If the characteristic polynomial of does not have distinct roots, then
may not have linearly independent eigenvectors. For example, the matrix

has only two distinct eigenvalues and
and two linearly independent eigenvectors, which we take to be

Indeed, the characteristic polynomial is

therefore the eigenvalues are and .

(a) Let . We use row operations:

hence

is the eigenvector of corresponding to . In particular, if , then

is an eigenvector of .

(b) Let . We use row operations:

hence

is the eigenvector of corresponding to . In particular, if , then

is an eigenvector of .

Consequently, the differential equation has only two linearly independent solutions

of the form . Our problem, in this case, is to find a third linearly independent solution. More generally, suppose that the matrix has only
linearly independent eigenvectors. Then, the differential equation
has only linearly independent solutions of the form
. Our
problem is to find an additional linearly independent solutions.

We approach this problem in the following ingenious manner. Recall
that is a solution of the scalar differential equation , for
every constant . Analogously, we would like to say that

is a solution of the vector differential equation

(1)

for every constant vector . However, is not defined if is an
matrix. This is not a serious difficulty, though. There is a very natural way
of defining so that it resembles the scalar exponential .

Since

simply set

(2)

It can be shown that the infinite series (2) converges for all , and can be
differentiated term by term. In particular

This implies that is a solution of (1) for every constant vector , since

REMARK: The matrix exponential and the scalar exponential satisfy
many similar properties. For example,

However,

We now show how to find linearly independent vectors for which
the infinite series can be summed exactly.

(1) Find all the eigenvalues and eigenvectors of . If has linearly independent
eigenvectors, then the differential equation has linearly
independent solutions of the form . (Observe that the infinite series
terminates after one term if is an eigenvector of
with eigenvalue .)

(2) Suppose that has only linearly independent eigenvectors.
Then, we have only linearly independent solutions of the form .
To find additional solutions we pick an eigenvalue of and find all vectors
for which

, but

For each such vector ,

is an additional solution of .
We do this for all the eigenvalues of .

(3) If we still do not have enough solutions, then we find all vectors
for which

, but

For each such vector ,

is an additional solution of .

(4) We keep proceeding in this manner until, hopefully, we obtain linearly
independent solutions.

Observe that

Indeed,

Moreover,

So, since and ,
it follows that

Next, we make the crucial observation that if satisfies

for some integer , then the infinite series terminates after
terms.

Indeed, if , then
is also zero, for every positive integer
, since

Consequently,

and

EXAMPLE: Solve the system

(3)

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

(4)

by setting

Indeed, if we rewrite the second equation of system (3) as

and plug it into the first equation, we get

To find two linearly independent solutions of (4)
we note that the characteristic equation is

Thus, is a repeated root. Consequently,

are two solutions of (4). It follows that

and

are solutions of (3).
The two solutions and
are linearly independent since their initial values

are linearly independent vectors in . Therefore, the general solution
of (3) is

Solution 2: We solve the first equation of system (3). We have

therefore

We have

so

We can easily verify that the function is also a solution of .
So we can write the general solution in the form

where is an arbitrary constant ( , or , or ).

We now substitute
into the second equation of system (3):

This is a first-order linear differential equation (see Section 1.2). Here so that

Multiplying both sides of the equation by we obtain the equivalent
equation

Hence

so

Therefore

Solution 3: We first note that (3) can be rewritten in the following matrix form

where

The characteristic polynomial of the matrix is .
Hence is an eigenvalue of with multiplicity two.
Then

is an eigenvector of .

We use row operations:

hence

is the eigenvector of corresponding to . In particular, if , then

is an eigenvector of .

Hence

is one solution of . Since has only one linearly
independent eigenvector we look for all solutions of the equation

We get

We have

Obviously, every vector is a solution of this equation.
Putting here and , we get

This vector satisfies
, but
since only multiples of
satisfy .

Hence,

is a second solution of .

Therefore,

EXAMPLE: Find three linearly independent solutions of the differential
equation

Solution: The characteristic polynomial of the matrix

is

Hence is an eigenvalue of with
multiplicity two, and is an eigenvalue of with multiplicity one.

(a) Let . Then

is an eigenvector of .

We use row operations:

hence

is the eigenvector of corresponding to . In particular, if , then

is an eigenvector of .

Consequently,

is one solution of . Since has only one linearly independent eigenvector with eigenvalue 1, we look for all solutions of the equation

We get

We have

This implies that and both and are arbitrary:

Putting here and , we get

This vector satisfies , but
since only multiples of
satisfy .
Hence,

is a second linearly independent solution.

(b) Let . Then

is an eigenvector of .

We use row operations:

hence

is the eigenvector of corresponding to . In particular, if , then

is an eigenvector of .

Consequently,

is a third linearly independent solution.

EXAMPLE: Solve the initial-value problem

Solution: The characteristic polynomial of the matrix

is . Hence is an eigenvalue of with multiplicity three. Then

is an eigenvector of .

We use row operations:

hence

is the eigenvector of corresponding to . In particular, if , then

is an eigenvector of .

Hence

is one solution of . Since has only one linearly
independent eigenvector we look for all solutions of the equation

We get

We have

This implies that and both and are arbitrary:

Putting here and , we get

This vector satisfies
, but
since only multiples of
satisfy . Hence,

is a second solution of .

Since the equation

has only two linearly independent solutions,
we look for all solutions of the equation

We get

We have

Obviously, every vector is a solution of this equation.
Putting here

we get

This vector satisfies

but

since the last entry is . Hence,

is a third linearly independent solution. Therefore,

The constants , and are determined from the initial conditions