NUMERICAL METHODS
with COMPUTER
APPLICATIONS
NEE 411 – EE1
NUMERICAL METHODS
Boundary Value and Eigen Value Problems of Ordinary Differential
Equations
Introduction of Eigen Values and Eigen Vectors
Diagonalization
Definition 1:
Let L: V → V be a linear transformation of an n-dimensional vector space
V into itself. We say that L is diagonizable or can be diagonalized if
there exists a basis S for V such that L is represented with respect to
S by a diagonal matrix D.

 1

0
D = 



 0
0

1
0
...



0 



n 
0
NUMERICAL METHODS
Introduction of Eigen Values and Eigen Vectors
Diagonalization
Definition 2:
Let L: V → V be a linear transformation of an n-dimensional vector space
V into itself (a linear operator on V). The real number  is called an
eigenvalue of L if there exists a nonzero vector (x) in V such that
L(x) = x.
Every nonzero vector x satisfying this equation is then called an
eigenvector of L associated with the eigenvalue . Eigenvalues are
also called proper, characteristic, or latent values, and eigenvectors
are also called proper, characteristic, or latent vectors.
Note that x = 0V always satisfies (1), but 0V is not an eigenvector,
since we insist that an eigenvector be a nonzero vector.
NUMERICAL METHODS
Introduction of Eigen Values and Eigen Vectors
Diagonalization
Theorem 1:
Let L: V → V be a linear transformation of an n-dimensional vector space
V into itself. Then L is diagonalizable if and only if V has a basis S of
eigenvectors of L. Moreover, if D is the diagonal matrix representing L
with respect to S, then the entries on the main diagonal D are the
eigenvalues of L.
NUMERICAL METHODS
Introduction of Eigen Values and Eigen Vectors
Diagonalization
Example 1:
Let L: V→V be the linear operator defined by L(x) = 2x. We can see
that the only eigenvalue of L is  = 2 and that every nonzero vector in
V is an eigenvector of L associated with the eigenvalue  = 2.
NUMERICAL METHODS
Introduction of Eigen Values and Eigen Vectors
Diagonalization
Example 2:
Let L: R2→R2 be the linear transformation defined by
 a   a 
L  1  =  1 
  a2    a2 
Then we can see that
 a  
a 
L     = 1 
a 
 a  
and
a 
Thus any vector of the form  
a 
 a 
a
L     = −1  
 −a 
  −a  
, where a is any nonzero real number,
NUMERICAL METHODS
Introduction of Eigen Values and Eigen Vectors
Diagonalization
for example,
1
x1 =  
1
, is an eigenvector of L associated with the
a 
eigenvalue  = 1; any vector of the form   , where a is any nonzero
 −a 
1
real number, such as x2 =   , is an eigenvector of L associated with
 −1
the eigenvalue  = -1.
NUMERICAL METHODS
Diagonalization
An eigenvalue  can be associated with it many different eigenvectors. In
fact, if x is an eigenvector of L associated with the eigenvalue 
[i.e., L(x) = x], then
L(rx) = rL(x) = r(x) = (rx),
for any real number r. Thus, if r  0, then rx is also an eigenvector of L
associated with  so that eigenvectors are never unique.
NUMERICAL METHODS
Diagonalization
Theorem 2:
An n x n matrix A is similar to a diagonal matrix D if and only if Rn has a
basis of eigenvectors of A. Moreover, the elements on the main diagonal
D are the eigenvalues of A.
To use Theorem 2, we need only to show that there is a set of n
eigenvectors of
A that are linearly independent, since n linearly
independent vectors in Rn form a basis for Rn.
NUMERICAL METHODS
Example 3:
Let
1
A=
 −2
1
4 
. We wish to find the eigenvalues of A and their
associated eigenvectors. Thus we wish to find all scalars  and all
nonzero vectors
 a1 
x= 
 a2 
Solution:
1
 −2

a1 + a2 =  a1
−2a1 + 4a2 =  a2
or
1   a1 
 a1 
=

→ (1)





4  a2 
 a2 
( − 1)a1 − a2 = 0
2a1 + ( − 4)a2 = 0
NUMERICAL METHODS
which gives us system of two equations in two unknowns has a nontrivial
solution if and only if the determinant of the coefficient matrix is zero. Thus
−1 
 − 1
=0
 2

 − 4

which gives us the characteristic equation of
 2 − 5 + 6 = 0 = ( − 3)( − 2)
therefore,  = 2 and  = 3 are the eigenvalues of A. To find all the eigenvectors
of A associated with  =2, we substitute  = 2 in equation (1)
(2 − 1)a1 − a2 = 0
2a1 + (2 − 4)a2 = 0
a1 − a2 = 0
or
2a1 − 2a2 = 0
NUMERICAL METHODS
which gives use all the eigenvectors associated with  = 2.
a1 = a2
a2 = any real number r.
hence all eigenvectors associated with the eigenvalue 1 = 2 are given by
r 
x1 =  
r 
and r is any nonzero real number. In particular, for r = 1,
1
x1 =  
1
NUMERICAL METHODS
To find all the eigenvectors of A associated with  =3, we substitute  = 3 in
equation (1)
(3 − 1)a1 − a2 = 0
2a1 + (3 − 4)a2 = 0
2a1 − a2 = 0
or
2a1 − a2 = 0
which gives use all the eigenvectors associated with  = 3.
a1 =
1
a2
2
a2 = any real number r.
hence all eigenvectors associated with the eigenvalue 2 = 3 are given by
 r/2 
x2 =   and r is any nonzero real number. In particular, for r = 2,
 r 
1
x2 =  
2
NUMERICAL METHODS
Problem:
Find the characteristic polynomial, the eigenvalues and their
associated eigenvectors for each of the following matrices.
Answer:
1
(a) A = 
2
-1
4 
2
(b) A = 0
0
-2
3
-1
3
-2 
2 
2
(c) A =  1
 2
2
2
-2
3
1 
1 
(a) f(  ) =  2 - 5  + 6;  = 2 and  = 3
1
1
x1 =   and x2 =  
 -1
 -2 
(b) f(  ) =  3 - 7  2 + 14 - 8; 1 = 1, 2 = 2 and 3 = 4
 -1
1
7 
x1 =  1  , x 2 = 0  and x 3 =  -4 
 1 
0 
 2 
(c) f(  ) =  3 - 5  2 + 2 + 8; 1 = -1, 2 = 2 and 3 = 4
1
 -2 
8 
x1 =  0  , x 2 =  -3  and x 3 =  5 
 -1
 2 
 2 
NUMERICAL METHODS
Problem:
Find the characteristic polynomial, the eigenvalues and their
associated eigenvectors for each of the following matrices.
4
(d) A = 
2
0
-4 
 -2
(f) A =  2
 -1
2
1
-2
-3 
-6 
0 
6
(g) A =  2
 -2
2
5
0
-2 
0 
7 
2
(h) A =  0
 -2
0
0
-2
-2 
-2 
1 
1
(i) A =  2
6
0
4
4
0
0 
2 
5
(e) A = 
9
-2 
-6 