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

VIETNAM NATIONAL UNIVERSITY HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY FACULTY OF APPLIED SCIENCE -*** Linear Algebra (MT1008) Group Assignments Topic: Group: 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 14th, 2022 0 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 f as below, find the dimension and one basic of Im(f): f(x1 ,x2 ,x3) = (x1 + x2;x2 + x3;x1 - ax3), (a = 9) I Theory: The set of all vectors in F that are images under f of at least one vector in E is If : E F is a linear called the range of f: ↦ f II.Solution details: Manual solving: 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 �㔼�㔼 0 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 = elseif (r == 2) Basic = [e1;e2]; disp('Basic of Imf = ');disp([e1',e2']) dimImf = elseif (r == 3) Basic = [e1;e2;e3]; disp('Basic of Imf = ');disp([e1',e2',e3']) dimImf = end b Results: e1 e2 e3 Basic of Imf = = [a b c]=[ = [a b c]=[ = [a b c]=[0 1 -9 dimImf = 3 Question 2: Given f : �㔼 ³ �㔼 and the matrix representation of f in E = {(1; 1; 1); (1; 0; 1); (1; 1; 0)}, and F = {(1; 1); (2; 1)} is �㔼 = I Theory: - Every linear transformation from �㔼 Ā × represented by an m m n represents a i.e, we can always write a linear transformation in the form - ÿ If we are given vector images of a basis E in �㔼 , that is we know f(E), then where�㔼 ÿ×Ā matrix 0 where f(E), E are two matrices formed by vector of basis, vector of basis in Ā = {ÿ1, ÿ2, ÿ., ÿĀ} - Let f : �㔼 Ā basis for �㔼 Then is called the matrix representation of f with respect to E and F ý,ỵ= II Solution details: Manual solving: E= F= ýỵ= ý = 21 21 * ý,ỵ = () = 21 = ý,ỵ [(ý)] = �㔼 ý 㔼 ⇔ f(1;2;9) = MATLAB code and explanation: 2.1 MATLAB code: 0 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 Line 6: Find f(1;2;9) by using the formula [f(x)] T = A Question 3: Let F = {(2; -1; 3); (1; 1; 2); (3; 0; 1); (-1; -4; a)} be a subspace of R3 with the inner product < x; y >= 3x1y1 - x1y2 - x2y1 + 4x2y2 + 4x3y3 a) Find a basis and the dimension of F⊥ b) Find the vector projection of w = (3; -2; 1) onto F I Theory: Definition 1: Two vectors x , y V in an inner product space V is called orthogonal =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 0 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 , x2, …, xn} is called orthogonal all pairs of distinct vectors in the set are orthogonal An orthogonal set in which each vector has norm is said to be orthogonal xk = 1, ( k =1, 2, , 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: Scheme of finding F⊥ of a subspace F: Find a basis of F Assume that basic of F contains vectors {e1 , e2, …, en} ü ⊥ F ÿ = ý ( y1 , y , , y n ) ỵ Find the dim and a basic of this null space Scheme of finding y = projFx - Find a basis of F, let it be S = {e1 , e2, …, em} - Since y F => y = 1e1 + 2e2 + + - x = y + z, then: y = 1e1 + 2e + + m em m em + z ü x , e1 ÿ x e ÿ , ý II x,e ỵ Solution details: Manual solving: a) Find a basis and the dimension of F⊥ A= Find the rank of the subspace F: m 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) if abs(result(i,j)) > 0.0001 rresult = rresult + 1; break; end end end for i = 1:n for j = 1:k if abs(A(i,j)) > 0.0001 rA = rA + 1; break; end 0 end end if rresult ~= rA disp('No solution !'); elseif rresult == rA if rresult == n 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: 0 Question 5: Generate a random matrix A which has the size of ×4 and is diagonalizable Then, return its eigenvalues, the corresponding eigenvectors and compute A100 I Theories summary What is a diagonalizable matrix? A square matrix is said to be diagonalizable if it is similar to a diagonal matrix That is, A is diagonalizable if there is an invertible matrix P and a diagonal matrix D such that A=PDP-1 What are eigenvalues and the corresponding eigenvectors of a matrix? Many problems present themselves in terms of=an eigenvalue problem: �㔼 �㔼 �㔼 �㔼 In this equation A is an n-by-n matrix, v is a non-zero n-by-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 A It is sometimes also called the characteristic value 0 The vector, v, which corresponds to this value is called an eigenvector The eigenvalue problem can be rewritten as: �㔼 �㔼 �(㔼 �2 㔼 2 =0 =0 �㔼 �㔼 �㔼 )�.㔼 �㔼=0 �㔼 �㔼 �㔼 �㔼 If v is non-zero, this equation will |only 2have a solution|= 0if �㔼 �㔼 �㔼 This equation is called the characteristic equation of A, and is an nth order polynomial in λ with n roots These roots are called the eigenvalues of A We will 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: If �㔼 = [ Then the characteristic equation is And the two eigenvalues are: All that's left is to find the two associated with the eigenvalue, λ1=-1, first so clearly from the Note that if we took the In either case we which the two elements have equal magnitude and opposite sign where k could have used any two quantities of equal magnitude and opposite sign Going through the same procedure for the second eigenvalue: So 10 0 = 222].[ +1 Again, the choice of +1 and -2 for�the eigenvector� was arbitrary; only their ratio is 㔼 㔼 important This is demonstrated in the MatLab code below 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, because the � = ( 21 � which is easy if the matrix is diagonal But if A=PDP-1, then �㔼 Ā 㔼 Solution details Choose the ma trix �㔼 The unit matrix is � 㔼 Then we have So that 11 0 Solve this equation we have: ,which are the eige nva lue s of the matrix �㔼 � 㔼 With With � 㔼 With �㔼 With � 㔼 So we have �㔼 Then �㔼 100 = 20.4234 [ 20.5217 20.4531 20.5858 12 0 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 13 0 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 �㔼 100 directly: 14 0 c Note: The result is strange because it is too big to display in decimal number Also, the eigenvectors are concluded in4 = 0.0000 �㔼 㔼 is shown in Figure The result of � 100 15 FIGURE Howeve r, if we input a ma tr ix tha t is not dia gona liz able �㔼 The result will be: 16 0 ... transformation f as below, find the dimension and one basic of Im (f) : f( x1 ,x2 ,x 3) = (x1 + x2;x2 + x3;x1 - ax 3), (a = 9) I Theory: The set of all vectors in F that are images under f of at least one vector... 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: Scheme of finding F? ?? of a subspace F: Find a basis of. .. basis of F Assume that basic of F contains vectors {e1 , e2, …, en} ü ⊥ F ÿ = ý ( y1 , y , , y n ) ỵ Find the dim and a basic of this null space Scheme of finding y = projFx - Find a basis of F,