Here we use the the principles of Cayley-Hamilton Theorem to find Eigenvalues, Eigenvectors, Trace of a Matrix, Determinant and Eigenspace.

1. Cayley–Hamilton theorem - Eigenvalues,
Eigenvectors and Eigenspaces
Isaac Amornortey Yowetu
NIMS-GHANA
March 17, 2021

2. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
Content
1 Cayley–Hamilton theorem
2 Finding Eigenvalues
3 EigenVectors and EigenSpaces

3. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
Cayley–Hamilton theorem
Every square matrix satisfies its own characteristic equation.
Thus:
p(λ) = det(λI − A)
Substituting the matrix A for λ in the polynomial,
p(λ) = det(λI − A) results in a zero matrix.
Thus:
p(A) = 0

4. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
Example : Using Non-singular Matrix
Consider the given matrix:
A =


2 1 −1
1 2 −1
− −1 2


Find the characteristic equation of the square matrix and
hence find the eigenvalues, eigenvectors and eigenspaces.

5. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
Solution
p(λ) = det(λI − A) (1)
=

10. λ


1 0 0
0 1 0
0 0 1

 −


2 1 −1
1 2 −1
−1 −1 2



15. (2)
=

20. 

λ − 2 −1 1
−1 λ − 2 1
1 1 λ − 2



25. (3)
= (λ − 2)

29. λ − 2 1
1 λ − 2

33. − (−1)

37. −1 1
1 λ − 2

41. (4)
+1

45. −1 λ − 2
1 1

50. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
Solution Continue
p(λ) = (λ − 2)[(λ − 2)2
− 1] + [(−1)(λ − 2) − 1] (5)
+[(−1)(1) − 1(λ − 2)]
= (λ − 2)(λ2
− 4λ + 3) − 2(λ − 1) (6)
= (λ − 2)(λ − 3)(λ − 1) − 2(λ − 1) (7)
= (λ − 1)[(λ − 3)(λ − 2) − 2] (8)
= (λ − 1)[(λ2
− 5λ + 4] (9)
= (λ − 1)(λ − 1)(λ − 4) (10)
p(λ) = λ3
− 6λ2
+ 9λ − 4 (11)

51. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
Considering our matrix:
A =


2 1 −1
1 2 −1
−1 −1 2


Finding the Eigenvalues
λ1 = 1, λ2 = 1 and λ3 = 4
Finding the trace of a matrix A
tr(A) =
3
X
i=1
λi =
3
X
i=1
aii = 6

52. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
Determinant of the of a matrix A
det(A) =
3
Y
i=1
λi = 4

53. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
Eigenvectors
Considering the eigenvalues, λ1 = 1 and λ2 = 1 and λ3 = 4
and the matrix.


λ − 2 −1 1
−1 λ − 2 1
1 1 λ − 2


When λ = 1: 

−1 −1 1 0
−1 −1 1 0
1 1 −1 0


Using row reduction method, we have
−1 −1 1 0
We consider expressing the argument matrix in terms of
equation as;

54. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
−x − y + z = 0 (12)
x = −y + z (13)
The eigenspace Λ1 for λ = 1 becomes
Λ1 =


x
y
z

 =


−s + t
s
t

 (14)
=



s


−1
1
0

 + t


1
0
1





(15)
Let z = t and y = s and where t 6= 0 and s 6= 0

55. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
Eigenvectors
. 

λ − 2 −1 1
−1 λ − 2 1
1 1 λ − 2


When λ = 4: 

2 −1 1 0
−1 2 1 0
1 1 2 0


Using row reduction method, we have


2 −1 1 0
3 3 0
−3 −3 0



56. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
2 −1 1 0
1 1 0
We consider expressing the argument matrix in terms of
equation as;
2x − y + z = 0 (16)
y + z = 0 (17)
y = −z (18)
y = −t (19)
Let z = t
x = −t (20)

57. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
The eigenspace Λ3 for λ = 4 becomes
Λ3 =


x
y
z

 =


−t
−t
t


=



t


−1
−1
1





Where t 6= 0

58. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
Conclusion
Eigenvectors
P =


−1 1 −1
1 0 −1
0 1 1


Eigenvalues
D =


1 0 0
0 1 0
0 0 4


A = PDP−1

59. Cayley–Hamilton theorem Finding Eigenvalues EigenVectors and EigenSpaces
End
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