skmm 3023 applied numerical methods · outline 1 introduction 2 engineering applications 3...

SKMM 3023 Applied Numerical Methods

Solution of Matrix Eigenproblems

ibn ‘Abdullah

Faculty of Mechanical Engineering

ibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 1 / 31

Outline

1 Introduction

2 Engineering Applications

3 Definitions and Basic Facts

Eigenvalue & Eigenvector

Characteristic Equation

Standard & General Eigenvalue Problems

4 Power Method

5 Faddeev-Leverrier Method

6 Matlab eig Function

7 Bibbliographyibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 2 / 31

Introduction

In many engineering problems, it may became necessary to formulate clearly thedifference between

the state of something that is uncertain (e.g. an electron in a probability cloud, mass indamped oscillations), ANDthe state of something having a definite value (e.g. having a definite position, a definitemomentum, a definite measured value, and a definite time of occurrence)

When an object can definitely be “pinned down” in some respect, it is said to

possess an eigenstate.

The word eigenstate is derived from the German/Dutch word eigen, meaning

inherent or characteristic.

An eigenstate is the measured state of some object possessing quantifiable

characteristics such as position, momentum, etc. The state being measured and

described must be observable (i.e. something such as position or momentum that

can be experimentally measured either directly or indirectly), and must have a

definite value, called an eigenvalue.

Source: http://en.wikipedia.org/wiki/Introdu tion_to_eigenstatesibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 3 / 31

Introduction

Eigenvalues play an important role

in situations where the matrix is a

transformation from one vector

space onto itself.

Systems of linear ordinary

differential equations are the

primary examples.

The eigenvalues can correspond to

critical values of stability

parameters, or energy levels of

atoms, or different characteristic

polynomials (of eigenvalue)

describing geometric

transformations (scaling, unequal

scaling, rotation, horizontal shear,

hyperbolic rotation) or frequencies

of vibration–see Figure 1.

Figure 1: Vibration of mass-spring systemwith three degress of freedom.

The equations of motion for a system of masses andspring shown in Figure 1 are

m1q̈1 + (k1 + k2 + k4)q1 − k2q2 − k4q3 = 0

m2 q̈2 − k2q1 + (k2 + k3)q2 − k3q3 = 0

m3 q̈3 − k4q1 − k3q2 + (k3 + k4)q3 = 0(1)ibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 4 / 31

Engineering ApplicationsExample 1

Figure 2: Forging hammer.ibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 5 / 31


Problem Statement:

A forging hammer of mass m1 is mounted on a concrete foundation block of mass m2.

The stiffness of the springs underneath the forging hammer and the foundation block

are given by k2 and k1, respectively (see Figure 2). The system undergoes simple

harmonic motion at one of its natural frequencies ω, which are given by

»

k1 + k2 −k2

−k2 k2

–

x1

x2

ff

= ω2

»

m1 0

0 m2

–

x1

x2

ff

(a)

where ω2 is the eigenvalue and ~XT = {x1 x2} is the eigenvector or mode shape

(displacement pattern) of the system. Determine the natural frequencies and mode

shapes of the system for the following data:

Solution:

Work through the example.



Problem Statement:

The pin-ended column shown in

Figure 3(a) is subjected to a compressive

(axial) force, P. If the column is

perturbed slightly as shown in

Figure 3(b), as might occur from a slight

vibration of the supports, it may not

return to horizontal position even after

removal of the distrubances; instead, the

deflection might grow if the load is

sufficiently large. Such load is called the

buckling load.

Solution:

Work through the example. Figure 3: Pin-ended column.


Definitions and Basic Facts

A matrix is called symmetric if it is equal to its transpose, i.e.

A = AT

or aij = aji (2)

Example:

A =

2

4

2 7 3

7 9 4

3 4 7

3

5 (3)

A Hermitian matrix (or self-adjoint matrix) is a square matrix with complex entries

that is equal to its own conjugate transpose, A∗–i.e. the element in the i-th row and

j-th column is equal to the complex conjugate of the element in the j-th row and

i-th column, for all indices i and j, i.e.

A = AT = A∗

or aij = aji (4)

Example:

A =

2

4

2 (2 + i) 4

(2 − i) 3 i

4 −i 1

3

5ibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 8 / 31

Definitions and Basic Facts

A matrix is termed orthogonal if its transpose equals its inverse, i.e.

AT = A

−1which entails A

T · A−1 = A

−1 · AT = 1 (5)

Example: A 2 × 2 rotation matrix, say for θ = 16.26◦,

A =

»

cos θ − sin θ

sin θ cos θ

–

A complex square matrix A is normal if

A∗ · A = A · A

∗

(6)

where A∗ is the conjugate transpose of A, e.g.

A =

2

4

1 1 0

0 1 1

1 0 1

3

5 (7)


Definitions and Basic FactsEigenvalue & Eigenvector

An eigenvalue and eigenvector of a square matrix [A] are a scalar λ, and a nonzero

vector ~X so that

[A]~X = λ~X

Consider matrix A and three vectors x1, x2, and x3,

A =

2

4

1 2 16 −1 0

−1 −2 −1

3

5 (8)

x1 =

2

4

16

−13

3

5 x2 =

2

4

−121

3

5 x3 =

2

4

23

−2

3

5 (9)

If we pre-multiply each of these vectors with matrix A we get

Ax1 =

2

4

000

3

5 Ax2 =

2

4

4−8−4

3

5 Ax3 =

2

4

69

−6

3

5 (10)

in other words, we have

Ax1 = 0x1 Ax2 = −4x2 Ax3 = 3x3

In this case we say 0, -4 and 3 are eigenvalues of the matrix A, and x1, x2 and x3 areeigenvectors of A.ibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 10 / 31


To find eigenvalues and eigenvectors of a given matrix we proceed as follows:

Form the matrix A − λI, i.e. substract λ from each diagonal element of A.

Solve the characteristic equation |A − λI| = 0 for λ.

Take each value of λ in turn, substitute into . . .



Standard eigenvalue problem is defined by the homogeneous equations

(a11 − λ)x1 + a12x2 + a13x3 + . . . + a1nxn = 0

a21x1 + (a22 − λ)x2 + a23x3 + . . . + a2nxn = 0

. . .

an1x1 + an2x2 + an3x3 + . . . + (ann − λ)xn = 0 (11)

and in matrix form

[A]~X = λ~X (12)



In Eq. (12),

1 [A] is a known square, (n × n),matrix

[A] =

2

6

4

a11 a12 . . . a1n

a21 a22 . . . a2n

. . .

an1 an2 . . . ann

3

7

5

(13)

2 ~X is the unknown n-componentvector,

~X =

8

>

<

>

:

x1

x2

. . .

xn

9

>

=

>

;

(14)

3 λ is an unknown scalar, a.k.a.eigenvalue.

Eq. (12) can be re-written as

[[A] − λ[I]]~X = ~0 (15)

where [I] is the identity matrix of

order n.

Eq. (15) has nontrivial solution

because it represents a system of n

homogeneous equation in n + 1

unknowns. Thus the determinant of

the coefficient matrix of ~X in

Eq. (15) must be zero:

|[A] − λ[I]| = 0 (16)


Definitions and Basic FactsCharacteristic Equation

If we expand Eq. (16) we get

˛

˛

˛

˛

˛

˛

˛

˛

(a11 − λ) a12 . . . a1n

a21 (a22 − λ) . . . a2n

. . .

an1 an2 . . . (ann − λ)

˛

˛

˛

˛

˛

˛

˛

˛

= 0 (17)

which, upon further expansion, gives an nth order polynomial in λ,

Pn(λ) = anλn + an−1λ

n−1 + . . . + a2λ2 + a1λ + a0 = 0 (18)

called the characteristic equation, where a0, a1, a2 . . . an are coefficients of

polynomial. Assuming λ1, λ2, . . . , λn are roots of the characteristic equation, then

there will be n solutions to Eq. (12).


Definitions and Basic FactsCharacteristic Equation

Corresponding to each distinct eigenvalue λi, a nontrivial solution of linear

equations in Eq. (12) can be determined through

[A]~X(i) = λi~X(i) (19)

where

~X(i) =

8

>

>

>

<

>

>

>

:

x(i)1

x(i)2

. . .

x(i)n

9

>

>

>

=

>

>

>

;

(20)

is called the eigenvector corresponding to eigenvalue λi.


Definitions and Basic FactsStandard & General Eigenvalue Problems

Standard Eigenvalue Problem

Eigenvalue problem as expressed by Eq. (12),repeated below,

(a11 − λ)x1 + a12x2 + . . . + a1nxn = 0

a21x1 + (a22 − λ)x2 + . . . + a2nxn = 0

. . .

an1x1 + an2x2 + . . . + (ann − λ)xn = 0

and in matrix form

[A]~X = λ~X

is known as the standard eigenvalue problem.

General Eigenvalue Problem

Many physical problems, however, areexpressed as the general eigenvalue problemgiven by

[A]~X = λ[B]~X (21)

General eigenvalue problem of Eq. (21) can bereduced to the standard eigenvalue problem ofEq. (12). We shall deal with this conversionlater!


Definitions and Basic FactsExample 1: Finding Eigenvalues & Eigenvectors of a 2 × 2 Matrix

If

A =

»

0 1−2 −3

–

then the characteristic equation is

|A − λ · I| =

˛

˛

˛

˛

»

0 1−3 −3

–

−

»

λ 00 λ

–˛

˛

˛

˛

= 0

˛

˛

˛

˛

»

−λ 1−2 −3 − λ

–˛

˛

˛

˛

= λ2 + 3λ − 2 = 0

and the two eigenvalues are

λ1 = −1, λ2 = −2

Eigenvector, v1, associated with the eigenvalue, λ1 = −1, is

A · v1 = λ1 · v1

(A − λ1) · v1 = 0»

−λ1 1−2 −3 − λ1

–

· v1 = 0

»

1 1−2 −2

–

· v1 =

»

1 1−2 −2

–

·

»

v1,1

v1,2

–

= 0ibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 17 / 31

Power Method

Since each eigenvector is associated with an eigenvalue, we often refer to an

eigenvector X and eigenvalue λ that correspond to one another as an eigenpair.

Many of the “real world” applications are primarily interested in the dominant

eigenpair.

The dominant eigenvector of a matrix is an eigenvector corresponding to the

eigenvalue of largest magnitude (for real numbers, largest absolute value) of that

matrix. The method that is used to find this eigenvector is called the power

method.


Power MethodSequencing the Iteration

Like the Jacobi and Gauss-Seidel methods, the power method for approximating

eigenvalues is iterative.

1 First we assume that the matrix A has a dominant eigenvalue with corresponding

dominant eigenvectors.

2 Then we choose an initial approximation of one of the dominant eigenvectors of A.

This initial approximation must be a nonzero vector in Rn.

3 Finally we form the sequence given by

x1 = A x0

x2 = A x1 = A(A x0) = A2x0

x3 = A x2 = A(A2x0) = A

3x0

...

xk = A xk−1 = A(Ak−1x0) = A

kx0 (22)ibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 19 / 31

Power MethodSequencing the Iteration–Example 1

We demonstrate those sequences on the matrix

A =

»

3 61 4

–

and, for Step 1, arbitrarily choose

x0 =

»

11

–

x1 = A x0 =

»

3 61 4

– »

11

–

=

»

95

–

=⇒ x′

1 =

»

1.81

–

x2 = A x1 =

»

3 61 4

– »

1.81

–

=

»

11.45.8

–

=⇒ x′

2 =

»

1.9655172411

–

x3 = A x2 =

»

3 61 4

– »

1.9655172411

–

=

»

11.896551725.96551724

–

=⇒ x′

3 =

»

1.9942196531

–

x4 = A x2 =

»

3 61 4

– »

1.9942196531

–

=

»

11.982658965.99421965

–

=⇒ x′

4 =

»

1.999035681

–

It looks like an eigenvector is

x =

»

21

–


Power MethodSequencing the Iteration–Example 1 (continued)

The corresponding eigenvalue is

xTAx

xTx=

ˆ

2 1˜

»

3 61 4

– »

21

–

ˆ

2 1˜

»

21

– =[30]

[5]=⇒ λ = 6


Power MethodAlgorithm & Matlab Implementation

The power method iteration in effect

takes successively higher powers of

matrix times initial starting vector

Algorithm: Power Method

x0 = arbitrary nonzero vector

for k = 1, 2, . . .

yk = A xk−1

xk = yk/||yk||∞

end

Implementing the algorithm in Matlab:

Matlab Session>> A = hilb(5) % matrix A>> x0 = ones(5,1) % nonzero ve tor>> yk = A*x0>> xk = yk/norm(yk)Compare the new value of x with the original.

Repeat the last two lines (hint: use the scrollup button).

Compare the newest value of x with theprevious one and the original. Notice thatthere is less change between the second two.

Repeat the last two commands over and overuntil the values stop changing.


Faddeev-Leverrier Method

Let A be and n × n matrix. The determination of eigenvalues and eigenvectors

requires the solution of

AX = λX

where λ is the eigenvalue corresponding the eigenvactor X. The values λ must

satisfy the equation

det(A − λI) = 0

Hence λ is a root of an nth degree polynomial P(λ) = det(A − λI), which we write

in the form

P(λ) = λn + c1λ

n−1 + c2λn−2 + . . . + cn−2λ

2 + cn−1λ + cn (23)

The Faddeev-Leverrier algorithm is an efficient method for finding the coefficients

ck of the polynomial P(λ). As an additional benefit, the inverse matrix A−1 is

obtained at no extra computational expense.ibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 23 / 31

Faddeev-Leverrier MethodTrace of Matrix A

The trace of the matrix A, written Tr[A], is

Tr[A] = a1,1 + a2,2 + . . . + an,n (24)

The algorithm generates a sequence of matrices Bnk=1 and uses their traces to

compute the coefficients of P(λ),

B1 = A and p1 = Tr[B1]

B2 = A(B1 − p1I) and p2 =1

2Tr[B2]

... . . ....

Bk = A(Bk−1 − pk−1I) and pk =1

kTr[Bk]

... . . ....

Bn = A(Bn−1 − pn−1I) and pn =1

nTr[Bn] (25)ibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 24 / 31

Faddeev-Leverrier MethodCharacteristic Polynomial

Then the characteristic polynomial, Eq. 23, can be re-written as

P(λ) = λn + p1λ

n−1 + p2λn−2 + . . . + pn−2λ

2 + pn−1λ + pn (26)

and the inverse matrix is given by

A−1 =

1

pn(Bn−1 − pn−1I) (27)


Faddeev-Leverrier MethodExample 1

Problem Statement:

Use Faddeev’s method to find the characteristic polynomial and inverse of the matrix

2

4

2 −1 1

−1 2 1

1 −1 2

3

5

Solution:

A =

2

4

2 −1 1

−1 2 1

1 −1 2

3

5

B1 =

2

4

2 −1 1

−1 2 1

1 −1 2

3

5

p1 = Tr[B1] = 6ibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 26 / 31


Solution (continued):

B2 = A(B1 − p1I)

B2 =

2

4

2 −1 1

−1 2 1

1 −1 2

3

5

0

@

2

4

2 −1 1

−1 2 1

1 −1 2

3

5 − 6

2

4

1 0 0

0 1 0

0 0 1

3

5

1

A

=

2

4

−6 1 −3

3 −8 −3

−1 1 −8

3

5

p2 =1

2Tr[B2] =

1

2(−22) = −11




B3 = A(B2 − p2I)

B3 =

2

4

2 −1 1

−1 2 1

1 −1 2

3

5

0

@

2

4

−6 1 −3

3 −8 −3

−1 1 −8

3

5 − (−11)

2

4

1 0 0

0 1 0

0 0 1

3

5

1

A

=

2

4

6 0 0

0 6 0

0 0 6

3

5

p3 =1

3Tr[B3] =

1

3(18) = 6

The characteristic polynomial is thus

P(λ) = λ3 −

3X

i=1

piλn−i = −6 + 11λ− 6λ

2 + λ3ibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 28 / 31



The inverse matrix is

A−1 =

1

p3(B3−1 − p3−1I)

=1

6

0

@

2

4

−6 1 −3

3 −8 −3

−1 1 −8

3

5 − (−11)

2

4

1 0 0

0 1 0

0 0 1

3

5

1

A

=1

6

0

@

2

4

−6 1 −3

3 −8 −3

−1 1 −8

3

5 −

2

4

−11 0 0

0 −11 0

0 0 −11

3

5

1

A

=1

6

2

4

5 1 −3

3 3 −3

−1 1 3

3

5 =

2

6

6

4

56

16

− 12

12

12

− 12

− 16

16

12

3

7

7

5


Matlab eig FunctionExample 1

Problem Statement:

2

4

−149 −50 −154

537 180 546

−27 −9 −25

3

5

8

<

:

x1

x2

x3

9

=

;

=

2

4

λ1 0 0

0 λ2 0

0 0 λ3

3

5

8

<

:

x1

x2

x3

9

=

;

Solution:

Matlab Session>> A = [-149 -50 -154; 537 180 546; -27 -9 -25℄>> [X,lambda℄ = eig(A)>> lambda1 = lambda(1,:)>> lambda2 = lambda(2,:)>> lambda3 = lambda(3,:)>> X1 = X(:,1)>> X2 = X(:,2)>> X3 = X(:,3)>> LHS1 = A*X1>> RHS1 = lambda1*X1>> % Che k if LHS=RHSibn.abdullah�dev.null 2014 SKMM 3023 Applied Numerical Methods Matrix Eigenproblems 30 / 31

Bibliography

1 STEVEN C. CHAPRA & RAYMOND P. CANALE (2009): Numerical Methods for Engineers, 6ed,ISBN 0-39-095080-7, McGraw-Hill

2 SINGIRESU S. RAO (2002): Applied Numerical Methods for Engineers and Scientists, ISBN0-13-089480-X, Prentice Hall

3 DAVID KINCAID & WARD CHENEY (1991): Numerical Analysis: Mathematics of ScientificComputing, ISBN 0-534-13014-3, Brooks/Cole Publishing Co.

4 STEVEN C. CHAPRA (2012): Applied Numerical Methods with MATLAB for Engineers andScientists, 3ed, ISBN 978-0-07-340110-2, McGraw-Hill

5 JOHN H. MATHEWS & KURTIS D. FINK (2004): Numerical Methods Using Matlab, 4ed, ISBN0-13-065248-2, Prentice Hall

6 WILLIAM J. PALM III (2011): Introduction to MATLAB for Engineers, 3ed, ISBN978-0-07-353487-9, McGraw-Hill


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