Lecture 01

FURTHER MATRIX ALGEBRA
FURTHER INTEGRATION
DIFFERENTIAL EQUATIONS
SERIES OF NUMBERS
TEGM3592: ENGINEERING MATHEMATICS II
University of Namibia
Faculty of Engineering & IT
Department of Mechanical Engineering
Mr. Naftali Indongo (MSc. Applied Mathematics (Stellenbosch))

Email: nindongo2@unam.na
12 August 2020

FURTHER INTEGRATION
SERIES OF NUMBERS
Outline
1 FURTHER MATRIX ALGEBRA
2 FURTHER INTEGRATION
3 DIFFERENTIAL EQUATIONS
4 SERIES OF NUMBERS

FURTHER INTEGRATION
SERIES OF NUMBERS
THE EIGENVALUE PROBLEM AND ITS APPLICATIONS

We study the problem
Av = λv (1)
where A is an n × n matrix, v is an n × 1 vector of unknowns
x1 , . . . , xn and λ is an unknown scalar. Equation (1) is called the
Eigenvalue Problem.

1 2
E.g. For the matrix A = the eigenvalue problem Av = λv
2 1
can be written as:

1 2 x1 x
=λ 1
2 1 x2 x2
| {z } |{z} |{z}
2×2 2×1 2×1

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SERIES OF NUMBERS
If we express Equation (1) as
Av = λIv (2)
where I is an n × n identity matrix. Thus,
(A − λI)v = 0 (3)
which is the homogeneous system of n equations in n unknowns
xi0 s where the coefficient matrix A − λI contains the parameter λ.

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Given the n × n matrix A, we find the values of λ (if any)

such that Equation (3) admits non-trivial solutions and find
those solutions.
The λi 0 s that lead to non-trivial solutions of vi are called

eigenvalues or characteristics values.
The corresponding non-trivial solutions for vi 0 s are called the

eigenvectors of A.

FURTHER INTEGRATION
SERIES OF NUMBERS
APPLICATIONS IN ENGINEERING
The eigenvalue problem (1) has numerous applications in different
engineering disciplines. e.g.
1 Study information transmission in Telecommunication

Engineering.
2 Designing bridges in Structural Engineering - The natural

frequency of the bridge is the eigenvalue of the smallest
magnitude of a system that models the bridge.
3 Solving decoupling three-phase systems through symmetrical

component transformation in Electrical Engineering.
4 Frequently used in Mining Engineering to explore land for

oil.
FURTHER INTEGRATION
SERIES OF NUMBERS
SOLUTION PROCEDURE
The eigenvalues problem (3) has the unique trivial solution v = 0

iff
det(A − λI) 6= 0 (4)
and a non-trivial solution iff
det(A − λI) = 0 (5)
Equation (5) is called the characteristic equation corresponding

to matrix A.

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SERIES OF NUMBERS
Let λ1 , λ2 , . . . , λk (1 ≤ k ≤ n) be k eigenvalues of an n × n matrix

A, then
Since det(A − λi I) = 0, then (A − λi I)x = 0 has a non-trivial
solution.
The solution can be designated by vi , where vi is called the

eigenvector corresponding to the eigenvalue λi .
The vi solution space is called eigenspace corresponding to

the eigenvalue λi .
For λ1 , λ2 , . . . , λk we have {v1 , v2 , . . . , vk } the

k − eigenspaces.

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SERIES OF NUMBERS
EXAMPLE 1.1
Determine all eigenvalues and eigenspaces of the matrix

 
2 2 1
A = 1 3 1
1 2 2

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SERIES OF NUMBERS
SOLUTION 1.1
(1.) Eigenvalues: To find the λi ’s, use det(A − λI) = 0. Thus
2−λ 2 1
1 3−λ 1 = 0 =⇒ −λ2 + 7λ2 − 11λ + 5 = 0
1 2 2−λ
=⇒ −(λ − 5)(λ − 1)2 = 0
∴ The eigenvalues of A are λ1 = 5 and λ2 = 1(twice).
λ2 = 1 is called an eigenvalue of multiplicity 2 because it is a

double root of the characteristic equation.

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SERIES OF NUMBERS
(1.) Eigenvectors: For λ1 = 5 we have that (A − λ1 I)v1 = 0. So,

      
−3 2 1 x1 0 −3 2 1 0
 1 −2 1  x2  = 0 =⇒  1 −2 1 0
1 2 −3 x3 0 1 2 −3 0
 
−3 2 1 0
=⇒  0 −4 4 0
0 8 −8 0
 
−3 2 1 0
=⇒  0 1 −1 0
0 0 0 0
We have the following system of equations:

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−3x1 + 2x2 + x3 = 0
x2 − x3 = 0
Let x3 = α where α ∈ R, then x2 − x3 = 0 =⇒ x2 = x3 = α and

−3x1 + 2x2 + x3 = 0 =⇒ x1 = − 13 (−2x2 − x3 ) = α. Thus,
     
x1 α 1
x2  = α = α 1
x2 α 1
 
1
∴ v1 = 1 is the eigenvector corresponding to the eigenvalue
1
λ1 = 5
FURTHER INTEGRATION
SERIES OF NUMBERS
For λ2 = 1 we have that (A − λ2 I)v2,3 = 0. So,

      
1 2 1 x1 0 1 2 1 0
1 2 1 x2  = 0 =⇒ 1 2 1 0
1 2 1 x3 0 1 2 1 0
 
1 2 1 0
=⇒ 0
 0 0 0
0 0 0 0
We have the following system of equations:
x1 + 2x2 + x3 = 0
Let x2 = α and x3 = β where α, β ∈ R, then x1 = −2α − β. So,

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SERIES OF NUMBERS
           
x1 −2α − β −2α −β −2 −1
x2  =  α  =  α  + 0 =α 1 +β 0 
    
x3 β 0 β 0 1
   
−2 −1
∴ v2 = 1 and v3 = 0  are the eigenvectors corresponding
  
0 1
to the eigenvalue λ2 = 1.
(3.) Eigenspaces: The eigenspaces corresponding to eigenvalues

λ1 = 5 and λ2 = 1 are:
     
 1   −2 −1 
s1 = 1 and s2 =  1  ,  0 
1 0 1
   
respectively.
FURTHER INTEGRATION
SERIES OF NUMBERS
HERMITIAN MATRICES
A complex square matrix A is said to be Hermitian or Self-adjoint
if and only if it is equals to its own conjugate transpose. i.e.
A is Hermitian ⇐⇒ A∗ = A
NB: Common notations use are AH and A†

Theorem 1.1
If A is a hermitian matrix, then all its eigenvalues are real.

1 2+i
For example, matrix A = is hermitian since
2−i 3
T
∗ 1 2−i 1 2+i
A = = =A
2+i 3 2−i 3

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SYMMETRIC MATRICES
A square matrix A is said to be symmetric if and only if AT = A.
Theorem 1.2: Real Eigenvalues Theorem

If A is symmetric, then all its eigenvalues are real.
Theorem 1.3
If an eigenvalue λi of a symmetric matrix A is of multiplicity k,
then eigenspace corresponding λi is of dimension k.
Theorem 1.4
If A is symmetric, then the eigenvectors corresponding to distinct
eigenvalues are orthogonal.

FURTHER INTEGRATION
SERIES OF NUMBERS
Exercise 1.1
For the symmetric matrix

 
2 1 1
A = 1 2 1
1 1 2
Find the eigenvalues, eigenvectors and eigenspaces of A.

FURTHER INTEGRATION
SERIES OF NUMBERS
PROPERTIES OF EIGENVALUES AND EIGENVECTORS

For an n × n matrix A;
1 The characteristic equation det(A − λI) = 0 has n distinct
roots (not necessarily unique) and if A is symmetric then the

roots are real (otherwise complex).
Pn
i=1 λi = trace(A)
2
Qn
i=1 λi = det(A)
3
4 If any λi = 0, then A is singular (det(A) = 0).

5 If vi is an eigenvector of A corresponding to the eigenvalue λi ,
the kv is also an eigenvector of A for any k ∈ R/{0}.
6 A and AT have the same eigenvalues, however there is no
general connection between the eigenvectors.
FURTHER INTEGRATION
SERIES OF NUMBERS
CAYLEY-HAMILTON (C-H) THEOREM
Named after Arthur Cayley (British 1821-1895; Lawyer) and

William Rowan Hamilton (Irish, Physician & Mathematician)
Let A be an n × n matrix and I be an n × n identity matrix, and let
D(λ) = det(A − λI)

= (−1)n λn + (−1)n−1 Cn−1 λn−1 + · · · + C0 (C1)
be the characteristics polynomial of A, then A satisfies its own

characteristics polynomial. Thus
D(A) = (−1)n An + (−1)n−1 Cn−1 An−1 + · · · + C0 I = [0] (C2)
where [0] is an n × n zero matrix.

FURTHER INTEGRATION
SERIES OF NUMBERS
From Equation (C1) we have that
Cn−1 = trace(A)
Cn−2 = ni=j cij = c11 + c22 + · · · + cnn - sum of the cofactors

P
along the main diagonal
..
.
C0 = det(A)
For example
1 D(λ) = λ2 − trace(A)λ + det(A) for a 2 × 2 matrix.

2 D(λ) = −λ3 + trace(A)λ2 − (c11 + c22 + c33 )λ + det(A) For a
3 × 3 matrix.

FURTHER INTEGRATION
SERIES OF NUMBERS
EXAMPLE 1.2
Show that the matrix

1 2
A=
3 4
satisfy the Cayley-Hamilton Theorem.

FURTHER INTEGRATION
SERIES OF NUMBERS
Solution 1.2

1−λ 2
= det
3 4−λ
= λ2 − 5λ − 2
The Cayley-Hamilton Theorem States that D(A) = [0]. Thus,
D(A) = A2 − 5A − 2I

7 10 5 10 2 0
= − −
15 22 15 20 0 2

0 0
=
0 0
∴ Matrix A satisfy the Cayley-Hamilton Theorem.
FURTHER INTEGRATION
SERIES OF NUMBERS
APPLICATIONS OF THE C-H THEOREM

1 FINDING THE INVERSE OF AN n × n MATRIX
Given
(−1)n An + (−1)n−1 Cn−1 An−1 + · · · + C0 I = [0]
then
(−1)n An + (−1)n−1 Cn−1 An−1 + · · · − C1 A = −C0 I
by multiplying both sides with A−1 we get
(−1)n An A−1 +(−1)n−1 Cn−1 An−1 A−1 +· · ·−C1 AA−1 = −C0 I A−1
Since An A−1 = An−1 , An−1 A−1 = An−2 , An−2 A−1 = An−3

AA−1 = I and I An−1 = A−1 , then
FURTHER INTEGRATION
SERIES OF NUMBERS
(−1)n An−1 + (−1)n−1 Cn−1 An−2 + · · · − C1 I = −C0 A−1

Therefore,
1
A−1 = (−1)n−1 An−1 + (−1)n−2 Cn−1 An−2 + · · · − C1 I

C0
1
= [Adj(A)] (C3)
det(A)

FURTHER INTEGRATION
SERIES OF NUMBERS
EXAMPLE 1.3

1 2
Using Matrix A = in Example 1.2 we have
3 4
A2 − 5A − 2I = [0]
Thus,
1
2A−1 = A − 5I =⇒ A−1 = [A − 5I ]
2
1 1 2 5 0
= −5
2 3 4 0 5

1 −4 2 −2 1
= = 3
2 3 −1 2 − 12

FURTHER INTEGRATION
SERIES OF NUMBERS
DIAGONALIZATION
Theorem 1.5: Diagonalization Theorem

Let A be an n × n matrix:
1 A is diagonalizable if and only if it has n linearly
independent eigenvectors (i.e α1 v1 + α2 v2 + · · · + αn vn = 0
for αi = 0 for all i = 1, . . . , n)
2 If A has n linearly independent eigenvectors v1 , v2 , . . . , vn

and we can make these to be columns of matrix Q, that is
Q = [v1 , v2 , . . . , vn ] then Q −1 AQ = D is a diagonal matrix
whose i th diagonal element is the i th eigenvalue of A

FURTHER INTEGRATION
SERIES OF NUMBERS
Theorem 1.6
If an n × n matrix A has distinct eigenvalues λ1 , λ2 ,. . . , λn then
the corresponding eigenvectors v1 , v2 , . . . , vn are linearly
independent.
Theorem 1.7
If an n × n matrix A has n distinct eigenvalues, then it is
diagonalizable .
Theorem 1.8
Every symmetric matrix is diagonalizable.

FURTHER INTEGRATION
SERIES OF NUMBERS
EXAMPLE 1.4
Consider the matrix

 
4 −3 −3
A =  3 −2 −3
−1 1 2
Show that A is diagonalizable, and find the non-singular matrix Q

such that Q −1 AQ = D

FURTHER INTEGRATION
SERIES OF NUMBERS
SOLUTION 1.4
(1.) Find the eigenvalues of A
The characteristics polynomial of A is

4−λ −3 −3
= 3 −2 − λ −3
−1 1 2−λ
= −(λ − 1)2 (λ − 2)
∴ The eigenvalues of A are λ1 = 1 (twice) and λ2 = 2.

FURTHER INTEGRATION
SERIES OF NUMBERS
(2.) Find the eigenvectors of A
For λ1 = 1 , we have (A − λ1 I)v1,2 = 0. So,

   
3 −3 −3 0 1 −1 −1 0
 3 −3 −3 0 = 0 0 0 0
−1 1 1 0 0 0 0 0
Let x2 = α and x3 = β where α, β ∈ R, then
x1 = x2 + x3 = α + β. So,
       
x1 α+β 1 1
x2  =  α  = α 1 + β 0
x3 β 0 1
   
1 1
∴ v1 = 1 and v2 = 0 are the eigenvectors corresponding
  
0 1
to the eigenvalue λ1 = 1
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For λ2 = 2 , we have (A − λ2 I)v3 = 0. So,

   
2 −3 −3 0 1 −1 −1 0
 3 −3 −3 0 = 0 1 3 0
−1 1 0 0 0 0 0 0
Let x3 = α where α ∈ R, then x2 = −3x3 = −3α and

x1 = −3x3 = −3α. So,
     
x1 −3α −3
x2  = −3α = α −3
x3 α 1
 
−3
∴ v3 = −3 is the eigenvector corresponding to λ2 = 2.
1

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(3.) Determine whether v1 , v2 and v3 are linearly independent

 
1 1 −3
Let Q = [v1 , v2 , v3 ] = 1 0 −3.
0 1 1
Since det(Q) = −1 6= 0, then v1 , v2 , and v3 are linearly
independent.
Since v1 , v2 , and v3 are linearly independent, then from

Theorem 1.5 matrix A is diagonalizable
(4.) Diagonalizing A
   
1 1 −3 1 0 0
Q −1 AQ = D where Q = 1 0 −3 and D = 0 1 0
0 1 1 0 0 2
FURTHER INTEGRATION
SERIES OF NUMBERS
QUADRATIC FORMS
A function of the form
f (x1 , x2 ) = a11 x12 + a22 x22 + 2a12 x1 x2 (6)
is called a quadratic form in variables x1 and x2 .
Similarly,
f (x1 , x2 , x3 ) = a11 x12 +a22 x22 +a33 x32 +2a12 x1 x2 +2a23 x2 x3 +2a13 x1 x3
(7)
is a quadratic form in x1 , x2 and x3 . etc. Thus,
f (x1 , x2 , . . . , xn ) = xT Ax (8)
 
x1
 x2 
. . ., A = {aij } is an n × n matrix.
where x =  
xn
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EXAMPLE 1.5
Find the matrix representation of the following quadratic functions
(a) Suppose that n = 2 and f (x1 , x2 ) = 2x12 − x22 + 6x1 x2

(b) Suppose n = 4 and
f (x1 , x2 , x3 , x4 ) = x22 + 5x32 + 6x1 x3 − x2 x3 + 10x3 x4

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SOLUTION 1.5
(a) We have that
xT Ax = a11 x12 + a22 x22 + (a12 + a21 )x1 x2
a11 = 2, a22 = −1, a12 + a21 = 6. Since A is symmetric, then
2 3
a12 = a21 . Thus a12 = a21 = 62 = 3. ∴ A =
3 −1
(b) We have that; a11 = coef (x12 ) = 0,
a12 = a21 = 12 coef (x1 x2 ) = 0, a13 = a31 = 12 coef (x1 x3 ) = 3,
a14 = a41 = 12 coef (x1 x4 ) = 0, a22 = coef (x22 ) = 1,
a23 = a32 = 12 coef (x2 x3 ) = − 12 , a24 = a42 = 21 coef (x2 x4 ),
a33 = coef (x32 ) = 5, a34 = a43 = 12 coef (x3 x4 ) = 5 and
 
0 0 3 0
0 1 − 1 0 
a44 = coef (x42 ) = 0. ∴ A =  3 − 1 5
2 
2 5
0 0 5 b0
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CANONICAL FORM
A quadratic form is said to be canonical if all its mixed terms

(such as x1 x2 , x2 x3 , x2 x4 ,. . . , etc.) are absent. i.e. aij = 0 for all
i 6= j. e.g.
f (x1 , x2 , . . . , xn ) = a11 x12 + a22 x22 + · · · + ann xn2
is canonical and its associated matrix is

 
a11 0 . . . 0
 0 a22 . . . 0 
A= .
 
.. .. ..
 ..

. . . 
0 0 ... ann
is diagonal.

FURTHER INTEGRATION
SERIES OF NUMBERS
CHANGE FROM QUADRATIC TO CANONICAL FORMS
Consider a linear change of variables
x = Qx̃ “Orthogonal transformation”
From x1 , . . . , xn to x˜1 . . . , xñ . Since f (x) = xt Ax, then
f (x̃) = (Qx̃)T AQx̃

= x̃T Q T AQx̃
= x̃T (Q T AQ)x̃

FURTHER INTEGRATION
SERIES OF NUMBERS
If Q = [v1 , vn , . . . , vn ] is a model matrix for a symmetric matrix A,

then
 
λ1 0 ... 0
 0 λ2 ... 0
Q T AQ = Q −1 AQ =  .
 
.. .. .. 
 .. . . .
0 0 ... λn
where λj ’s are the eigenvalues of A. ∴ f (x̃) = λ1 x˜1 2 + · · · + λn xñ 2 .
Theorem 1.9
A quadratic form f (x) = xT Ax can be reduced to canonical form
f (x̃) = λ1 x˜1 2 + · · · + λn xñ 2 by introducing the change of variables
x = Qx̃, where the λi ’s are the eigenvalues of A and the columns
of Q are the corresponding eigenvectors of A.
The reverse transformation is given x̃ = Q T x.

FURTHER INTEGRATION
SERIES OF NUMBERS
EXAMPLE 1.6
Reduce
f (x1 , x2 ) = 3x12 + 3x22 + 2x1 x2
to canonical form.

FURTHER INTEGRATION
SERIES OF NUMBERS
SOLUTION 1.6
We have that
f (x) = xT Ax

x1 3 1
where x = and A =
x2 1 3
To change to canonical form, let x = Qx̃, where Q = [v1 , v2 ]
where v1 and v2 are the eigenvectors of A.
So,
det(A − λI ) = λ2 − 6λ + 8
= (λ − 2)(λ − 4)
∴ λ1 = 2 and λ2 = 4.

FURTHER INTEGRATION
SERIES OF NUMBERS

−1 1
For λ1 = 2, v1 = and for λ2 = 4, v2 = .
1 1

−1 1
∴ Q = [v1 , v2 ] = and f (x̃) = x̃T (Q T AQ)x̃ = x̃T Dx̃,
1 1

2 0
where D = .
0 4
∴ f (x̃) = 2x˜1 2 + 4x˜2 is the canonical form of the quadratic form

f (x1 , x2 ) = 3x12 + 3x22 + 2x1 x2 .
1 1
−
Withx = Q x̃ = 12 21 or x1 = − x˜21 + x2˜2 and x2 = x˜21 + x˜22
∗
2 2

FURTHER INTEGRATION
SERIES OF NUMBERS
Theorem 1.10
A quadratic form xT Ax is classified as positive definite (i.e
definitely positive) if xT Ax > 0 for x 6= 0, and negative definite if
xT Ax < 0 for all x 6= 0.
Likewise, A is classified as positive (or negative) if the quadratic

form xT Ax is positive (negative) definite.
Theorem 1.11
Let A be symmetric, then A and its quadratic form xT Ax are
positive (negative) definite if every eigenvalue of A is positive
(negative).

FURTHER INTEGRATION
SERIES OF NUMBERS
EXERCISE 1.2
Determine whether the quadratic form
f (x1 , x2 ) = x12 + x22 + x1 x2
is positive or negative definite.

FURTHER INTEGRATION
SERIES OF NUMBERS
COMPLEX CONJUGATE OF A MATRIX
Given an n × n matrix
A = {aij }
we define the complex conjugate of matrix A as
Ā = {a¯ij }

2+i 3
For example, the complex conjugate of A = is
2−i 1−i

2−i 3
Ā =
2+i 1+i

FURTHER INTEGRATION
SERIES OF NUMBERS
The End of CHAPTER 1

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FURTHER MATRIX ALGEBRA

TEGM3592: ENGINEERING MATHEMATICS II

Mr. Naftali Indongo (MSc. Applied Mathematics (Stellenbosch))

FURTHER MATRIX ALGEBRA

1 FURTHER MATRIX ALGEBRA

FURTHER MATRIX ALGEBRA

THE EIGENVALUE PROBLEM AND ITS APPLICATIONS

FURTHER MATRIX ALGEBRA

If we express Equation (1) as

where I is an n × n identity matrix. Thus,

xi0 s where the coefficient matrix A − λI contains the parameter λ.

FURTHER MATRIX ALGEBRA

Given the n × n matrix A, we find the values of λ (if any)

The λi 0 s that lead to non-trivial solutions of vi are called

The corresponding non-trivial solutions for vi 0 s are called the

FURTHER MATRIX ALGEBRA

1 Study information transmission in Telecommunication

2 Designing bridges in Structural Engineering - The natural

3 Solving decoupling three-phase systems through symmetrical

4 Frequently used in Mining Engineering to explore land for

The eigenvalues problem (3) has the unique trivial solution v = 0

det(A − λI) = 0 (5)

Equation (5) is called the characteristic equation corresponding

FURTHER MATRIX ALGEBRA

Let λ1 , λ2 , . . . , λk (1 ≤ k ≤ n) be k eigenvalues of an n × n matrix

The solution can be designated by vi , where vi is called the

The vi solution space is called eigenspace corresponding to

For λ1 , λ2 , . . . , λk we have {v1 , v2 , . . . , vk } the

FURTHER MATRIX ALGEBRA

Determine all eigenvalues and eigenspaces of the matrix

FURTHER MATRIX ALGEBRA

(1.) Eigenvalues: To find the λi ’s, use det(A − λI) = 0. Thus

∴ The eigenvalues of A are λ1 = 5 and λ2 = 1(twice).

λ2 = 1 is called an eigenvalue of multiplicity 2 because it is a

FURTHER MATRIX ALGEBRA

(1.) Eigenvectors: For λ1 = 5 we have that (A − λ1 I)v1 = 0. So,

We have the following system of equations:

FURTHER MATRIX ALGEBRA

Let x3 = α where α ∈ R, then x2 − x3 = 0 =⇒ x2 = x3 = α and

For λ2 = 1 we have that (A − λ2 I)v2,3 = 0. So,

We have the following system of equations:

Let x2 = α and x3 = β where α, β ∈ R, then x1 = −2α − β. So,

FURTHER MATRIX ALGEBRA

(3.) Eigenspaces: The eigenspaces corresponding to eigenvalues

NB: Common notations use are AH and A†

FURTHER MATRIX ALGEBRA

Theorem 1.2: Real Eigenvalues Theorem

FURTHER MATRIX ALGEBRA

For the symmetric matrix

Find the eigenvalues, eigenvectors and eigenspaces of A.

FURTHER MATRIX ALGEBRA

PROPERTIES OF EIGENVALUES AND EIGENVECTORS

roots (not necessarily unique) and if A is symmetric then the

4 If any λi = 0, then A is singular (det(A) = 0).

CAYLEY-HAMILTON (C-H) THEOREM

Named after Arthur Cayley (British 1821-1895; Lawyer) and

Let A be an n × n matrix and I be an n × n identity matrix, and let

D(λ) = det(A − λI)

be the characteristics polynomial of A, then A satisfies its own

D(A) = (−1)n An + (−1)n−1 Cn−1 An−1 + · · · + C0 I = [0] (C2)

Show that the matrix