Mathematics I

[Canonical Forms]

Dr. Anirban Lakshman

Assistant Professor

Department of Mathematics
IIIT Kalyani

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 1 / 35

 O UTLINE

1 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

2 M INIMAL P OLYNOMIAL

3 R EFERENCES

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 2 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

Definition

Let V be a vector space over the field F and let T be a linear

operator on V . A Eigen value(or, characteristic value) of T is a
scalar λ ∈ F such that there is a non-zero vector α ∈ V with
T (α) = λα.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 3 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

Definition of Eigen-vector
If λ is a eigen value of T , then any non-zero vector α such that
T (α) = λα is called a Eigen vector of T associated with the
eigen value λ.

Definition of Eigen-space
Collection of all α such that T (α) = λα forms a subspace of V .
This subspace is called the Eigen-space associated with the eigen
value λ.
Eigen space is the null space of the linear transformation
(T − λI).

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 4 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

Geometry

F IGURE : Eigen Value and Eigen Vector

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 5 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

Geometry

Geometrically, Ax = λx says that under transformation by A,

eigenvectors experience only changes in magnitude or sign
but the orientation of Ax in Rn is the same as that of x.
The eigenvalue λ is simply the amount of stretch or shrink to
which the eigenvector x is subjected when transformed by A.
The above depicts the situation in R2

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 6 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

Definition of Eigen value of a matrix

If A is an n × n matrix over the F , a eigen value of A in F is a scalar
λ ∈ F such that the matrix A − λI is singular (not invertible).

The equation Ax = λx can be written in the form

(A − λI)x = 0 − (i)
Thus, λ is an eigenvalue of A if and only if (i) has a non trivial
solution.
The set of solutions to (i) is N(A − λI), which is a subspace of R n .
The subspace N(A − λI) is called the eigenspace corresponding to
the eigenvalue λ.
Any nonzero vector in N(A − λI) is an eigenvector corresponding to
λ.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 7 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

Equation (A − λI)x = 0 will have a non trivial solution if and only if

A − λI is singular or, equivalently, det(A − λI) = 0. − (ii)
If the determinant in (ii) is expanded, we obtain an nth-degree
polynomial in the variable λ: P(λ) = det(A − λI)
This polynomial is called the Characteristic Polynomial, and
equation (ii) is called the Characteristic Equation, for the matrix A.

The roots of the characteristic polynomial are the eigenvalues of

A.
An n × n matrix will have n eigenvalues, some of which may be
repeated.
A root λ of characteristic polynomial of a matrix A is said to be
r-fold eigen value of A if the multiplicity of λ is r.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 8 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

T HEOREM
Let T be a linear operator on a finite-dimensional vector space V
and let λ be a scalar. Then the followings are equivalent:
(a) λ is a eigen value of T .
(b) The operator (T − λI)is singular (not invertible).
(c) det(T − λI) = 0

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 9 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

E XAMPLE
   
4 −2 2
Let A = and x =
1 1 1
      
4 −2 2 6 2
Since Ax = = =3 = 3x.
1 1 1 3 1
So, λ = 3 is an eigenvalue of A and x = (2, 1)T is an eigenvector
corresponding to λ = 3.
Actually, any non-zero multiple of x will be an eigenvector, because
A(αx) = αA(x) = αλx = λ(αx).
For example, x = (4, 2)T is also an eigenvector corresponding to
      
4 −2 4 12 4
λ = 3. = =3
1 1 2 6 2

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 10 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

Example: Find the eigenvalues and the corresponding

 
3 2
eigenvectors of the matrix
3 −2
Solution: The characteristic equation is
 
3−λ 2
det = 0 =⇒ λ2 − λ − 12 = 0.
3 −2 − λ
Thus, the eigenvalues of A are λ1 = 4 and λ2 = −3.
Now we find eigen vectors corresponding to λ1 = 4.
 
−1 2
For this we determine the null space of (A − 4I) =
3 −6
    
−1 2 x1 0
From (A − 4I)x = θ we get =
3 −6 x2 0
Then we get two equations −x1 + 2x2 = 0, 3x1 − 6x2 = 0.
These two equations are similar.
Dr. Anirban Lakshman (IIIT Kalyani) MAC101 11 / 35
 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

From these two equations we get x1 = 2x2 .

   
T 2x2 2
So, we get the solution x = (2x2 , x2 ) = = x2 = x2 (2, 1)T
x2 1
Hence, any non zero multiple of (2, 1)T is an eigenvector
corresponding to λ1 = 4 and {(2, 1)T } is a basis for the eigenspace
corresponding to λ1 = 4.
Similarly, to find the eigenvectors for λ2 , we must solve
(A + 3I)x = 0.
In this case, {(−1, 3)T } is a basis for N(A + 3I) and non zero
multiple of {(−1, 3)T } is an eigen vector corresponding to λ2 = −3.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 12 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

Example: Find the eigenvalues and the corresponding

 
2 −3 1
eigenvectors of the matrix A = 1 −2 1
1 −3 2
Solution: The characteristic equation is
 
2−λ −3 1
det  1 −2 − λ 1  = 0 ⇒ −λ(λ − 1)2 = 0
1 −3 2−λ
Thus, the eigen values are λ1 = 0, λ2 = λ3 = 1.
The eigenspace corresponding to λ1 = 0 is N(A − 0I) i.e., N(A),
which we determine in the usual manner.
If x = (x1 , x2 , x3 )T ∈ N(A), then Ax = θ − (i)

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 13 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

From the augmented matrix of A we get,

   
2 −3 1 0 1 0 −1 0
 1 −2 1 0  →  0 1 −1 0 
1 −3 2 0 0 0 0 0
So, the system (i) is equivalent to
x1 − x3 = 0 − (ii)
x2 − x3 = 0 − (iii)
From (ii) and (iii) we get x1 = x2 = x3
   
x1 1
So, the solution of (i) is x = (x1 , x1 , x1 )T = x1  = x1 1
x1 1
Hence, any non zero multiple of (1, 1, 1)T is an eigenvector
corresponding to λ1 = 0.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 14 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

Consequently, the eigenspace corresponding to λ1 = 0 consists of

all vectors of the form α(1, 1, 1)T , where α is a scalar number.
Now we find the eigenspace corresponding to λ2 = λ3 = 1.
For this we solve the system (A − I)y = θ;
    
1 −3 1 y1 0
i.e., 1 −3 1 y2  = 0 − (iv )
1 −3 1 y3 0
From the augmented matrix of the system (iv ) we get
   
1 −3 1 0 1 −3 1 0
 1 −3 1 0  →  0 0 0 0 
1 −3 1 0 0 0 0 0
So, the system (iv ) is equivalent to x1 − 3x2 + x3 = 0
i.e, x1 = 3x2 − x3

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 15 / 35

 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

So, the solution of the system of equations (iv ) is

x = (x1 , x2 , x3 ) = (3x2 − x3 , x2 , x3 )
   
3 −1
= x2 1 + x3  0 
0 1
So, the eigen vectors corresponding to λ2 = λ2 = 1 are
(3, 1, 0), (−1, 0, 1).
The eigen space corresponding to λ2 = λ3 = 1 is
L{(3, 1, 0), (−1, 0, 1)}

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 16 / 35

 O UTLINE

1 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

2 M INIMAL P OLYNOMIAL

3 R EFERENCES

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 17 / 35

 M INIMAL P OLYNOMIAL

Cayley-Hamilton theorem
T HEOREM
Every square matrix satisfies its own characteristic equation.

This theorem states that if A be an n × n matrix and the characteristic

equation of A be c0 x n + c1 x n−1 + . . . + cn = 0, then
c0 An + c1 An−1 + . . . + cn In = 0.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 18 / 35

 M INIMAL P OLYNOMIAL

E XAMPLE
 
2 1
Let A = Then the characteristic equation of A is
3 5
 
2−x 1
det = 0 or, x 2 − 7x + 7 = 0
3 5−x
Then by Cayley-Hamilton theorem A2 − 7A + 7I2 = 0

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 19 / 35

 M INIMAL P OLYNOMIAL

Annihilating Polynomials

Let A be any square matrix. Let J(A) denote the collection of all
polynomials f (t) for which A is a root of the polynomial f (t),
i.e., for which f (A) = 0.
The set J(A) is not empty, because the Cayley-Hamilton theorem
tells us that the characteristic polynomial of A belongs to J(A).
If f (t) is an element of J(A), then we say that f (t) annihilates A
In this case, f (t) is called the Annihilating Polynomials of A.

A polynomial f (t) ̸= 0 is monic if its leading coefficient equals one.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 20 / 35

 M INIMAL P OLYNOMIAL

Definition of Minimal Polynomial

Let A be any square matrix and J(A) denote the collection of all
polynomials f (t) which annihilates A. Then the monic polynomial of
lowest degree in J(A) is called the minimal polynomial m(t) of the
matrix A.

Minimal Polynomial m(t) of a matrix is unique.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 21 / 35

 M INIMAL P OLYNOMIAL

T HEOREM
The minimal polynomial m(t) of a matrix A divides every polynomial
which annihilates A. In particular, m(t) divides the characteristic
polynomial of A.

T HEOREM
The characteristic polynomial and the minimal polynomial m(t) of a
matrix A have the same irreducible factors.

T HEOREM
A scalar λ is an eigenvalue of the matrix A ⇐⇒ λ is a root of the
minimal polynomial m(t) of A.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 22 / 35

 M INIMAL P OLYNOMIAL

 
2 2 −5
Prob: Find the minimal polynomial m(t) of A = 3 7 −15
1 2 −4
=⇒ Characteristic polynomial of A is
 
2−t 2 −5
det  3 7−t −15  = 0 or, (t − 1)2 (t − 3) = 0.
1 2 −4 − t
The minimal polynomial m(t) must divide Characteristic polynomial
of A. Also, each irreducible factor of Characteristic polynomial
[ i.e., (t − 1) and (t − 3)] must also be a factor of m(t).
Thus, m(t) is exactly one of the following:
f (t) = (t − 1)(t − 3) and g(t) = (t − 1)2 (t − 3).

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 23 / 35

 M INIMAL P OLYNOMIAL

Since g(t) is the Characteristic polynomial of A, so g(A) = 0.

Now we check whether f (t) is satisfies by A or not.
  
1 2 −5 −1 2 −5
f (A) = (A − I)(A − 3I) = 3 6 −15  3 4 −15
1 2 −5 1 2 −7
 
0 0 0
= 0 0 0
0 0 0
So, f (t) annihilates A. Also g(t) annihilates A.
But f (t) is the lowest degree polynomial which annihilates A.
So, f (t) = (t − 1)(t − 3) is the minimal polynomial of A.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 24 / 35

 M INIMAL P OLYNOMIAL

Minimal Polynomial of a Linear Operator

The minimal polynomial m(t)of a linear operator T is defined to be

the monic polynomial of lowest degree for which T is a root.
However, for any polynomial f (t), we have f (T ) = 0 ⇐⇒ f (A) = 0,
where A is any matrix representation of T .

T and A have the same minimal polynomial.

The above theorems on the minimal polynomial of a matrix also
hold for the minimal polynomial of a linear operator.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 25 / 35

 D IAGONALISATION

Algebraic multiplicity and Geometric multiplicity of an eigen value

For an r-fold eigen value λ, r is called the algebraic multiplicity

(A.M.) of λ.
Rank of the eigen space corresponding to λ is called the
Geometric multiplicity (G.M.) of λ.
1 ≤ Geometric multiplicity ≤ algebraic multiplicity

Definition: An eigen value λ is said to be regular if

algebraic multiplicity = Geometric multiplicity for λ.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 26 / 35

 D IAGONALISATION
 
1 1 1
Example: Let A = −1 −1 −1 .
0 0 1
The characteristic equation of A is x 2 (1 − x) = 0.
So, the eigen values of A are 0, 0, 1
0 is an eigen value of algebraic multiplicity 2 and 1 is a simple
eigen value of A (i.e., of algebraic multiplicity 1 ).
 
1
The eigen vectors corresponding to the eigen value 0 are c −1 ,
0
where c is a non-zero real number.
∴ So, The rank of eigen space corresponding to the eigen value 0
is 1.
Hence the geometric multiplicity of the eigen value 0 is 1.
So, in this case G.M.≤ A.M.
Dr. Anirban Lakshman (IIIT Kalyani) MAC101 27 / 35
 D IAGONALISATION

 
1
The eigen vectors corresponding to the eigen value 1 are c −1 ,
1
where c is a non-zero real number.
Hence the geometric multiplicity of the eigen value 1 is 1.
So, in this case G.M. = A.M.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 28 / 35

 D IAGONALISATION

Definition
An n × n matrix A is said to be diagonalisable if A is similar to an
n × n diagonal matrix.

If A is similar to a diagonal matrix D = diag(λ1 , λ2 , . . . , λn ), then

λ1 , λ2 , . . . , λn are the eigen values of A.
If A is similar to a diagonal matrix D, then D = P −1 AP, where P is
an non singular matrix.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 29 / 35

 D IAGONALISATION

T HEOREM
An n × n matrix A over a field F is diagonalisable iff ∃ n eigen
vectors of A which are linearly independent.

T HEOREM
Let A be an n × n matrix over a field F . If the eigen values of A be
all distinct and belong to F , then A is diagonalisable.
The condition stated in this theorem is not necessary for matrix A
to be diagonalisable
 
3 2 1
If we consider the matrix A = 2 3 1, then we show that
0 0 1
matrix A is diagonalisable but all the eigen values are not distinct.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 30 / 35

 D IAGONALISATION

Defiition
An eigen value λ is said to be regular if the G.M. of λ is equal to its
A.M.

T HEOREM
An n × n A is diagonalisable ⇐⇒ all its eigen values are regular.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 31 / 35

 D IAGONALISATION
Example:Find −1
 a matrix Psuch that P AP is a diagonal matrix
1 1 −2
where A = −1 2 1 
0 1 −1
=⇒ The eigen values of A are 1, 2, −1
The eigen vector
  corresponding
    to the eigen values 1,2,-1 are
3 1 1
respectively c1 2 , c2 3 , c3 0 ; where c1 , c2 , c3 are arbitrary
1 1 1
non-zero real numbers.
Since the eigen values are distinct, there are three linearly
independent eigen vectors of A, one corresponding to each eigen
value.      
3 1 1
The eigen vectors 2 , 3 and 0 are linearly independent.
1 1 1
Dr. Anirban Lakshman (IIIT Kalyani) MAC101 32 / 35
 D IAGONALISATION

 
3 1 1
Let P = 2 3 0 . Then P is a non singular matrix.
1 1 1
Now,      
1 1 −2 3 1 1 3 1 1 1 0 0
AP = −1 2 1  2 3 0 = 2 3 0 0 2 0  = PD
0 1 −1 1 1 1 1 1 1 0 0 −1
⇒ P −1 AP = D = diag(1, 2, −1)
So, A is a diagonal matrix.

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 33 / 35

 O UTLINE

1 E IGEN - VALUES ,E IGEN - VECTOR AND E IGEN SPACE

2 M INIMAL P OLYNOMIAL

3 R EFERENCES

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 34 / 35

 R EFERENCES

Gilbert Strang, Linear Algebra and Its Applications.

Steven J. Leon, Linear Algebra with Applications, Pearson.
Carl D. Meyer, Matrix Analysis and Linear Algebra,Siam
Kenneth Hoffman; Ray Kunze, Linear Algebra, Pearson

Dr. Anirban Lakshman (IIIT Kalyani) MAC101 35 / 35