matrix, number one,
we determine the eigenvalues values of the matrix,
which in our case will be lambda sub one and lambda sub two.
We use these eigenvalues to construct the diagonal matrix,
where we have the eigenvalues along the main diagonal
and the other entries are zero.
The next step is to determine N,
linearly independent eigenvectors, which in our case
will be vector P sub one and vector P sub two.
And then we construct matrix P
by using the eigenvectors as the columns.
And the last step is to determine the inverse of matrix P.
So to begin, we need to determine the eigenvalues
of the given two by two matrix by solving the equation.
The determinant of the difference
of matrix A and the product of lambda
in the identity matrix equals zero.
So I've already set this up below.
We have the determinant of the difference of matrix A
and lambda times the identity matrix equals zero.
Simplifying inside the parentheses, we have the matrix
where the first entry is six minus lambda.
The second entry is 10.
In row two, we have negative two
and then negative three minus lambda.
Now we find the determinant,
set it equal to zero, and solve.
The determinant is equal to the product of six minus lambda
and negative three minus lambda
minus a product of 10 and negative two.
And now we multiply out the left side and simplify.
Multiplying the two binomials, we have negative 18
minus six lambda plus three lambda plus lambda squared,
which gives us negative 18 minus three lambda
plus lambda squared.
And then we have plus 20 equals zero.
Combining like terms and writing the terms
in descending order, we have lambda squared
minus three lambda plus two equals zero.
And now we solve the characteristic equation.
In this case, it is factorable.
The two factors are lambda minus one and lambda minus two.
Because our product must equal zero, we have lambda sub one
equals one and lambda sub two equals two.
And now we use these eigenvalues to form
the diagonal matrix D, where we have one and two
along the main diagonal.
The other entries are zero.
The next step is to find the corresponding eigenvectors,
which we will use as the columns of matrix P.
So beginning with lambda sub one equals one,
to find the corresponding eigenvectors
we solve the equation.
The difference of matrix A and the product of lambda
and the identity matrix times vector X equals zero,
where vector X is an eigenvector.
So again, I've already set this up below.
We have matrix A, we have one for lambda,
and then we have the two by two identity matrix.
Let's write the corresponding matrix equation.
Let's check this coefficient matrix.
The first entry is six minus one times one, which is five.
Next, in row one, we have 10 minus one
times zero, which is 10.
In the second row we have negative two minus one
times zero, which is negative two.
The last entry is negative three, minus one times one,
which is negative four.
I've also written vector X as the vector X one, X two
and the zero vector as the vector zero, zero.
Next, we write the corresponding augmented matrix
and write the augmented matrix and reduce to echelon form.
Notice there's no pivot in column two,
which indicates P two is a free variable.
The first row indicates X one plus two X two equals zero.
And because X two is a free variable,
we have X two equals X two.
Next, we will solve the first equation for X one
and then prioritize the solution by letting X two equal T,
which gives us X one equals negative two T
and X two equals T.
This indicates that all the eigenvectors
corresponding to lambda sub one equals one
are in the form of the vector negative two TT
or T times the vector negative two, one.
This indicates all the eigenvectors
corresponding to lambda equals one
are all the scalar multiples of the vector
negative two, one except the zero vector
because remember, an eigenvector can't be the zero vector.
So we can select any vector as the eigenvector
corresponding to lambda sub one.
Let's just let T equal one
and use the vector negative two, one,
which will be the first column of matrix P.
And now for the second column, we need to determine
an eigenvector corresponding to lambda sub two equals two.
So we go through the same process again.
Let's check the coefficient matrix.
The first entry is six minus two times one or four.
And then we have 10 minus two times zero, which is 10.
In the second row, we have negative two
minus two times zero, which is negative two.
And finally we have negative three minus two times one,
which is negative five.
Again, we have the augmented matrix
as well as the augmented matrix in reduce to echelon form.
The first row indicates X one plus 5/2 X two equals zero.
Again, X two is a free variable and therefore
X two equals X two.
And let's again prioritize the solution by letting X two
equal S, which gives us X one equals at negative 5/2 S.
X two equals S.
All the eigenvectors corresponding to lambda sub two
equals two are scalar multiples of the vector
negative 5/2 one, again excluding the zero vector.
To avoid fractions for the second eigenvector,
let's let S equal two and use the eigenvector
of negative five two for the second column of matrix P.
So now that we have matrix D and matrix P,
the last step is to determine the inverse of matrix P.
So there's a couple ways to find the inverse of matrix P.
I'm going to go ahead and use the formula shown above.
Substituting negative two for A, negative five for B,
one for C, and two for D, we have the inverse of matrix P
is equal to the two by two matrix two, five,
negative one, negative two.
And now we have all the pieces we need
for the diagonalization of the given two by two matrix.
The matrix six, 10, negative two, negative three
is equal to again matrix P times matrix D
times the inverse of matrix P.
These three two by two matrices
form the diagonalization of the given matrix.
And of course, if we wanted to we could check this product
to make sure we do get the given matrix.
So you may want to take the time and do that.
I hope you found this helpful.