Chapter 1
1
Chapter 2
POLYNOMIAL OF MATRICES.
Definition 2.1. Let P (x) = a0 + a1 x + .... + an xn be a polynomial over a field F. If A is
a square matrix over F, then P (A) is defined as P (A) = a0 I + a1 A + .... + an An .
Further, if P (B) = 0 for some matrix B then we say that B is a root (solution) of
P (x) = 0.

Example:f (x) = x2 + 2x + 3 and A = 

1 1
 , find f (A).
0 1
f (A) = 
A2 + 2A+ 
3I
=
1 1

0 1

1 1

+ 2 
0 1

1 1
Example: Show that A = 
1 2
3 4
Considerf (A) =A− 5A −
2I 
=
3 4


1 0

=
0 1

6 4
.
0 6

 is a solution of f (x) = x2 − 5x − 2.
2
1 2
+ 3 
0 1

1 2

-5 
3 4
1 2
3 4



 −2 
1 0
0 1


=
0 0
=0
0 0
Hence A is a solution of f (x) = x2 − 5x − 2.
Theorem 2.1. If A is a square matrix and f (x), g(x) are polynomials over F, then
1. (f + g)(A) = f (A) + g(A)
2. (f.g)(A) = f (A).g(A)
2
3. (kf )(A) = k.f (A), ∀k ∈ F.
where f (x) = a0 + a1 x + a2 x2 + .... + an xn , g(x) = b0 + b1 x + b2 x2 + .... + bn xm .
Proof: Exercise.
Theorem 2.2.
(Cayley - Hamilton theorem)
Every square matrix A is a zero of its characteristic polynomial.
ie. ψA (A) = 0.
Proof. Let A = (aij ) be an n × n matrix and let
ψA (λ) = |λI − A| = λn + an−1 λn−1 + .... + a1 λ + a0 , where

λ − a11 −a12 . . . −a1n


 −a21 λ − a22 . . . −a2n
λI − A = 

..

.

−an1
−an2 . . . λ − ann




.



Let B(λ) be the adjoint of the matrix (λI − A).
The elements of B(λ) are co-factors of (λI − A) and hence are polynomial in λ of degree
not exceeding (n − 1).
Thus B(λ) = λn−1 Bn−1 + λn−2 Bn−2 + .... + λB1 + B0 , Where Bn−1 , Bn−2 , ...., B1 , B0 are
square matrices of order n which are independent of λ.
We have (λI − A) B(λ) = |λI − A| I.
⇒ (λI − A)(λn−1 Bn−1 + λn−2 Bn−2 + .... + λB1 + B0 ) = (λn + an−1 λn−1 + .... + a1 λ + a0 )I.
by equating the coefficient of λr r = 0, n we can get,
r = n ⇒ Bn−1 = I.
.......(1)
r = n − 1 ⇒ Bn−2 − ABn−1 = an−1 I.
..
.
......(2)
r = 1 ⇒ B0 − AB1 = a1 I.
.......(n)
r = 0 ⇒ −AB0 = a0 I.
.......(n + 1)
Now, An × (1) + An−1 × (2) + ..... + A × (n) + (n + 1) ⇒
0 = An + an−1 An−1 + ..... + a1 A + a0 I
3
ie. A is the zero of its characteristic polynomial ψA (λ).
ie. ψA (A) = 0.


Exercise: Find the polynomial for which A = 
1 2
 is a solution.
3 2
Solution: Consider the characteristic polynomial of A
|λI − A| =
λ−1
−2
−3
λ−2
= (λ − 1)(λ − 2) − 6 = λ2 − 3λ − 4
By Cayley - Hamilton theorem, ψA (A) = A2 − 3A − 4I = 0.

2




1 2
1 2
1 0
 − 3
 − 4
=0
ie. 
3 2
3 2
0 1
2
The polynomial is p(x) = x − 3x − 4.
Definition 2.2. Minimum Polynomial of a matrix.
Among all the non-zero polynomials for which the matrix A is a zero, the lowest degree
monic polynomial (leading coefficient is 1) is called the minimum polynomial of A.
Theorem 2.3. For any square matrix A the minimum polynomial exists and is unique.
Proof. Let A be an n × n matrix.
By Cayley Hamilton theorem, ψA (A) = 0.
Thus, there is a non-zero polynomial for which A is a solution.
Let n be the lowest degree for which the polynomial f (t) exists such that f (A) = 0.
By dividing the polynomial f (t) by its leading coefficient, we obtain the monic polynomial
m1 (t) of degree n which has A as a zero.
So minimum polynomial exists for any matrix.
Suppose m1 (t) and m2 (t) be two minimum polynomials of A of degree n,
Then m1 (A) = 0 and m2 (A) = 0.
Assume that m1 (t) ̸= m2 (t).
Consider g(t) = m1 (t) − m2 (t) ̸= 0.
Then g(A) = m1 (A) − m2 (A) = 0.
4
This is a contradiction (∵ degree of g(t) < degree of m1 (t) and m2 (t).
Thus, m1 (t) = m2 (t).
Hence the minimum polynomial is unique.
Theorem 2.4. Let m(t) be the minimum polynomial of a matrix A and f (t) be a polynomial such that f (A) = 0, then m(t) divides f (t).
Proof. By the division algorithm ∃ g(t), q(t) s.t f (t) = g(t).m(t) + q(t), where degree of
m(t) > degree of q(t).
Suppose q(t) ̸= 0.
since f (A) = 0, g(A).m(A) + q(A) = 0
⇒ q(A) = 0
(∵ m(A) = 0.)
This is a contradiction.
(∵ degree of q(t) < degree of m(t).)
∴ q(t) = 0.
⇒ m(t) divides f (t).
Theorem 2.5. If m(t) is the minimum polynomial of an n × n matrix A and χA (t) is
the characteristic polynomial of A then χA (t) divides [m(t)]n .
Proof. Let m(t) = tr + a1 tr−1 + a2 tr−2 + ..... + ar−1 t + ar .
Consider the following matrices,
B0 = I.
B1 = A + a1 I.
B2 = A2 + a1 A + a2 I.
..
.
Br−1 = Ar−1 + a1 Ar−2 + .... + ar−1 I.
.....(1)
Then
B0 = I.
B1 − AB0 = a1 I.
B2 − AB1 = a2 I.
..
.
Br−1 − ABr−2 = ar−1 I.
5
Now multiply the equation (1) by −A,
−ABr−1 = −[Ar + a1 Ar−1 + ... + ar−1 A]
= −[Ar + a1 Ar + .... + ar−1 A + ar I] + ar I
ie. −ABr−1 = −m(A) + ar I = ar I
(∵ m(A) = 0)
Now let B(t) = tr−1 B0 + tr−2 B1 + .... + tBr−2 + Br−1 .
Then (tI − A)B(t) = (tI − A)(tr−1 B0 + tr−2 B1 + ..... + tBr−2 + Br−1 )
= tr B0 +tr−1 B1 +.....+t2 Br−2 +tBr−1 −(tr−1 AB0 +tr−2 AB1 +...+tABr−2 +ABr−1 )
= tr B0 + tr−1 (B1 − AB0 ) + tr−2 (B2 − AB1 ) + .... + t(Br−1 − ABr−2 ) − ABr−1
= tr I + a1 Itr−1 + .... + ar−1 It + ar I
= (tr + a1 tr−1 + .... + ar−1 t + ar ) I
ie. (tI − A)B(t) = m(t)I.
⇒ |(tI − A)B(t)| = |m(t)I|
⇒ |tI − A||B(t)| = [m(t)]n
ie. ψA (t)|B(t)| = [m(t)]n
Since |B(t)| is a polynomial, ψA (t) divides [m(t)]n .
Definition 2.3. irreducible polynomial A Polynomial f (x) over the field F of positive
degree is said to be irreducible polynomial over F, if f (x) cannot be expressed as a product
of two other polynomials over F, both of positive degree. Any polynomial of positive
degree which is not an irreducible polynomial over F is said to be reducible polynomial.
Example: (x2 + 1), (x2 − 2) are irreducible over Z.
For: x2 + 1 = (x + i)(x − i), i ∈ C,
√
√
√
x2 − 2 = (x − 2)(x + 2),
2 ∈ Qc .
Theorem 2.6. Characteristic and minimum polynomials of a matrix A have the same
irreducible factors.
Proof. Let f (t) be an irreducible factor of m(t).
⇒ f (t) | m(t),
6
But m(t) | χA (t)
(by Theorem 2.4 )
⇒ f (t) | χA (t).
Thus f (t) is an irreducible factor for χA (t).
Now let f (t) be an irreducible factor of χA (t).
⇒ f (t) | χA (t).
But χA (t) | [m(t)]n
(by Theorem 2.5)
⇒ f (t) | [m(t)]n
⇒ f (t) | m(t).
Thus f (t) is an irreducible factor of m(t).
Hence m(t) and χA (t) have the same irreducible factors.

2 1 0


 0 2 0
Exercise:Find the minimum polynomial of the matrix A = 

 0 0 1

0 0 −2

0


0 
.

1 

4
Solution: Consider the characteristic polynomial of A,
χA (λ) = |λI − A|
=
λ−2
−1
0
0
0
λ−2
0
0
0
0
λ−1
−1
0
0
2
λ−4
λ−2
0
0
0
λ−1
−1
0
2
λ−4
= (λ − 2)
−0
0
0
0
λ−1
−1
0
2
λ−4
+1
= (λ − 2)[(λ − 2)(λ2 − 5λ + 4 + 2)] = (λ − 2)[(λ − 2)(λ2 − 5λ + 6)]
= (λ − 2)3 (λ − 3)
Therefore, the minimum polynomial of A must be one of the following:
7
m1 (λ) = (λ − 2)(λ − 3)
m2 (λ) = (λ − 2)2 (λ − 3)
m2 (λ) = (λ − 2)3 (λ − 3)
By Cayley Hamilton Theorem, m3 (A) = 0.
Further consider m1 (A) = (A − 2I)(A − 3I)

0 1 0 0


 0 0 0 0
=

 0 0 −1 1

0 0 −2 2








−1 1
0 0


 0 −1 0 0


 0
0 −2 1

0
0 −2 1



0 −1 0 0






 0 0 0 0 

 =
 ̸= 0



 0 0 0 0 




0 0 0 0
m2 (A) = (A − 2I)2 (A − 3I)

0 1 0 0


 0 0 0 0
=

 0 0 −1 1

0 0 −2 2

0 0 0 0


 0 0 0 0
=

 0 0 −1 1

0 0 −2 2

0 1 0 0


 0 0 0 0


 0 0 −1 1

0 0 −2 2

−1 1
0 0


 0 −1 0 0


 0
0 −2 1

0
0 −2 1

−1 1
0 0


  0 −1 0 0


 0
0 −2 1

0
0 −2 1










0





 0
 =



 0


0

0 0 0


0 0 0 
=0

0 0 0 

0 0 0
Since the degree of m2 (λ) < degree of m3 (λ), it follows that m2 (λ) = (λ − 2)2 (λ − 3) is
the minimum polynomial of A.

−1 0 0


Exercise: Find the minimum polynomial of A =  0 −1 0

0
0 1
Exercise: Find the minimum polynomial of
8



.



2 5 0


 0 2 0


A= 0 0 4


 0 0 3

0 0 0
0 0


0 0 


2 0 .


5 0 

0 7


P
0
0


Solution: We have A =  0 Q 0

0 0 R
R = (7).





2 5
4 2

, Q=
 and
 , where P = 

0 2
3 5
For simplicity we are blocking the matrices and we will calculate the minimum polynomial of each P, Q and R. Finally we obtained the minimum of A by finding the least
common multiple of these minimum polynomials.
Consider the characteristic polynomial of P
χP (λ) = |λI − P |
=
λ−2
−5
0
λ−2
= (λ − 2)2
The minimum polynomial of P exactly one of the following:
m1 (λ) = (λ − 2),
m2 (λ) = (λ − 2)2 .
By Cayley - Hamilton Theorem m2 (P ) = 0
Consider m
(P − 2I)
1 (P ) = 
0 5
 ̸= 0
=
0 0
Therefore m2 (λ) is the minimum polynomial of P.
Consider |λI − Q|
=
λ−4
−2
−3
λ−5
9
= (λ − 4)(λ − 5) − 6 = λ2 − 9λ + 14
= (λ − 2)(λ − 7)
Let m1 (λ) = (λ − 2), m2 (λ) = (λ − 2)(λ − 7).
m2 (Q) = 0
(by Cayley Hamilton Theorem)
Consider m1 (Q) = (Q − 2I)

=

2 2
3 3
 ̸= 0
⇒ m2 (λ) is a minimum polynomial of Q.
Consider |λI − R| = (λ − 7)
By Cayley Hamilton Theorem m(R) = 0 ⇒ m(λ) = (λ − 7)
⇒ (λ − 7) is the minimum polynomial of R
Let m(λ) be the minimum polynomial of A.
⇒ m(λ) = l.c.m[(λ − 2)2 , (λ − 7)(λ − 2), (λ − 7)]=(λ − 2)2 (λ − 7).










Exercise: Find the minimum polynomial of 







2 8 0 0 0 0 0
0 2 0 0 0 0 0
0 0 4 2 0 0 0
0 0 1 3 0 0 0
0 0 0 0 3 0 0
0 0 0 0 0 0 0








.







0 0 0 0 0 0 5
Answer:λ(λ − 2) (λ − 3)(λ − 5).
2
Exercise: Find the characteristic polynomial and minimum polynomial of each matrices
10

3 1 0


 0 3 0


A= 0 0 3


 0 0 0

0 0 0

−1 1


C =  −1 1

0 1

0 0




0 0 
1 2 3






B
=


1 0
0 1 2 




3 1 
0 0 1

0 3



0
0 0 1






D
=
 1 0 1 
0 



−1
0 1 1
11