Group assignments given a linear transformation as below, find the f dimension and one basic of im( )

ĀĀ = �㔴Ā ⇔ �㔴 = ĀĀĀ21 where fE, E are two matrices formed by putting image vector of basis, vector of basis in column, respectively.. A square matrix is said to be diagonalizable if it i

VIETNAM NATIONAL UNIVERSITY HO CHI MINH CITY

UNIVERSITY OF TECHNOLOGY FACULTY OF APPLIED SCIENCE -*** -

Linear Algebra (MT1008)

Group Assignments

Topic: 1

Group: 9

Team members:

1 Trần Quốc Thái - 2153795

2 Bùi Văn Nhật Thanh - 2152280

3 Hà Huy Thành - 2152967

4 Lê Thành - 2052706

5 Trịnh Hoài Thanh - 2053427

Semester: HK212

Lecturer: Phan Thi Huong

Submission date: Saturday, May 14 , 2022 th

REQUIREMENTS:

● Students work on your assigned groups

● Detailed explanations must be provided to get full scores

● To solve the following questions, please let the constant a be your group ID

● All given exercises must be done by both methods: manual solving and

using Matlab or Python

TABLE OF CONTENT Project details for each question:

I Theories summary

II Solution details

III Coding details

Question 1: Given a linear transformation as below, find the dimension and one f

basic of Im( ): f

f(x ,x ,x1 2 3) = (x + x 1 2;x2 + x3;x1 - ax3), ( = 9) a

I Theory:

The set of all vectors in F that are images under of at least one vector in E is f

called the range of : f

�㔼�㕚(Ā) = {þ ∈ ā: ∃ý ∈ Ā, þ = Ā(ý)} = Ā(Ā)

If f: E F is a linear transformation, then: Im( ) of is a subspace of ↦ f f f

II Solution details:

1 Manual solving:

2 MATLAB code and explanation:

a Coding details:

clc;

syms e1 e2 e3

%y=f(x)=x1(e1)+x2(e2)+x3(e3)

%a,b,c are constants

e1 = input('e1 = [a b c]=');

e2 = input('e2 = [a b c]=');

e3 = input('e3 = [a b c]=');

A = [e1', e2', e3'];

%Find the rank of A

r = rank(A);

%Find basics of Im(f)

if (r == 1)

Basic = [e1];

disp('Basic of Imf = ');disp([e1])

dimImf = 1

elseif (r == 2)

Basic = [e1;e2];

disp('Basic of Imf = ');disp([e1',e2'])

dimImf = 2

elseif (r == 3)

Basic = [e1;e2;e3];

disp('Basic of Imf = ');disp([e1',e2',e3'])

dimImf = 3

end

b Results:

e1 = [a b c]=[1 0 1]

e2 = [a b c]=[1 1 0]

e3 = [a b c]=[0 1 -9]

Basic of Imf =

1 1 0

0 1 1

1 0 -9

dimImf =

3

Question 2: Given f : �㕅3³ �㕅2 and the matrix representation of f in E = {(1; 1; 1);

(1; 0; 1); (1; 1; 0)}, and F = {(1; 1); (2; 1)} is �㔴ýþ= ; find f(1; 2; 9)

I Theory:

- Every linear transformation from �㕅Ā ³ �㕅ÿ can be

represented by an m n matrix Conversely, every matrix ×

m×n represents a linear transformation from �㕅Ā ³ �㕅ÿ,

i.e, we can always write a linear transformation in the

form

[Ā(ý)]� 㕇 = �㔴ý�㕇 where �㔴ÿ×Ā matrix

- If we are given vector images of a basis E in �㕅ÿ, that is we

know f(E), then

Ā(Ā) = �㔴Ā ⇔ �㔴 = Ā(Ā)Ā21 where f(E), E are two matrices formed by putting image

vector of basis, vector of basis in column, respectively

- Let f : �㕅Ā ³ �㕅ÿ be a linear transformation and

Ā = ÿ{ 1, ÿ2, ., ÿĀ} be a basis for �㕅Ā, ā = Ā{ 1, Ā2, ,ĀĀ} be a

basis for �㕅ÿ Then

is called the matrix representation of f with respect to E

and F

�㔴 = ā Ā(Ā)ý,þ 21

II Solution details:

1 Manual solving:

E =

F =

�㔴ýþ=

ý� 㕇 =

*�㔴 = ā Ā(Ā)ý,þ 21 = ā21�㔴Ā

[Ā(ý)]� 㕇 = �㔴ý�㕇 ⇔ f(1;2;9) = =

2 MATLAB code and explanation:

2.1 MATLAB code:

2.2 Explanation:

Line 1, 2, 3, 4: Create the matrix E, F, x, AEF respectively

Line 5: Find the matrix A through the formula A = F A E-1× EF×

Line 6: Find f(1;2;9) by using the formula [f(x)] = A xT × T

Question 3: Let = {(2; -1; 3); (1; 1; 2); (3; 0; 1); (-1; -4; )} be a subspace of R F a 3 with the inner product < x; y >= 3x 1y - x1y2 1 - x 2y1 + 4x 2y2 + 4x 3y3

a) Find a basis and the dimension of F⊥

b) Find the vector projection of = (3w ; - 2; 1) onto F

I Theory:

Definition 1:

Two vectors x y, Vin an inner product space V is called orthogonal

 <x, y>=0 We denote it by x⊥y (x perp y)

Vector x is orthogonal to the set M V if x is orthogonal to every vector in M

We denote it by x⊥M

Theorem 1: Vector x is orthogonal to a subspace F if and only if x is orthogonal with a basic of F

Definition 2:

A set of two or more vectors in a real inner product space {x , x1 2, …, xn} is called orthogonal  all pairs of distinct vectors in the set are orthogonal

An orthogonal set in which each vector has norm 1 is said to be orthogonal

1, ( 1, 2, , )

k

x = k= n

Definition 3: If F is a subspace of a real inner product space V, then the set F of ⊥ all vectors in V that are orthogonal to F is called the orthogonal complement of F

Theorem 2: Let R be a subspace of a real inner product space V then F is a ⊥

subspace of V, and:

dim F + dim F = dim V ⊥ Scheme of finding F of a subspace F: ⊥

Find a basis of F Assume that basic of F contains vectors {e , e1 2, …, en}

1

( , , , ) , 0

n n

e y

F y y y e y

e y

⊥

 Find the dim and a basic of this null space

Scheme of finding y = proj x F

- Find a basis of F, let it be S = {e , e1 2, …, em}

- Since y  F => y = 1 1e + 2 2e + + m m e

- x = y + z, then: y =  1 1e +  2 2e + + m m e + z



1 1 2 2

x e e e e e e e

x e e e e e e e

y proj x e e e

x e e e e e e e

ý

ÿ

þ

II Solution details:

1 Manual solving:

a) Find a basis and the dimension of F⊥

1 4 0

A

−

Find the rank of the subspace :F

3 3 2

2 2 1

3 3 1

3 2

2 3

r r r

r r r

r r r

( ) 3

dim 3

dim 0

r F

F

F⊥

A basis of F is S = {(2, -1, 3), (1, 1, 2), (3, 0, 1)}

F⊥ null SA

 A basis of F is {(0, 0, 0)} ⊥

b) Find the vector projection of = (3w ; - 2; 1) onto F

Since y  F => y = 1 1e + 2 2e + + m m e

1 2 3

T T T

x e e e e e e e

x e e e e e e e

x e e e e e e e

x x e x e x e

A e e e

ý

þ

ö ö

÷ ÷

÷ ÷

÷ ÷

5 33

25 21 14

33 14 31

1

3

5 6

7

6

5 6





ÿ

−

= ÿþ

1 1 2 2 3 3 (3, 2,1)

F

y proj x e e e

2 MATLAB code and explanation:

a Coding details:

clc;

syms x y z

A =[3 -1 0;-1 4 0; 0 0 4];

%Input the matrix of spanning set of F

e1 = [2 -1 3];

e2 = [1 1 2];

e3 = [3 0 1];

e4 = [-1 -4 9];

F = [e1;e2;e3;e4]'

r = rank(F)

if (r == 1)

S = e1

elseif (r == 2)

S = [e1;e2]

elseif (r == 3)

S = [e1;e2;e3]

end

%Input the matrix of the inner product

%Fo: Orthogonal F

Fo =(S')*A

kerf=null(Fo);

dimFo = size(ker,2)

% b)Find the vector projection of w = (3; -2; 1) onto F

x = [3;-2;1];

E = [dot(e1,e1) dot(e1,e2) dot(e1,e3); dot(e2,e1) dot(e2,e2) dot(e2,e3); dot(e3,e1) dot(e3,e2) dot(e3,e3)]

D = [dot(x,e1); dot(x,e2) ; dot(x,e3)];

X = mldivide(sym(E), sym(D)) ;

fprintf('The projection of w onto F: ')

(X(1,1)*e1 + X(2,1)*e2 + X(3,1)*e3)

b Results:

F =

2 1 3 -1

-1 1 0 -4

3 2 1 9

r =

3

S =

2 -1 3

1 1 2

3 0 1

Fo =

5 2 12

-4 5 0

7 5 4

dimFo =

0

E =

14 7 9

7 6 5

9 5 10

The projection of w onto F:

ans=

[ 3, -2, 1]

Question 4: Determine the currents I , I , I and for the given electrical network 1 2 3

1 Solution:

3 2 4�㔼12 3�㔼2 = 0 4�㔼1+ 3�㔼2 = 3 ³ (1)

4 2 1�㕅32 3�㔼2 = 0

�㕅 + 3�㔼3 2 = 4 ³ (2)

3 2 4�㔼1+ 1�㕅32 4 = 0 4�㔼12 �㕅3 = 1 ³ (3) Eauation (1), (2) & (3) in matrix form as,

[0 3 14 3 0

4 0 21

3 4

1] ´ �㕅3 ³ �㕅 2 �㕅3 1 [4 3 00 3 1

0 3 21

3 4

22] ´ �㕅3³ �㕅 + �㕅3 2 [4 3 00 3 1

0 0 0

3 4

2] ´ �㕅3 ³ �㕅 × 03 [4 3 00 3 1

0 0 0

3 4

0] 3�㔼2+ �㔼3 = 4 3�㔼2 = 4 2 �㔼3

�㔼23 2= 4�㔼3 3 4�㔼1 = 3 2 3�㔼2

4�㔼4�㔼1 = 3 2 3 [1 = 3 2 4 + �㔼4 2 �㔼3 ] 33 4�㔼1 = 21 + �㔼3

�㔼1 = 21

4 +

�㔼4 3 Dependent solution if �㔼3 = 1 �㕎�㕚þ

�㔼1 = 0 �㕎�㕚þ

�㔼2 = 1 �㕎�㕚þ

If �㔼3 = 2 amp

�㔼4�㕎�㕚þ 1 = 1

�㔼23=�㕎�㕚þ 2

2 Matlab

a) The code:

function ex3

clc, clear, close all;

disp('Solve the systems AX = B using gaussian elimination');

n = input('Input number of equations n = ');

k = input('Input number of variables k = ');

result = zeros(0,0);

disp('Example for input equation: [1 2 3 4] means x1 + 2*x2 + 3*x3 = 4');

for i = 1:n

T = input(['Input equation ' num2str(i) ' : ']);

result = [result; T];

end

disp('The system of equations :'); disp(result);

for i = 1:(n-1)

disp(' -');

disp(['Step ' num2str(i)]);

for j = (i+1):n

result(i,i) == 0 if

temp = result(j,:);

result(j,:) = result(i,:);

result(i,:) = temp;

else

result(j,:) = result(j,:) -(result(j,i)/result(i,i))*result(i,:);

end

end

disp(result);

disp('Press Enter to continue');

pause;

end

disp(' -');

A = result(:,1:k);

rresult = 0; rA = 0;

for i = 1:n

for j = 1:(k+1)

abs(result(i,j)) > 0.0001 if

rresult = rresult + 1;

break;

end

end

end

for i = 1:n

for j = 1:k

abs(A(i,j)) > 0.0001 if

rA = rA + 1;

break;

end

end

end

rresult ~= rA if

disp('No solution !');

elseif rresult == rA

rresult == n if

disp('Unique Solution');

x = zeros(k,1);

for s = 1:n

i = n+1-s;

sum = 0;

for j = (i+1):k

sum = sum + result(i,j)*x(j);

end

x(i) = (result(i,k+1) - sum)/result(i,i);

end

disp('Solution: ');disp(x);

else

disp('Infinity Solution');

disp([num2str(k - rresult) ' free variable']);

end

end

end

b) On the screen:

Question 5: Generate a random matrix which has the size of 4 4 and is A ×

diagonalizable

Then, return its eigenvalues, the corresponding eigenvectors and compute A100

I Theories summary

1 What is a diagonalizable matrix?

A square matrix is said to be diagonalizable if it is similar to a diagonal matrix

That is, is diagonalizable if there is an invertible matrix P and a diagonal matrix A

D such that A=PDP -1

2 What are eigenvalues and the corresponding eigenvectors of a matrix?

Many problems present themselves in terms of an eigenvalue problem:

�㔴 �㕣 = �㔆 �㕣

In this equation A is an n-by-n matrix, is a non-zero n-by- v 1 vector and λ is a scalar (which may be either real or complex) Any value of λ for which this

equation has a solution is known as an eigenvalue of the matrix It is sometimes A

also called the characteristic value

The vector, , which corresponds to this value is called an eigenvector The v

eigenvalue problem can be rewritten as:

�㔴 �㕣 2 �㔆 �㕣 = 0

�㔴 �㕣 2 �㔆 �㔼 �㕣 = 0 (�㔴 2 �㔆.�㔼) �㕣 = 0

If v is non-zero, this equation will only have a solution if

|�㔴 2 �㔆 �㔼| = 0 This equation is called the characteristic equation of , and is an n order A th

polynomial in λ with n roots These roots are called the eigenvalues of We will A

only deal with the case of n distinct roots, though they may be repeated For each eigenvalue there will be an eigenvector for which the eigenvalue equation is true This is most easily demonstrated by example:

Find Eigenvalues and Eigenvectors of a 2x2 Matrix

If �㔴 = [ 022 23]1

Then the characteristic equation is |�㔴 2 �㔆 �㔼| = | [ 022 23] 2 [�㔆 00 �㔆]1 |=0 |[2�㔆22 23 2 �㔆]| =1 �㔆2+ 3�㔆 + 2 = 0 And the two eigenvalues are: {�㔆�㔆 = 211

All that's left is to find the two eigenvectors Let's find the eigenvector, �㕣1, associated with the eigenvalue, λ1=-1, first

�㔴 �㕣1 = �㔆 �㕣1

(�㔴 2 �㔆1).�㕣1= 0 [222 23 2 �㔆�㔆1 1

1] �㕣1 = 0 [ 122 22] 1 �㕣1 = [ 122 22] [1 �㕣�㕣1,11,2] = 0

so clearly from the top row of the equations we get

�㕣1,1+ �㕣1,2 = 0 ý�㕟 �㕣1,1 = 2�㕣1,2

Note that if we took the second row we would get

(22).�㕣1,1+ ( )22 �㕣1,2 = 0 ý�㕟 �㕣1,1 = 2�㕣1,2

In either case we find that the first eigenvector is any 2 element column vector in which the two elements have equal magnitude and opposite sign

�㕣1 = �㕘21]1.[+1 where k is an arbitrary constant Note that we didn't have to use +1 and -1, we 1

could have used any two quantities of equal magnitude and opposite sign

Going through the same procedure for the second eigenvalue:

�㔴 �㕣 = �㔆 �㕣2 2

(�㔴 2 �㔆2) �㕣22 23 2 �㔆2=[2�㔆2 1

2] �㕣222 21] [= [ 2 1�㕣�㕣2,12,2] = 0

So 2 �㕣2,1+ 1 �㕣 = 0 2,2

2.�㕣 = 2�㕣2,1 2,2

�㕣 = �㕘 [2 2 +1

22]

Again, the choice of +1 and -2 for the eigenvector was arbitrary; only their ratio is

important This is demonstrated in the MatLab code below

3 The application of diagonal matrix

Diagonal matrices are relatively easy to compute with, and similar matrices share

many properties, so diagonalizable matrices are well-suited for computation

In particular, many applications involve computing large powers of a matrix,

which is easy if the matrix is diagonal But if A=PDP -1, then

�㔴Ā = (�㕃ÿ�㕃21)Ā = (�㕃ÿ�㕃21)(�㕃ÿ�㕃21)(& )(�㕃ÿ�㕃21)= �㕃ÿ

because the �㕃21�㕃 terms in the middle all collapse

4 Solution details

Choose the matrix �㔴 = [

1 2 3 4

2 4 5 2

4 2 1 4

2 3 4 5

] The unit matrix is �㔼 = [

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

]

Then we have

|�㔴 2 �㔆 �㔼| = 0

So that

|[

1 2 3 4

2 4 5 2

4 2 1 4

2 3 4 5

]2 �㔆.[

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

]| = 0

1 224� 4 2㔆2� 1 22㔆� 35㔆 424

]| = 0 Solve this equation we have:

�㔆1 = 12.2111

�㔆2 = 22.2111

�㔆3= 1.0000

�㔆4= 0.0000 ,which are the eigenvalues of the matrix �㔴 = [

1 2 3 4

2 4 5 2

4 2 1 4

2 3 4 5

] With �㔆1 = 12.2111 the eigenvector is [

20.4234 20.5217 20.4531 20.5858

]

With �㔆2 = 22.2111 the eigenvector is [

20.2156 20.5249 20.8059 20.1688

] With �㔆3 = 1.0000 the eigenvector is [

20.3455 0.7775 20.5183 0.0864

] With �㔆4 = 0.0000 the eigenvector is [

0.2641 20.8805 0.3522 0.1761

]

So we have �㕃 = [

20.4234 20.2156 20.3455 0.2641

20.5217 20.5249 0.7775 20.8805

20.4531 20.8059 20.5183 0.3522

20.5858 20.1688 0.0864 0.1761

] and

ÿ = [

] Then �㔴100 = �㕃ÿ �㕃100 21 =

[

20.4234 20.2156 20.3455 0.2641

20.5217 20.5249 0.7775 20.8805

20.4531 20.8059 20.5183 0.3522

20.5858 20.1688 0.0864 0.1761

] [

] 10

20.20.20.423452174531 20.20.20.2156 20.3455 0.264152498059 20.5183 0.35220.7775 20.8805

20.5858 20.1688 0.0864 0.1761

] 21

= 10108 [

0.76070.9375 1.1102 1.2984 1.54810.8142 0.9642 1.1276 1.34460.9008 1.0535 1.2562 1.0526 1.2466 1.4578 1.7383

]

5 Coding details

a Full MATLAB code:

%Clear all text from the Command Window

clear all;

clc;

%Input the matrix 4x4 that we need to use in this problem

A=input('Enter the matrix A: ')

%Check the matrix if it is already diagonal?

check=isdiag(A);

if check==true

disp('The matrix is already diagonal');

%Calculate A power 100

fprintf('A power of 100 = A^100:')

A100 = A^100;

ANSWER = real(A100)

else

F=size(A,1);

%Check the matrix whether it is diagonalizable or not?

syms lambda

I = eye(F);

lambda1 = solve(det(A - I*lambda)==0);

numrows=size(lambda1,1);

V=[];

for i = 1:numrows

v = null(A-lambda1(i,1)*I);

V=[V,v];

end

B=transpose(V);

C=rank(B);

E=size(B,1);

if C==F

disp("The matrix is diagonalizable")

p =poly(A);

eigs = roots(p);

M=[];

for i=1:4

m= eigs(i);

M=[M,m];

end

disp("The eigenvalues are:")

disp(M)

%Show the output of the problem: V is the matrix concluding all

%eigenvectors and D is a diagonal matrix with eigenvalues located on the

%main diagonal

fprintf('Eigenvectors are:\n')

[V,D] = eig(A)

%Calculate A power 100

fprintf('A power of 100 = V*D^100*inv(V):')

A100 = V*D^100*inv(V);

ANSWER = real(A100)

else

%Ask the user to input another matrix

disp("The matrix is not diagonalizable Let's try another matrix")

end

end

b Results and explanation:

If we input a diagonal matrix, this code will help us to find �㔴100 directly: