eigen value and eigen vector

FINDING EIGENVALUES AND EIGENVECTORS
EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix
A =?
?
1 ?3 3
3 ?5 3
6 ?6 4
??
.
SOLUTION:
• In such problems, we first find the eigenvalues of the matrix.
FINDING EIGENVALUES
• To do this, we find the values of  which satisfy the characteristic equation of the
matrix A, namely those values of  for which
det(A ? I) = 0,
where I is the 3×3 identity matrix.
• Form the matrix A ? I:
A ? I =?
?
1 ?3 3
3 ?5 3
6 ?6 4
??
??
?
 0 0
0  0
0 0 
??
=?
?
1 ?  ?3 3
3 ?5 ?  3
6 ?6 4 ? 
??
.
Notice that this matrix is just equal to A with  subtracted from each
entry on the main diagonal.
• Calculate det(A ? I):
det(A ? I) = (1 ? )
?5 ?  3
?6 4 ? ? (?3)
3 3
6 4 ? 
+ 3
3 ?5 ? 
6 ?6 
= (1 ? ) ((?5 ? )(4 ? ) ? (3)(?6)) + 3(3(4 ? ) ? 3 × 6) + 3(3 × (?6) ? (?5 ? )6)
= (1 ? )(?20 + 5 ? 4 + 2 + 18) + 3(12 ? 3 ? 18) + 3(?18 + 30 + 6)
= (1 ? )(?2 +  + 2) + 3(?6 ? 3) + 3(12 + 6)
= ?2 +  + 2 + 2 ? 2 ? 3 ? 18 ? 9 + 36 + 18
= 16 + 12 ? 3.
• Therefore
det(A ? I) = ?3 + 12 + 16.
REQUIRED: To find solutions to det(A ? I) = 0 i.e., to solve
3 ? 12 ? 16 = 0. (1)
* Look for integer valued solutions.
1
* Such solutions divide the constant term (-16). The list of possible integer
solutions is
±1,±2,±4,±8,±16.
* Taking  = 4, we find that 43 ? 12.4 ? 16 = 0.
* Now factor out  ? 4:
( ? 4)(2 + 4 + 4) = 3 ? 122 + 16.
* Solving 2 + 4 + 4 by formula1 gives
 = ?4 ± ?42 ? 4.1.4
2
= ?4 ± 0
2
,
and so  = ?2 (a repeated root).
• Therefore, the eigenvalues of A are  = 4,?2. ( = ?2 is a repeated root of the
characteristic equation.)
FINDING EIGENVECTORS
• Once the eigenvalues of a matrix (A) have been found, we can find the eigenvectors
by Gaussian Elimination.
• STEP 1: For each eigenvalue , we have
(A ? I)x = 0,
where x is the eigenvector associated with eigenvalue .
• STEP 2: Find x by Gaussian elimination. That is, convert the augmented matrix
A ? I
...
0
to row echelon form, and solve the resulting linear system by back substitution.
We find the eigenvectors associated with each of the eigenvalues
• Case 1:  = 4
– We must find vectors x which satisfy (A ? I)x = 0.
1To find the roots of a quadratic equation of the form ax2 + bx + c = 0 (with a 6= 0) first compute

= b2 ? 4ac, then if
? 0 the roots exist and are equal to x = ?
b
?
p

2a and x = ?
b+p

2a .
2
– First, form the matrix A ? 4I:
A ? 4I =?
?
?3 ?3 3
3 ?9 3
6 ?6 0
??
.
– Construct the augmented matrix A ? I
...
0 and convert it to row echelon
form
??
?3 ?3 3 0
3 ?9 3 0
6 ?6 0 0
??
R1
R2
R3
R1!?1/3×R3
?? ?
?
1 1 ?1 0
3 ?9 3 0
6 ?6 0 0
??
R1
R2
R3
R2!R2?3×R1
R3!R3?6×R1
?? ?
?
1 1 ?1 0
0 ?12 6 0
0 ?12 6 0
??
R1
R2
R3
R2!?1/12×R2
?? ?
?
1 1 ?1 0
0 1 ?1/2 0
0 ?12 6 0
??
R1
R2
R3
R3!R3+12×R2
?? ?
?
1 1 ?1 0
0 1 ?1/2 0
0 0 0 0
??
R1
R2
R3
R1!R1?R2
?? ?
?
1 0 ?1/2 0
0 1 ?1/2 0
0 0 0 0
?? R1
R2
R3
– Rewriting this augmented matrix as a linear system gives
x1 ? 1/2x3 = 0
x2 ? 1/2x3 = 0
So the eigenvector x is given by:
x =?
?
x1 = x3
2
x2 = x3
2
x3
??
= x3?
?
1
21
2
1
??
For any real number x3 6= 0. Those are the eigenvectors of A associated
with the eigenvalue  = 4.
• Case 2:  = ?2
– We seek vectors x for which (A ? I)x = 0.
– Form the matrix A ? (?2)I = A + 2I
A + 2I =?
?
3 ?3 3
3 ?3 3
6 ?6 6
??
.
3
– Now we construct the augmented matrix A ? I
...
0 and convert it to row
echelon form
??
3 ?3 3 0
3 ?3 3 0
6 ?6 6 0
??
R1
R2
R3
R1!1/3×R3
?? ?
?
1 ?1 1 0
3 ?3 3 0
6 ?6 6 0
??
R1
R2
R3
R2!R2?3×R1
R3!R3?6×R1
?? ?
?
1 1 ?1 0
0 0 0 0
0 0 0 0
??
R1
R2
R3
– When this augmented matrix is rewritten as a linear system, we obtain
x1 + x2 ? x3 = 0,
so the eigenvectors x associated with the eigenvalue  = ?2 are given by:
x =?
?
x1 = x3 ? x2
x2
x3
??
– Thus
x =?
?
x3 ? x2
x2
x3
??
= x3?
?
1
0
1
??
+ x2?
?
?1
1
0
??
for any x2, x3 ? R\{0}
are the eigenvectors of A associated with the eigenvalue  = ?2.
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