082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

304 Chapter 8 Matrix algebra
Notethattheeigenvectorhasbeendeterminedtowithinanarbitraryscalar,t.Thusthere
isan infinity of solutions corresponding to λ = 1.
We now consider λ = 5 and seek solutions of the systemequation:
(A−λI)X =0
[( ) ( ( ) ( 4 1 1 0 x 0
−5 =
3 2 0 1)]
y 0)
( ( ) ( −1 1 x 0
=
3 −3)
y 0)
Written as individual equations we have
−x+y =0
3x−3y =0
Wenotethatthesecondequationissimplyamultipleofthefirstsothatinessencethere
isonlyoneequation.Solving −x+y = 0givesy =xforanyx.Sowewritex ( =t,y =t.
1
Hencetheeigenvectorcorrespondingto λ = 5isX =t .Againtheeigenvectorhas
1)
been determined towithin an arbitraryscaling constant.
Sometimesthearbitraryscalingconstantsarenotwrittendown;itisunderstoodthat
they arethere. Insuch a case wesay the eigenvectors of the systemare
( )
( 1 1
X = and X=
−3 1)
Example8.47 Determine the eigenvectors of
( ( ) ( 3 1 x x
= λ
−1 5)
y y)
Solution InExample8.43wefoundthatthereisonlyoneeigenvalue, λ = 4.Weseekthesolution
of (A−λI)X =0.Withλ =4wehave
( ) ( ) ( )
3 1 1 0 −1 1
A−λI= −4 =
−1 5 0 1 −1 1
Hence (A − λI)X = 0is the same as
( ( ) ( −1 1 x 0
=
−1 1)
y 0)
Thus there isonly one equation, namely
−x+y=0
which has an infinity of solutions:x =t,y =t. Hence thereisone eigenvector:
( 1
X =t
1)