The Eigenvalue-Eigenvector Method of Finding Solutions

EXAMPLE: Let be a linear operator such that

Let also

Find and

Solution: We have

EXAMPLE: Let be a linear operator such that

Find all nonzero vectors and all numbers such that

Solution: Suppose there is a vector and a scalar such that

Since we rewrite this as

We have

So, we should find such that has a nontrivial solution. Recall that
has a nontrivial solution if and only if (see the Corollary to Theorem 11 from Section 3.7).
From this it follows that

if and only if

Note that

therefore we can rewrite as

Expanding this determinant, we obtain

(a) Let To solve the homogeneous system we use row operations:

hence

We get

(b) Let To solve the homogeneous system
we use row operations:

hence

We get

Conclusion: The equation

has a nonzero solution if and only if

and

and

where is any nonzero real number.

DEFINITION: An eigenvector of an matrix is a nonzero
vector such that

for some scalar A scalar is called
an eigenvalue of The set of all
solutions of the equation

is called the eigenspace of corresponding to

DEFINITION: Let

then

is called the characteristic polynomial of and
is called the characteristic equation of

EXAMPLE: Let

Then

is a characteristic polynomial,

is a characteristic equation; and are the eigenvalues of and

are the eigenvectors of where is any nonzero real number;

is the eigenspace of corresponding to

is the eigenspace of corresponding to .

EXAMPLE: Let

Find all eigenvalues, eigenvectors and bases for the corresponding eigenspaces.

Solution: We first solve the following equation:

Expanding this determinant, we obtain

hence

are the eigenvalues of

(a) Let To solve the homogeneous system
we use row operations:

hence

We get

is the eigenvector of corresponding to
The 1-dimensional eigenspace corresponding to is

and
is a basis for

(b) Let To solve the homogeneous system
we use row operations:

hence

We get

is the eigenvector of corresponding to
The 1-dimensional eigenspace corresponding to is

and
is a basis for

EXAMPLE: Let

Find all eigenvalues, eigenvectors and bases for the corresponding eigenspaces.

Solution: We first solve the following equation:

Expanding this determinant, we obtain

hence

are the eigenvalues of

(a) Let To solve the homogeneous system
we use row operations:

hence

We get

is the eigenvector of corresponding to
The 1-dimensional eigenspace corresponding to is

and
is a basis for

(b) Let To solve the homogeneous system
we use row operations:

hence

We get

is the eigenvector of corresponding to
The 1-dimensional eigenspace corresponding to is

and
is a basis for

EXAMPLE: Let

The eigenvalues are and . Find bases for the corresponding eigenspaces.

Solution:

(a) Let . We use row operations:

hence

We get

is the eigenvector of corresponding to

To find a basis for the eigenspace corresponding to we note that

therefore the 2-dimensional eigenspace corresponding to is

and

is a basis for

(b) Let . We use row operations:

hence

We get

is the eigenvector of corresponding to The 1-dimensional eigenspace corresponding to is

and
is a basis for

THEOREM 12: Any eigenvectors of with distinct eigenvalues respectively, are linearly independent.

We return now to the first-order linear homogeneous differential equation

(1)

Our goal is to find linearly independent solutions
. One can show that

is a solution of (1) if, and only if,
is an eigenvalue and is an eigenvector of .

Recall that both the first-order and second-order linear homogeneous scalar
equations have exponential functions as solutions. This suggests that we try

where is a constant vector, as a solution of (1). To this end, observe that

and

Hence, is a solution of (1) if, and only if,

Dividing both sides of this equation by gives

Thus,

is a solution of (1) if, and only if,
is an eigenvalue and is an eigenvector of .

EXAMPLE: Find the general solution of the system

(2)

Solution 1: The system can be derived from the second-order differential equation

(3)

by setting

Indeed, if we rewrite the first equation of system (2) as

and plug it into the second equation, we get

To find two linearly independent solutions of (3)
we note that the characteristic equation is
with the roots and . Consequently,

are two solutions of (3).

We see that

We have

and

are two solutions of (2). To determine whether and are linearly dependent
or linearly independent, we check whether their initial values

are linearly dependent or linearly independent vectors in .
Since and are not multiples of each other, they are linearly independent. Consequently, by the Theorem (Test for Linear Independence) from Section 3.4,
and are linearly independent solutions of (2), and every solution
of (2) can be written in the form

Solution 2: Note that system (2) can be rewritten as

where

The characteristic polynomial of the matrix is

so the eigenvalues of are and .

(a) Let . We use row operations:

hence

We get

is the eigenvector of corresponding to Consequently,

is a solution of the differential equation for any constant . For simplicity,
we take

(b) Let . We use row operations:

hence

We get

is the eigenvector of corresponding to Consequently,

is a solution of the differential equation for any constant . For simplicity,
we take

The solutions and must be linearly
independent, since has distinct eigenvalues. Therefore, every solution
must be of the form

EXAMPLE: Find all solutions of the equation

Solution: The characteristic polynomial of the matrix

is

We have

Note that

By inspection, is a root of , therefore

so the eigenvalues of are , and .

(a) Let . We use row operations:

hence

We get

is the eigenvector of corresponding to Consequently,

is a solution of the differential equation for any constant . For simplicity,
we take

(b) Let . We use row operations:

hence

We get

is the eigenvector of corresponding to Consequently,

is a solution of the differential equation for any constant . For simplicity,
we take

(c) Let . We use row operations:

hence

We get

is the eigenvector of corresponding to Consequently,

is a solution of the differential equation for any constant . For simplicity,
we take

The solutions must be linearly
independent, since has distinct eigenvalues. Therefore, every solution
must be of the form