Eigenvalue Problems part 1-Parallel Computing-Lecture Slides, Slides of Parallel Computing and Programming

This lecture was delivered by Dr. Hanif Durad at Pakistan Institute of Engineering and Applied Sciences, Islamabad (PIEAS) for Parallel Computing course. it includes: Eigenvalue, Problems, Existence, Uniqueness, Conditioning, Eigenvectors, Structural, Engineering, Stability, Analysis

Typology: Slides

2011/2012

Uploaded on 07/19/2012

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Dr. Hanif Durad 2
Lecture Outline
Eigenvalue Problems
Existence, Uniqueness and Conditioning
Computing Eigenvalues and Eigenvectors
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Download Eigenvalue Problems part 1-Parallel Computing-Lecture Slides and more Slides Parallel Computing and Programming in PDF only on Docsity!

Dr. Hanif Durad

Lecture Outline



Eigenvalue Problems



Existence, Uniqueness and Conditioning



Computing Eigenvalues and Eigenvectors

origins (1/2)Eigenvalue Problems. Their



Structural Engineering [K

x

2

2

M

x

]



mechanical system,..]Stability analysis [e.g., electrical networks,



Bifurcation analysis [e.g. in fluid flow]



equation..]Electronic structure calculations [Schrödinger



world-wide web.Application of new era: page ranking on the

Dr. Hanif Durad

LecN8_R.pdf,P-

Dr. Hanif Durad

Eigenvalues and Eigenvectors

Note:  may be complex even if A is real

Eigen_002.pdf, P-

Geometric Interpretation

Dr. Hanif Durad

Characteristic Polynomial (1/2)

Dr. Hanif Durad

PolynomialExample: Characteristic

Dr. Hanif Durad

Characteristic Polynomial (2/2)

Dr. Hanif Durad

Assignment -

Dr. Hanif Durad

Find the characteristic polynomial, eigenvalues

and eigen vectors of the matrix

A

AEM Alan Jeffrey.pdf, P-

DiagonalizabilityMultiplicity and

Dr. Hanif Durad

SubspacesEigenspaces and Invariant

Dr. Hanif Durad

(1/2)Examples: Matrix Properties

Dr. Hanif Durad

(II)

(I)

III

(a)

(I

V

(a)

(b) (b)

(2/2)Examples: Matrix Properties

Dr. Hanif Durad

V

I)

(b) (a)

V

(a) (b) (c)

Dr. Hanif Durad

20

Consider the following example.

4

2

3

2

5

8

1

0

3

A























C The circles that bound the Eigenvalues are:

1 : Center point (4,0) with radius r

1

= |2|+|3|=

C

2 : Center point (-5,0) with radius r

2 =|-2|+|8|=

C

3 : Center Point (3,0) with radius r

3 =|1|+|0|=

5

10

-15 -

  • 5

10

Union of the Circles

5

-15 -

  • 5 10

Eigenvalueslocation of thethe actualthe right mark The red dots to

Dr. Hanif Durad

1

0

7

2

5

0

4

4

3

A























C The circles that bound the Eigenvalues are:

1 : Center point (1,0) with radius r

1

= |0|+|7|=

C

2 : Center point (-5,0) with radius r

2 =|2|+|0|=

C

3 : Center Point (-3,0) with radius r

3 =|4|+|4|=

5

10

-15 -

  • 5 10

inside the union of all the circles. We see all the Eigenvalues lie

Consider another example.