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Lesson 3: Diagonalization, Apuntes de Administración de Empresas

Asignatura: corporate mathematics, Profesor: , Carrera: Business Administration and Management, Universidad: URJC

Tipo: Apuntes

2017/2018

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TOPIC 3. DIAGONALIZATION OF ENDOMORPHISMS
Mathematics – 2011/2012 - 21 -
LESSON 3
DIAGONALIZATION OF ENDOMORPHISMS
THE PROBLEM OF MATRIX DIAGONALIZATION
In this lesson we only consider square matrices. If A is an n x n matrix, then
for X in
n
, AX is also a vector in
n
. A question of considerable
importance in many applied problems is the determination of vectors X
such that X and AX are parallel.
Such questions arise in many economic applications, such as “Continuous
Growth Models”, etc.
A matrix diagonalization is the process of taking a square matrix and
converting it into a special type of matrix, the so-called diagonal matrix,
which shares the same fundamental properties of the underlying matrix.
Matrix diagonalization is equivalent to transforming the underlying system
of equations into a special set of coordinates axes in which the matrix takes
the canonical form. Diagonalizing a matrix is also equivalent to finding the
matrix’s Eigenvalues, which turn out to be precisely the entries of the
diagonalized matrix. Similarly, the eigenvectors make up the new set of
axis corresponding to the diagonal matrix.
As we already know, we can represent a linear map according to a matrix
shape. If the linear transformation maps a vector space of dimension n into
another vector space of dimension n, then the underlying matrix of a linear
map is a square matrix and it can be diagonalized.
nn
VVf :
( )
Y AX
x y f x
=
 =
So, diagonalizing is equivalent to changing basis in a vector space so that
the underlying matrix of the linear map is diagonal.
Consequently, we are looking for a diagonal matrix
Λ
, similar to
A
, which
represents
f
.
Remark: Two n-by-n matrices A and B are called similar if there exists an
invertible n-by-n matrix P such that
P
−1
AP = B.
So, we need to define two new topics, Eigenvalues and Eigenvectors.
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LESSON 3

DIAGONALIZATION OF ENDOMORPHISMS

THE PROBLEM OF MATRIX DIAGONALIZATION

In this lesson we only consider square matrices. If A is an n x n matrix, then

for X in ℜ n , AX is also a vector in ℜ n. A question of considerable

importance in many applied problems is the determination of vectors X

such that X and AX are parallel.

Such questions arise in many economic applications, such as “Continuous

Growth Models”, etc.

A matrix diagonalization is the process of taking a square matrix and

converting it into a special type of matrix, the so-called diagonal matrix,

which shares the same fundamental properties of the underlying matrix.

Matrix diagonalization is equivalent to transforming the underlying system

of equations into a special set of coordinates axes in which the matrix takes

the canonical form. Diagonalizing a matrix is also equivalent to finding the

matrix’s Eigenvalues, which turn out to be precisely the entries of the

diagonalized matrix. Similarly, the eigenvectors make up the new set of

axis corresponding to the diagonal matrix.

As we already know, we can represent a linear map according to a matrix

shape. If the linear transformation maps a vector space of dimension n into

another vector space of dimension n , then the underlying matrix of a linear

map is a square matrix and it can be diagonalized.

f : Vn → V n x  Y^ = AXy = f ( ) x

So, diagonalizing is equivalent to changing basis in a vector space so that

the underlying matrix of the linear map is diagonal.

Consequently, we are looking for a diagonal matrix Λ , similar to A , which

represents f.

Remark: Two n-by-n matrices A and B are called similar if there exists an

invertible n-by-n matrix P such that P −1 AP = B.

So, we need to define two new topics, Eigenvalues and Eigenvectors.

Definition: Eigenvectors or proper vectors

For all x ∈ Vn , x ≠ 0 is said to be an eigenvector of f if and only if

⇔ ∃λ∈ K / f ( x )= λ x , x ∈ V n

Y = AX =λ X = λ In X

Definition: Eigenvalues or proper values

For all λ ∈ K , λ ∈ K is said to be an eigenvalue of f if and only if

f ⇔ ∃ x ≠ 0 / f ( x )= λ x , x ∈ V n

Summarising, the real number λ is called an eigenvalue of A if there exists

a nonzero vector X in ℜ n such that Y = AX = λ X. Every nonzero vector X

satisfying the previous equation is called an eigenvector of A associated

with the eigenvalue λ. We might mention that the word “eigenvalue” is

hybrid (“eigen” in German means “self”). Eigenvalues are also called

proper values, characteristics values, and latent values ; and

eigenvectors are also called proper vectors, and so on.

HOW TO COMPUTE EIGENVALUES AND EIGENVECTORS

If x ≠ 0 is an eigenvector associated with the eigenvalue λ , then

n n^0 (^ n )^0 the eigenvectorsHOMOGENEOUS SYSTEMwhose solution is associated with

K Y AX X I X AX I X A I X

λ

HOW TO COMPUTE EIGENVALUES

If the previous system contains the eigenvectors ( x ≠ 0 ), then the matrix

associated with the homogeneous system should not admit inverse. So, it

must be satisfied that:

Rk A ( − λ In )< n n

Characteristic equation

⇒ 1 A 4 − 24 λ I 3 = 0

From the former condition we can get the characteristic polynomial of A of

degree “ n ”, and whose real roots are the eigenvalues.

6. If λ i is an eigenvalue of A , λ^2 is an eigenvalue of A^2 as well. And

besides, every eigenvector in A will also be an eigenvector in A^2.

7. The eigenvalues in A and in At are the same.

8. The eigenvalues of the inverse matrix are the inverse of the matrix

eigenvalues.

Example

Let f be an endomorphism defined in ℜ^3 , and whose associated matrix is:

^ −
A
  • Eigenvalues:

( 2 )[( 1 )( 4 ) 2 ] 0

P λ A λ I

Eigenvalues ^ λλ 12 ==^ 2 ( 3 (^^ double root simple root ))

  • Eigenvectors:

S ( λ= 2 )=^ { x^ ∈ℜ^3 / f ( x )= 2 x }^ ={ x^ ∈ℜ^3 /( A − 2 I ) X = 0 }

^ ⇒
^ =
^ −

2 3

2 3

2 3

2 3 1 x x

x x

x x

x

x x

⇒ ^ xx 32^^ == −^0 x 2 = 0 ⇒ S ( λ= 2) = { x ∈ℜ^3 / x =( x 1 ,0, 0 )}

Dim S ( ({ λ( 11 , 0 =, 0 2))) } = n − rk A ( − 2 ) I = 3 − 2 = 1 = 1 free var iable ( x 1 )

B λ= 2 =

- S ( λ 2 = 3 )={ x ∈ℜ^3 / f ( x )= 3 x } ={ x ∈ℜ^3 /( A − 3 I ) X = 0 }=

(^223) 1 → − − + = ⇒ − + = ⇒ = =

^ =
^ −

2 3

2 3 3 1 2 3

2 3 1 x x

x x

x x x x

x

x x

= { x ∈ℜ^3 / ^ xx 31 == x −^22 x 2 ={ x ∈ℜ^3 / x =( x 2 , x 2 ,− 2 x 2 )}

Dim S ( { (( λ 1 , 1 =, 3)) 2 ) } = n − rk A ( − 3 ) I = 3 − 2 = 1 = 1 free var iable ( x 2 )

B λ= 3 = −

PROPERTIES OF SIMILAR MATRICES

1. If A and B are matrices, such that ( ∃ P / B = PAP −^1 ), then both of them

have got the same characteristic polynomial and, consequently, the

same roots (eigenvalues), trace, determinant, and rank.

2. If A and B are similar, their power matrices B k and Ak are also

similar, for positive integer values k :

P / B = PAP −^1 ⇒ Bk = PAkP −^1

THEOREM OF EXISTENCE OF Λ

f is diagonalizable ⇔ there exists a vector basis in Vn made by

eigenvectors. When that is the case, the diagonal matrix Λ is the

eigenvalues matrix.

  • In practice:

1. Work out the eigenvalues of A.

2. Work out the dimensions of each vector system of eigenvectors:

- if ∑∀ iDim ( S ( λ i ))= n ⇒∃an eigenvectors’ basis ⇒ Λ exists.

- if ∑∀ iDim ( S ( λ i ))< n ⇒ Λ doesn’t exist

3. If f is diagonalizable, then:

Exercise

Study if matrix A is diagonalizable:

^ −
A

Which is the relationship between the matrix A and its similar one?