Linear Transformation Examples Matrix Eigenvalue problems

Md59
Multiple eigenvalues
⎡−2
2 −3⎤
−2−λ2 −3
Matrix, A= ⎢
2 1 −6
⎥
has characteristic polyomial D(
λ)
= 2 1−λ−6 = 0
⎢
⎥
⎣⎢
−1 −2
0 ⎦⎥
−1 −2 −λ
3 2
D(
λ) =− ( 2+ λ){( 1−λ)( −λ) −12} −2{ −2λ −6} −3{ − 4+ ( 1− λ)} =−λ − λ + 21λ + 45 = 0
2
which factorises to D(
λλ) = ( λ − 5)( λ + 3) = 0, whence roots λ1 = 5, λ2 = λ3
= −3
To find eigenvectors, we apply Gauss elimination to the system ( A− λI)
x = 0,
first with λ = 5 and then with λ = −3.
T
For λ = 5, we obtain an eigenvector, e1
= [ 1 2 −1]
For λ =−3
the characteristic matrix
⎡ 1
A − λI
= A+ 3I
=
⎢
2
⎢
⎣⎢
−1 2
4
−2
−3⎤
−6
⎥
⎥
3 ⎥⎦
⎡1
which row reduces to
⎢
0
⎢
⎣⎢
0
2
0
0
−3⎤
0
⎥
⎥
0 ⎦⎥
Hence it has rank 1. From x + 2x − 3x = 0 we have x = − 2x + 3x
.
Md60
1 2 3 1 2 3
Complex Eigenvalues
For the case λ =− 3, we have x1
=− 2x2 + 3x3
:
Choosing x2 = 1, x3 = 0 and then x2 = 0, x3
= 1 obtains two linearly
independent eigenvectors of A corresponding to λ =−3,
thus :
T T
e = −2
1 0 , and e 3 0 1;
so that µ e νe
is an eigenvector solution.
[ ] = [ ] +
2 3 2 3
● Complex eigenvalues example:
⎡
A= ⎢
⎣−
⎤
⎥
⎦
A− I =
j ie j j jx x
jx x x
T
e
T
j e j
−
0
1
1
0
λ
has det( λ )
−1 1 2
= λ + 1= 0
− λ
Eigenvalues ± , . . λ1 = , λ2
= − ; Eigenvectors from − 1 + 2 = 0
and 1 + 2 = 0 respectively, so e.g. choose 1 = 1 to obtain :
1 = [ 1 ] , 2 = [ 1 − ] . More generally, these are the eigenvectors of :
⎡ a
A = ⎢
⎣−b
b⎤
for real a b with eigenvalues a jb
a⎥
, , ± .
⎦