Example 3
(one repeated real eigenvalue giving only one eigenvector)
Find the eigenvalues and eigenvectors of the matrix
Or, using the eigenvals(
) command:
To compute the eigenvectors, we solve the system
So, we compute the row reduced echelon form of the matrix
A - 3 I
to get a second eigenvector
Note that the choices are made in such a way that we
get two linearly independent
vector.
Mathcad can compute the eigenvalues and eigenvectors
of a given matrix using the commands
eigenvals( ) and eigenvecs(
) as follows:
Mathcad returns the eigenvectors as columns of a matrix
and the eigenvector associated with the eigenvalue l = 2 + i is
If we repeat the process as in case 1, we get the eigenvector
Remark 1:
If the entries of the matrix A are all real, then the
characteristic equation will be a polynomial with real
coefficients. Thus, if A has a complex eigenvalue then
its conjugate is also an eigenvalue of A. Consequently,
the eigenvectors will also be conjugate of each other.
Remark 2:
If v is an eigenvector of a matrix A, then any constant
multiple of v is also an eigenvector of A (See textbook
for proof).
Since we only have one arbitrary variable, the eigenvalue
l
= 3 (with multiplicity 2) is able to generate only
one eigenvector. Such eigenvalue is called a deficient
eigenvalue with deficiency one.
Example 4 (Complex
eigenvalues)
Find the eigenvalues and eigenvectors of the matrix
Here, Mathcad has a problem finding the row reduced
echelon form of a complex matrix:
So, we will do the reduction manually
Therefore, the eigenvalues of the matrix A are l = 5 and l = 8.
Define I as the 2 x 2 identity matrix
Compute the determinant and use the solve command to
obtain the roots
Now that we have the eigenvalues, we are can find the
associated eigenvectors by solving the homogenous linear
system given in equation (1). Namely,
Recall that to solve such a homogenous system, we compute
the row reduced echelon form of the coefficient matrix
6.1 The Eigenvalue Problem
Definition
A number l
is said to be an eigenvalue of an n x n matrix A if
there is a nonzero
vector v such that A v = l
v. The vector v is called an eigenvector associated
with the eigenvalue l
.
So, the question is: given an n x n matrix A, how do
we find its eigenvalues and their associated eigenvectors?
The equation A v
= l
v can be rewritten as
or, taking v as a common factor
where I is the n x n identity matrix.
Note that we can look at the above equation as a homogenous
linear system with matrix
A - l I
as the coefficient matrix and v as the vector of unknowns.
Such a system will have a nontrivial solution (since
we are looking for a nonzero vector v) if and only
if the determinant of A - l I is zero. That is
Equation (2) is called the
characteristic equation
. The equation is an nth degree polynomial in l
whose roots are the eigenvalues of the matrix A.
Find the eigenvalues of the matrix
To obtain the eigenvalues, we compute characteristic
equation and then find its roots.
Thus, the characteristic equation is given by
Therefore, the eigenvector associated with the eigenvalue
l
= 8 is
Note:
The two eigenvectors associated the the two eigenvalues
are clearly linearly independent since neither is a
scalar multiple of the other.
The previous example illustrates the case where the
eigenvalues are distinct real numbers. The following
examples will illustrate other possibilities
Example 2
(one repeated real eigenvalue giving two eigenvectors)
Find the eigenvalues and the associated eigenvectors
of the matrix
Thus, the eigenvalue is l
= 2 with multiplicity 2.
To find the associated eigenvector, we solve the system
Or, we simply compute the row reduced echelon form of
the matrix A - 2 I (it is obvious but just to be systematic)
We can choose v2
to be any number except 0 because this will give us
the zero vector as a solution and we are looking for
a nonzero vector.
is the eigenvector associated with the eigenvalue l = 5
Mathad is computing A - 8 I
Compute the row reduce echelon form
