Solution of 3 × 3 systems when three
eigenvalues are equal and there exists unique
eigenvector
1. Solve the difference system:
xn+1 =
xn + 2yn + zn
yn+1 = −xn + 3yn + zn
zn+1 =
yn + 2zn
(1)
2. Solve the differential system:
x0 =
x + 2y + z
0
y = −x + 3y + z
z0 =
y + 2z
Solution.

1 2

A =  −1 3
0 1
Characteristic

(2)
1
1 

2
equation:


1−λ
2
1
2
3
3−λ
1 
det (A − λI) = det 
 −1
 = 8 − 12λ + 6λ − λ =
0
1
2−λ
3
− (λ − 2) = 0
λ1 = λ2 = λ3 = 2, or λ = 3 with multiplicity 3.
1.Eigenvector:
1
Av1 =λv1⇔ (A − λI) v1 = 0;
x


v1 =  y  ,
z




1−λ
2
1
x
0




3−λ
1 
 −1
 y  =  0
0
1
2−λ
z
0
λ
=
2
:




−1 2 1
x
−x + 2y + z




 −1 1 1   y  =  −x + y + z
0 1 0
z
y






0



= 0 
0
−x + 2y + z = 0
−x + y + z = 0
y = 0
, Solution is : {y =0, x= z,z = 
z}


x
z
1
1








v1 =  y  =  0  =  0  z ⇒ v1 =  0 
z
z
1
1
2. The first generalized eigenvector:
Av2 =λv2+ v1 , ⇔ (A − λI) v2 = v1 ;
x


v2 =  y 
z





1−λ
2
1
x
1





3−λ
1 
 −1
 y  =  0 
0
1
2−λ
z
1
λ=2:







−1 2 1
x
−x + 2y + z
1







 −1 1 1   y  =  −x + y + z  =  0 
0 1 0
z
y
1
−x + 2y + z = 1
−x + y + z = 0
y = 1
, Solution is : {y = 1, z = z, x = 1 + z}
2










x
1+z
1
1
1





 



v2 =  y  =  1
 =  1  +  0  z ⇒ v2 =  1 
z
z
0
1
0
3. The second generalized eigenvector:
Av3 =λv3+ v2 , ⇔ (A − λI) v3 = v2 ;
x

v3 =  y 

z





1−λ
2
1
x
1





3−λ
1 
 −1
 y  =  1 
0
1
2−λ
z
0
λ=2:







−1 2 1
x
−x + 2y + z
1







 −1 1 1   y  =  −x + y + z  =  1 
0 1 0
z
y
0
−x + 2y + z = 1
−x + y + z = 1
y = 0
, Solution is : {y =0, z = z, x= z− 1}   


x
−1 + z
−1
1
−1




 



v2 = 
 y = 0
 =  0  +  0  z ⇒ v3 =  0 
z
z
0
1
0
4.General solution of difference system (1).
Transition Matrix:
T = (v1 , v2 , v3 ) ;
Av1 = λv1 , Av2 = λv2 + v1 , Av3 = λv3 + v2
AT= A (v1 , v2, v3 ) = (Av1 , Av2 , Av3 ) = (λv1 , λv2 + v1 , λv3 + v2 ) =
λ 1 0


=  0 λ 1  (v1 , v2 , v3 ) = ∆T,
0 0 λ


λ 1 0


where ∆ =  0 λ 1  .
0 0 λ
3
We have
∆2
∆3
∆n

2


3

n


λ2 2λ 1
λ 1 0


2
2λ 
= 
 0 λ 1  = 0 λ

2
0 0 λ
0 0 λ
λ 1 0

=  0 λ 1
0 0 λ
...

λ 1 0

=  0 λ 1
0 0 λ



λ3 3λ2 3λ


λ3 3λ2 

 = 0
0
0
λ3
λn nλn−1


 = 0
λn
0
0

n(n−1) n−2
λ
2!
n−1
nλ
λn




x0
C1



−1 
Let (v1 , v2 , v3 )  y0  =  C2 
z0
C3
 n

n−2
λ nλn−1 n(n−1)
λ
2!


−1
An = T ∆n T −1 = (v1 , v2 , v3 )  0
λn
nλn−1  (v1 , v2 , v3 )
0
0
λn


 n



x0
x0
λ nλn−1 n(n−1)
λn−2
2!




−1 
An w0 = An 
 y0  = (v1 , v2 , v3 )  0
λn
nλn−1  (v1 , v2 , v3 )  y0  =
z0
z0
0
0
λn


C1
³
³
´
´
n(n−1) n−2
n
n−1
n
n−1
n 
= v1 λ , v1 nλ
+ v2 λ , v1 2! λ
+ v2 nλ
+ v3 λ  C2 
=
C
3
³
´
³
´
n(n−1) n−2
n
n−1
n
n−1
= C1 v1 λ + C2 v1 nλ
+ v2 λ + C3 v1 2! λ
+ v2 nλ
+ v3 λ n
Here

ωn

xn


=  yn  =
zn







1
1
1

 n

 n−1 
n
= c1  0  2 + c2  0  n2
+ 1 
2  +
1
1
0







1
1
−1

 n (n − 1) n−2 
 n
n−1
+c3 
2
+ 1 
+
 0 
 n2
 0 2 
2!
1
0
0
4
Answer:
³
n
n−1
xn = c1 2 + c2 n2
+2
n
´
Ã
+ c3
n (n − 1) n−2
2
+ n2n−1 − 2n
2!
!
yn = c2 2n + c3 n2n−1
zn = c1 2n + c2 n2n−1 + c3
n (n − 1) n−2
2
2!
5. Solution of differential
system
(2).




x0
C1



Again, let (v1 , v2 , v3 )−1 
 y0  =  C2  ,
z0
C3
At
e
=
∞
P
n=0
n
An tn!
=
∞
P
n=0

n
An tn!
=
∞
P
T∆
n
n
T −1 tn!
n=0
n−1 n(n−1) n−2
nλ
λ
2!
n
n−1
n
∞  λ
P
= (v1 , v2 , v3 )
 0
n=0
0
 ∞
P n tn
 n=0 λ n!


0
= (v1 , v2 , v3 ) 



λ
0
eλt teλt

= (v1 , v2 , v3 )  0 eλt
0
0
nλ
λn
∞
P
n
nλn−1 tn!
n=0
∞
P
n
λn tn!
n=0
0

0
t2
=T


µ∞
P
n=0
n
∆n tn!
2!
n!
n
nλn−1 tn!
n=0
∞
P
n
λn tn!
 n=0
C1
eλt
λt   C  =
te   2 
C3
eλt
2!


C1

t2
λt
λt
λt 
= v1 e , (v1 t + v2 ) e , v1 2! + v2 t + v3 e  C2  =
C3
´
³ 2
t
λt
λt
C1 v1 e + C2 (v1 t + v2 ) + C3 v1 2! + v2 t + v3 e .
Here
³
³

´

x (t)


 y (t)  =
z (t)
5
´
T −1 =
 tn 
−1
 n!  (v1 , v2 , v3 ) =
∞
P
n(n−1) n−2 tn
λ
n=0
∞
P
¶




 (v1 , v2 , v3 )−1 =









1
1
1

 λt



 λt
= C1  0  e + C2  0  t +  1 
 e +
1
1
0






1
−1
1
2



 λt

t
+C3  0  +  1  t +  0 
 e .
2!
0
0
1
6. Answer:
Ã
2t
2t
x (t) = C1 e + C2 (t + 1) e + C3
y (t) = C2 e2t + C3 te2t
z (t) = C1 e2t + C2 te2t + C3
6
t2 2t
e
2!
!
t2
+ t − 1 e2t
2!