applied maths day 1

Applied Mathematics Method I

Avid Farhoodfar

Lecture 1 Part 1

Four important matrices

(reviews of applied linear algebra)

Computational Science & Engineering

Gilbert Strang

Key points; Important Matrices

We are interested on their properties I will ask you about that?

We are interested on their meaningWhere they come from?

Why that matrix instead of some other ?

The numerical part is how to deal with themHow we solve a linear system with that coefficient

matrix?

What can we say about the solution?

Properties of Matrix K

2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

Consider Matrix K What are its properties

K=


2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

K=


1- K is symmetric

K = KT

2- K is sparse

3- K is tridiagonal

and those diagonals are

constant

If n=100

the matrix size is nXn=10000

And the number of non-zeros is 298


2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

K=


1- K is symmetric

K = KT

2- K is sparse

3- K is tridiagonal


constant Toeplitz

If n=100




2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

K=


1- K is symmetric

K = KT

2- K is sparseIf n=100



3- K is tridiagonal


constant Toeplitz


2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

K=


1- K is symmetric

K = KT

2- K is sparseIf n=100



3- K is tridiagonal


constant Toeplitz

boundary

boundary

Fixed-Boundaries


2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

K=


1- K is symmetric2- K is sparse3- K is Toeplitz

4- Is K invertible?


2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

K=



4- K invertible

There is an inverse matrix K-1

where KK-1=II is a unit matrix

I =

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1


2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

K=



4- K invertible

There two ways to see if a matrix is invertible

or not invertible

First way is elimination

Where we clean up below diagonal

and make our matrix upper triangular U


2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

K=



4- K invertible

First way elimination

2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

2 -1 0 0

0 3/2 -1 0

0 -1 2 -1

0 0 -1 2

2 -1 0 0

0 3/2 -1 0

0 0 4/3 -1

0 0 -1 2

2 -1 0 0

0 3/2 -1 0

0 0 4/3 -1

0 0 0 5/4

U =


2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

K=



4- K invertible

If an upper triangular matrix has a full set of

(non-zero) pivots it is invertible

2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

2 -1 0 0

0 3/2 -1 0

0 -1 2 -1

0 0 -1 2

2 -1 0 0

0 3/2 -1 0

0 0 4/3 -1

0 0 -1 2

2 -1 0 0

0 3/2 -1 0

0 0 4/3 -1

0 0 0 5/4

U =

pivot


2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

K=



4- K invertible

Matrix K is so important we will see it over and over again.

Part of the purpose of this lecture is to get matrices names

and to see if they are invertible or not invertible.

What we can change in K to make it non-invertible?

(one of the things we can do we get C matrix)


2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

K=



4- K invertible

2 -1 0 -1

-1 2 -1 0

0 -1 2 -1

-1 0 -1 2

C=

The C matrix is:

Properties of Matrix C

Consider Matrix C What are its properties

1- C is symmetric2- C is sparse3- C is Toeplitz

4- C is not invertible

2 -1 0 -1

-1 2 -1 0

0 -1 2 -1

-1 0 -1 2

C=





2 -1 0 -1

-1 2 -1 0

0 -1 2 -1

-1 0 -1 2

C=

But why C is not invertible?





2 -1 0 -1

-1 2 -1 0

0 -1 2 -1

-1 0 -1 2

C=


We can see it without eliminationC stands for Circulant.





2 -1 0 -1

-1 2 -1 0

0 -1 2 -1

-1 0 -1 2

C=


We can see it without eliminationC stands for Circulant.

It is because each -1 diagonal

Which have three elements

circled around to the forth.

The same for 0.

So they are not constants

but the loop around

Thats a periodic matrix

But can you find a solution

to get to zero?





2 -1 0 -1

-1 2 -1 0

0 -1 2 -1

-1 0 -1 2

C=

C is periodic

Cu= 02 -1 0 -1

-1 2 -1 0

0 -1 2 -1

-1 0 -1 2

We are looking for a u which

=

0

0

0

0

u can be

1

1

1

1

1

1

1

1

u=

But is it a solution?





2 -1 0 -1

-1 2 -1 0

0 -1 2 -1

-1 0 -1 2

C=

C is periodic

Cu= 02 -1 0 -1

-1 2 -1 0

0 -1 2 -1

-1 0 -1 2

=

0

0

0

0

1

1

1

1

If C-1 exists?

Multiply both sides

by C-1C-1Cu= C-10

I

u 0=This is the only solution which is not true here

u

So C is not invertible

A physical explanations about K

2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

K=

fixed

fixedsprings & masses

springs & masses are all the same Toeplitz

-1

-1Fixed ends

Here if there is no force

the system does not move

A physical explanation about C

2 -1 0 -1

-1 2 -1 0

0 -1 2 -1

-1 0 -1 2

C=

springs & masses are all the same Toeplitz

C is periodic

springs & masses

If there is no force

the system can still rotate

without compress of spring

No solution

because

Properties of Matrix L(T)

2 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 1

L=

Next Matrix is L What are its propertiesfixed

1- L and T are symmetric

2- L and T are sparse

3- L and T are not Toeplitz

are L and T invertible?

free

Lower end is free

L fixed-free boundary condition

T free-fixed boundary condition

free

fixed

L and T are tridiagonal

but those diagonals

are not constant

T=

fixed

free

Top end is free

1 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 2

Properties of Matrix B

1 -1 0 0

-1 2 -1 0

0 -1 2 -1

0 0 -1 1

B=

Last Matrix is B What are its properties

1- B is symmetric

2- B is sparse

is B invertible?free

Both ends are free

B free (open) boundary condition

free

free free

3- B is not Toeplitz

B is tridiagonal

but those diagonals

are not constant

Q : Is B invertible?

applied maths day 1

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