[hw2 ls ee571] akhtar rasool

8/12/2019 [HW2 LS EE571] Akhtar Rasool

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EE 571 Linear Systems Home Work 2 Page 1/ 28

EE 571 Linear Systems Spring 2014

Home Work # 2

Due Date: 19-03-2014

Submitted By: Akhtar Rasool

Student ID: 17197

Faculty of Engineering and Natural Sciences,

Sabanci University, Istanbul

Turkey


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Contents Page No.

Question # 01 3

Question # 02 3

Part (i) 3

Part (ii) 4

Part (iii) 6

Part (iv) 6

Question # 03 6

Question # 04 8

Question # 05 9

Question # 06 10

Part (a) 10

Part (b) 10

Part (c) 11

Question # 07 11

Part (a) 11

Part (b) 13

Part (c) 13

Question # 08 14

Question # 09 22

Question # 10 23

Part (a) 25

Part (b) 25

Part (c) 26


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Question # 01

Let

be a vector space and let

. Prove that

is a subspace of

if for all

, and for all

scalars ,, + .Solution-Proof:To show that is a subspace of , we need to prove that:

(a) If , , then is in .(b) If is any real number and is any vector in , then is in.

Thus let = + = + be vectors in

. Then

+ =( + ) + ( + )=(+ ) + (+ )which is in .Also, if is a scalar, then = ( + )=() + ()is in .Hence is a subspace of .Question # 02

Let =1 11 41 4 4221 1 0i. What is basis of the column span of.ii. Is this an orthonormal basis? If not determine an orthonormal basis for the column span of.iii. Determine the factorization of the matrix of.iv. Verify your result in part (iii) using MATLAB.

Solution (i)-Basis of the column span of A: The reduced Echelon form of

is determined as follows;

=1 10 50 5 4620 0 4 , , , = 1 10 10 5

46/520 0 4 , 15


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= 1 00 10 0

14/56/540 0 4

, + , 5= 1 00 10 0 14/56/510 0 4 ,

14 =1 00 10 0

0010 0 0 , 145, + 65 , + 4

The original vectors corresponding to all non-zero columns are a basis.

=[ ], =[ ], =[ ] Solution (ii):A set of vectors form an orthonormal set if all vectors in the set are mutually orthonogonal and all of unit

length.

For orthogonality: = ( )

=1111

1441= 1 + 4 + 4 1 = 6 0

Since the above one is not satisfied so this set is not an orthonormal set.

Now transform to an orthonormal basis:

= 1111 , 1441 ,

4220Let us call the three vectors above , . First, we will compute the norm of the vector ;

=1 + 1 + 1 + 1 =4 = 2Thus the first vector in the orthonormal basis is;= =

1/21/21/21/2Next we use and to find vectors orthogonal to .


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= =

1/21/21/21/2

Thus the orthonormal basis is;

1/21/21/21/2 , 1/21/21/21/2 ,

1/21/21/21/2 Solution-part(iii): QR factorization:

= 1111 ,

1441 ,

4220

= 1/2 1/2 1/21 / 2 1 / 2 1 / 21/21/2 1/21/2 1/21/2 0 .5 0.5 0 .50 . 5 0 . 5 0 . 50.50.5 0.50.5 0.50.5

= = 1/2 1/2 1/2 1/21/2 1/2 1/2 1/21/2 1/2 1/2 1/2 1 11 41 4

4221 1 0

= 2 3 20 5 20 0 4Solution-part(iv):QR factorization by MATLAB:

[, ]= ()= 2 3 20 5 20 0 4Gives the same result just the signs are opposite with all the elements. factorization of an matrixas the product of an matrixand matrix = , where = if > .Question # 03 (Problem 3.8 on page 80 from Chens Book)

Find the general solution of

=

How many parameters do you have?

Solution:

Let


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= 1 20 10 0

3 42 20 1

= 321

Also let

= 1 20 10 0 3 42 20 1 = [ ]

Since () = so , are linearly independent and ()= .Here : = () i.e; ([, ])= So this

lies in the range space of

and a solution

exists in

= .

Now we will find the solution, as below by using Gaussian Elimination Method.The augmented form is:1 2 30 1 20 0 0

4 32 21 1 = 1 2 30 1 20 0 0

4 3 2 21 1,(1)= 1 2 30 1 20 0 0

4 3 2 21 1 , 2Now writing the linear combination gives the following equations;

+ 8= 7, + 2 2= 2, = 1By backward substitution and letting = 0, the values are found to be;= 1, = 0, = 0, = 1So =[,,,]is the required solution.

As we know that

()= ()= = Which means the dimension of the null space ofis 1, the number of parameters in the general solutionwill be 1.Now we will find basis of the null space of, for this let;

= 1 20 10 0 3 42 20 1

The Reduced Echelon form is found to be;


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= = 1 00 10 0

1 82 20 1

, (1)& 2= = 1 00 10 0 1 02 00 1 , 8& + 2

Here is a free variable and , are basis variable.By solving = with = 1;

1 00 10 0 1 02 00 1

= 000

= 0 , + 2= 0 = 0

We get ; = 0, = 1, = 2 = 1So basis of the null space ofis; = [ ]Thus the general soution of = can be expressed as;

= + = +

, is the only parameter.

Question # 04 (Problem 3.10 on page 80 from Chens Book)

Find the solution in the following that has the smallest Euclidean norm.

=

Solution:

= + = + , ()Now of is;

=( 1)+ (2)+ + 1 =6 2 + 2The of is;


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= = = = =Substituting = in ()yields;

= /// Has the smallest .Question # 05

Lethave distinct eigenvalues with associated eigenvectors . Show that {} is linearlyindependent.Solution-Proof: Suppose that that {} is linearly independent.By reordering the if necessary,there exists some such that 1 and coefficients , , such that;

= ()And 0 . Let be the eigenvalue associated to for each . Then

=( )0

=( ) , ()= ( ) ,

= ( ) ,

= ( ) , = ( ) , =

= ( ) ,


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is mapped to

=

, , ,, ,

,, , ,

Where for each index ;= ,+ ,+ + ,=,

In general ifare not all zero, the vectorsandwill not be parallel. When they are parallel (i.e;when there is some real number such that = ) we say thatis an eigenvector of. In that case,the scale factor

is said to be the eigenvalue corresponding to that eigenvector.

Solution Part(c): Rank of matrix:Since hereis a matrix. So we can find its rank by the statement below. Ifis an matrix, then ()= iffis row equivalent to .For the proof of this statement,

If ()= thenis row equivalent to a matrix in reduced row echelon form, and ()=.Since

()= , we conclude that

has no zero rows. For this, as we know that;

If be an matrix in reduced row echelon form and ,thenhas a row consisting entirelyof zeros.This statement implies that = .Henceis row equivalent to .Conversely, ifis row equivalent to , then()= ()= Question # 07

Let be a Hermitian-symmetric matrix (i.e. = where represents complex conjugateoperation). Prove the following statements.(a) All eigenvalues of matrix A are real.(b) Eigenvectors corresponding to different eigenvalues are orthogonal. Then we can normalize the

length of all eigenvectors to unity in order to construct an orthonormal basis {} .(c) The model matrix of, namely =[, , , ], is unitary (.. = ).

Solution-Proof (a)Letbe a Hermitian Matrix. Then by definition;() = ()


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Let be an eigenvalue of.Let

be an eigenvector corresponding to the eigenvalue

.

By definition of eigenvector; = Pre-multiplying both sides by (), () =()=() = () ()Firstly note that both ()and ()are 1 1 matrices.Now from (ii) , consider Left side and taking conjugate transpose, we get;

()=()()() ()Asis Hermitian, and ()= , by double conjugate transpose is itself, it follows that;

()()()=()Substituting this result in (iii), we get;

()=()That is ()is also Hermitian. By product with conjugate transpose Matrix is Hermitian, ()isHermitian.

So both ()and ()are Hermitian1 1matrices. Now suppose that we have for some , () = []() = []Note that 0as a definition: eigenvector is non-zero.By definition of a Hermtian matrix: = = Where

denotes the complex conjugate of

.

By complex number equals conjugate iff wholly real, it follows that , , that is, are real.From equation (ii), []= []= =

= = , 0


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Hence , being a quotient of real numbers, is real.Solution-Proof (b)Suppose

is an eigenvalue of

with eigenvector

and

is an eigenvalue of

with

eigenvector

. Then

(. )=(). =(). , = = .(), = ()= . = . (), = = (.)= (. ), = . = = ,

So if the eigenvalues are different then . = .Hence proved..!

Suppose is an eigenvector corresponding to the eigenvalue of. Then= So

==|| () ( )

Butpreserves lengths becauseis here constructing orthonormal set, so= ()

Substituting ()in (), we get; =|| =||Hence proved..!

Solution-Proof (c):

Here = [ ]Suppose that = ,


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Then

= = =()

()

=()

By using if , are Hermitianthen their dot product is given by; =()So that; () = In particular, this holds when; = () So that;

() () = When; () = So, () = =() =

()=

Hence proved..!

Question # 08 (Problem 3.13 on page 81 Chens Book)

Find Jordan-form representations of the following matrices:

= , = , =

, =

Note that all exceptcan be diagonalized.Solution:

(a) := 1 4 1 00 2 00 0 3The characteristic polynomial ofis ;


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()= ( )= 1 4 1 00 2 00 0 3

=( 1)( 2)( 3)Since ()is a diagonal matrix so the eigenvalues ofare 1,2,3. They all are distinct so the ofwill be diagonal.The associated wtih = 1is any non-zero solution of

( 1)= 0 4 1 00 1 00 0 2 = 0= 0 4 1 00 1 00 0 2

= 000

Here we set

= ,now the above system can be written as;

4+ 10= 0, = 0,2= 0 = = 0, = 0Hence = ( )= ( 0 0)= (1 0 0) Where is any arbitrary number. We may also write as follows;= (1 0 0)The associated wtih = 2is any non-zero solution of

( 2)= 1 4 1 00 0 00 0 1 = 0= 1 4 1 00 0 00 0 1 =

000Here we set = , now the above system can be written as;+ 4+ 10= 0,2= 0= = 0, = , = 4Hence

= ( )= (4 0)= (4 1 0)Where is any arbitrary number. We may also write as follows;= (4 1 0)The associated wtih = 3is any non-zero solution of

( 3)= 2 4 1 00 1 00 0 0 = 0


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= 2 4 1 00 1 00 0 0

= 000

Here we set = , now the above system can be written as;2+ 4+ 10= 0, = 0= = , = 0, = 5Hence = ( )= (5 0 )= (5 0 1)Where is any arbitrary number. We may also write as follows;

= (5 0 1)

Now we have;

=[ ]= 1 4 50 1 00 0 1||= 1 4 50 1 00 0 1 = 1 1 00 1= 1

=()|| =11 1 00 1 0 00 1 0 10 0

4 50 1 1 50 1 1 40 04 51 0 1 50 0 1 40 1

= 1 0 0 4 1 0 5 0 1

= 1 4 50 1 00 0 1

Thus the ofwith respect to { } is;= = 1 4 50 1 00 0 1

1 4 1 00 2 00 0 3 1 4 50 1 00 0 1 =

1 4 50 1 00 0 1 1 8 1 50 2 00 0 3

= = = (b) := 0 1 00 0 1 2 4 3

The characteristic polynomial of is ;()= ( )= 1 00 12 4 + 3 = 14 + 3 + 0 12 + 3= (+ 3 + 4) +(2)


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= ()= ( )= + 3+ 4 + 2 = ( + 1)(+ 2 +2)

= ()= ( )=( + 1)(

+ 2 + 1 + 1)=( + 1)[( + 1)

(1)]

= ()= ( )=( + 1)[( + 1) ]=( + 1)( + 1 )( + 1 + )Thus the eigenvalues ofare 1 , 1 + , 1 . They all are distinct so the ofwill be diagonal.The associated wtih = 1is any non-zero solution of;

(+ 1)= 1 1 00 1 1 2 4 2 = 0

= 1 1 00 1 1 2 4 2

=

000

Here we set = , now the above system can be written as;+ = 0, + = 0, 2 4 2= 0 = = , = , = Hence = ( )= ( )= (1 1 1)Where is any arbitrary number. We may also write as follows;= ( )The

associated wtih

= 1 + is any non-zero solution of;

( (1 + ))= 1 1 00 1 1 2 4 2 = 0= 1 1 00 1 1 2 4 2

= 000

Here we set = , now the above system can be written as;(1 )+ = 0; (1 )+ = 0; 2 4+ ( 2 )= 0= = , =( 1 + ), = 2,

Hence = ( )= ( ( 1 + ) 2 )= (1 1 + 2 )Where is any arbitrary number. We may also write as follows;= ( + )The associated wtih = 1 is any non-zero solution of


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( (1 ))= 1 + 1 00 1 + 1 2 4 2 +

= 0= 1 + 1 00 1 + 1 2 4 2 + = 000

Here we set = ,now the above system can be written as;(1 + )+ = 0, (1 + )+ = 0, 2 4+ ( 2 + )= 0= = , =( 1 ), = 2Hence

= ( )

= ( ( 1 ) 2 )

= (1 1 2 )

Where is any arbitrary number. We may also write as follows;= ( )Now we have;

=[ ]= + ||= 1 1 1 1 1 + 1

1 2 2 = 1 1 + 1

2 2 + 1 1 1

2 2 + 1 1 1

1 + 1 = 2

=()|| = 12 1 + 1 2 2 1 1 1 2 1 1 + 1 2 1 12 2 1 11 2 1 11 2 1 1 1 + 1 1 1 1 1 1 1 1 1 +

= 12 2 2 2 + 2 2 1 2 + 1 2 2 2 1 2 + 11 + 1 1 + 1 1 + + 1 = 12 4 1 + 1 + 4 2 1 1 + 2 2

= =()|| = 12 4 4 21 + 2 1 1 + 1 + 2 = 2 2 1 12 2 1 2 12 12 +2 1 +2 12

Thus the ofwith respect to { } is;


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=

= 2 2 1 1

2

2 1

2 1

2 12 +2 1 +2 12

0 1 00 0 1 2 4 3

1 1 1 1 1 + 1 1 2 2

= = = 2 2 1 12 2 1 2 12 12 +2 1 +2 12

1 1 + 1 1 2 2 1 2 + 2 2 2 = = = +

(c)

:

= 1 0 10 1 00 0 2The characteristic polynomial ofis ;

()= ( )= 1 0 10 1 00 0 2 =( 1)( 1)( 2)=( 1)( 2)Since ()is a diagonal matrix so the eigenvalues ofare 1,1,2. The eigenvalue 1 has multiplicity 2,and nullity

( )= 3 ( )= 3 1 = 2.

Thehas two linearly independent associated wtih = 1.( 1)= 0 0 10 0 00 0 1 = 0= 0 0 10 0 00 0 1

= 000

So,

= ( )

Similarly, = ( )The associated wtih = 2is any non-zero solution of

( 2)= 1 0 10 1 00 0 0 = 0


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= 1 0 10 1 00 0 0

= 000

So, = ( )Now we have;

=[ ]= ||= 1 0 10 1 0

0 0 1 = 1 1 00 1= 1

=()|| = 11 1 00 1 0 00 1 0 10 0 0 10 1 1 10 1 1 00 00 11 0 1 10 0 1 00 1

= 1 0 00 1 01 0 1 = 1 0 10 1 00 0 1

Thus the ofwith respect to { } is;= = 1 0 10 1 00 0 1

1 0 10 1 00 0 2 1 0 10 1 00 0 1 =

1 0 10 1 00 0 1 1 0 20 1 00 0 2

= = = (d) :

= 0 4 30 2 0 1 60 2 5 2 0The characteristic polynomial ofis ;

()= (

)= 4 30 2 0 1 60 2 5 + 2 0 =

Clearlyhas only one disctinct eigenvalue 0with multiplicity 3, ( 0) = 3 2 = 1Thushas only one independent eigenvector associated with 0;=

= 0 4 30 2 0 1 60 2 5 2 0 =

000


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Here we set = 1, = 0now the above system can be written as;

4+ 3= 0 = = 0Hence = ( )= ( )We can compute generalized eigenvectorsoffrom equations below;=

= 0 4 30 2 0 1 60 2 5 2 0 =

100

4+ 3= 1,20+ 16= 0

Solving both of the above equations gives; = ( )Now, =

= 0 4 30 2 0 1 60 2 5 2 0 =

045

4+ 3= 0,20+ 16= 4

Solving both of the above equations gives; = ( )Now we have;

=[ ]= ||= 1 0 00 4 3

0 5 4 = 1 4 35 4= 1

=()|| =11 4 35 4 0 30 4 0 40 5 0 05 4 1 00 4 1 00 50 04 3 1 00 3 1 00 4

= 1 0 00 4 50 3 4 = 1 0 00 4 30 5 4

Thus the repreofwith respect to the basis { } is ;


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= = 1 0 00 4 30 5 4

0 4 30 2 0 1 60 2 5 2 0

1 0 00 4 30 5 4

= 1 0 00 4 30 5 4

0 1 00 0 40 0 5

= = =

Question # 09 (Problem 3.21 on page 82 from Chens Book)

Given

= 1 1 00 0 10 0 1Find

,

.

Solution:

The characteristic polynomial ofis;()= ( )= =( ) = ()= ( )=( )()( )=( ) = ( )

Let

()= + +

on the spectrum of, we have(i) Computation of: ()= ()(0)= (0), (0)= = = 0(1)= (1), (1)= + + = + = 1()= (), 10(1)= + 2= + 2= 10Solving above equations we have;

= 0, = 8 = 9= 8 + 9= 8 1 1 00 0 10 0 1 + 9 1 1 10 0 10 0 1 =

(ii) Computation of: ()= ()(0)= (0), (0)= = = 0


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(1)= (1), (1)= + + = + = 1

()

= ()

, 103(1)

= + 2= + 2= 103

Solving above equations we have; = 0, = 101 = 102= 101 + 102= 1011 1 00 0 10 0 1 + 1 0 2

1 1 10 0 10 0 1 =

(iii) Computation of :

()= ()

(0)= (0), = = = 1(1)= (1), = + + = + = 1()= (), = + 2= + 2= Solving above equations we have;= 1, = 2 2 = + = + +

= = 1 0 00 1 00 0 1 + (2 2 ) 1 1 00 0 10 0 1 + ( + ) 1 1 10 0 10 0 1

= = + Question # 10

(a) Show that norm of a vector , =[ || ]/ satisfies all the axioms of a norm for . (Hint: Use Minkowski Inequality when it is necessary)(b) Show that norm of a vector becomes = ,, ||when we set = in the above

expression.

(c) Determine the , , norms of the following matrices:= , = , = Solution:

(a) Show that norm of a vector , =[ || ]/ satisfies all the axioms of a norm for . (Hint: Use Minkowski Inequality when it is necessary)


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Solution-Proof (a): A norm on a vector space is a map . : such that for any , and any , where is a field.1.

, = = 2. = ||3. + + (Triangular Inequality)

Proof (a-1):

It is obvious as 1so

=

|| / = ( ) || = = , for each

Proof (a-2):

= ||

/

= ||

/

= ||

/

= ||

/

= Proof (a-3):

The case = 1is obtained by taking summation on,| + ||| +||If 1 < < , define from the equation + = 1.Then using Holders inequality, we get

|+ | = |+ ||+ | Using absolute property of inequality;

+ + |+ |||

Using Holders inequality,


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+ ()

/

/ + + ()

/

/Dividing the inequality by + () /andusing that ( 1) = and (1 = ), we get

+ /

+ /

i.e;

+ + Hence proved..!(b) Show that norm of a vector becomes = ,, ||when we set = in the above

expression.

Solution-Proof (b):Since

||

=

, 1We have Thus in particular; ()On the other had, we know that

=


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= (3.3772, 66.6228)

= = (

) =66.6228 = . : =4+ 3+ 3+ 6=70 = . Solution - Computation of Norms (c-2):

= ,= max(|1| + |3|), (|2| + |0|) = max(4,2)= = max(|1| + |2|), (|3| + |0|) = max(3,3)= =1 32 0=1 32 0 1 23 0 =10 22 4

= = 10 22 4=( 10)( 4) 4 = 14 + 36 = = ( 7) 13= 7 13 7 + 13 = 7 + 13,7 13 = (3.3944,10.6056)

= = () = 10.6056 = . : =1+ 2+ 3+ 0=14 = . Solution - Computation of Norms (c-3):

=1 20 1= max(|1| + |0|), (|2| + |1|) = max(1,3)=

= max(|1| + |2|), (|0| + |1|) = max(3,1)=

=1 02 1=1 02 1 1 20 1 =1 22 5 = = 1 22 5=( 1)( 5) 4 = 6 + 1 = = ( 3) 22= 3 22 3 + 22


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[hw2 ls ee571] akhtar rasool

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