We're given matrix A.
And that's to find the eigenvalues of A,
which are the values of lambda
that satisfy the equation.
The determinant of A minus
lambda I equals zero.
So for review.
If A is an n by n matrix, suppose that x
is a nonzero vector in R
n and lambda is a number
or scalar such that A times vector x
equals lambda times vector x.
This means that A times vector x
is a scalar multiple of vector x,
and vector x is called
an eigenvector of A.
And lambda is called an eigenvalue of A.
We say lambda is the eigenvalue associated
with or corresponding to the vector x.
And vector x is the eigenvector
associated with or
corresponding to lambda.
So the number lambda is an eigenvalue of A
if and only if it satisfies either
of these two equivalent equations.
We have the determinant of
lambda I minus A equals zero.
Or the determinant of A
minus lambda I equals zero.
When expanded the determinant is
a polynomial in lambda of degree n
that is called the
characteristic polynomial of A.
And the entire equation is called
the characteristic equation.
So, again for our example,
we'll only be doing number one.
We'll find the eigenvalues of A
by determining the values of lambda
that satisfy this equation here.
Later we'll also be
finding the eigenvectors
where the eigenvectors of
A corresponding to lambda
are the nonzero solutions
to this equation here.
So going back to our example.
To set this up, we'll
have the determinant of,
again, matrix A, which is given.
Minus lambda times the din-it-y matrix.
Where because A is a two by two matrix,
lambda times I would be this matrix here.
And this must be equal to zero.
Let's go ahead and write
this using vertical bars.
So the first row of the determinant
is going to be one minus lambda.
And then three minus zero which is three.
The second row is going to
be four minus zero over four.
And negative three minus lambda.
The value of the two by two determinant
is equal to this product,
minus this product.
So we have the quantity one minus lambda
times the quantity negative
three minus lambda.
Minus three times four.
And this must equal zero.
Let's go ahead and multiply this out
so we have negative three.
Minus one lambda, or minus lambda.
Plus three lambda.
Plus lambda squared.
Minus 12 equals zero.
Combining like terms,
we have lambda squared
plus two lambda
minus 15 equals zero.
So we have a quadratic
equation in terms of lambda
which does factor.
The left side factors
into two binomial factors.
The factors of lambda squared are
lambda and lambda.
The factors of negative 15
that add to positive two
are positive five and negative three.
So this product equals zero when
lambda equals negative five or
when lambda equals positive three.
So these are the eigenvalues
of the given vector A.
I hope you found this helpful.
