Exercises 6.3 - 7.3
Chapter 6. Linear Transformations
6.3 Kernel and Range
Exercise 6.4. (Invertible Matrix as a Bijective Linear Transformation)
Consider the matrix
\[A = \left[\begin{array}{rrrr} 3& \hspace{1mm} -5& \hspace{1mm} -2&\hspace{3mm} 2\\ -4 & 7 & 4 & 4 \\ 4 & -9 & -3 & 7 \\ 2 & -6 & -3 & 2 \end{array} \right].\]Referring to the Theorem 6.3.15 in Section 6.3, show that \(T_{A} : \mathbb{R}^{4} \rightarrow \mathbb{R}^{4}\) is onto in at least four different ways. You may use several related MATLAB commands.
Solution.
A = [3 -5 -2 2; -4 7 4 4; 4 -9 -3 7; 2 -6 -3 2];
format short;
% By (a) in Theorem 6.3.15, use the command rref of A.
rref(A)
% Since the reduced row echelon form of A is the identity matrix,
% the linear transformation T is onto.
% By (d) in Theorem 6.3.15, use the command null of A.
% Find a basis for the null space of A.
null(A, 'r')
% Since the null space contains only the zero vector,
% the linear transformation T is onto.
% By (g) in Theorem 6.3.15, use the command rank of A.
% Find the number of linearly independent columns of A.
rank(A)
% Since the column vectors of A are linearly independent,
% the linear transformation T is onto.
% By (i) in Theorem 6.3.15, use the command det of A.
% Find the determinant of A.
det(A)
% Since det(A) is nonzero,
% the linear transformation T is onto.
% By (j) in Theorem 6.3.15, use the command eig of A.
% Find the eigenvalues A.
eig(A)
% Since 0 is not an eigenvalue of A,
% the linear transformation T is onto.
6.4 Composition and Invertibility of Linear Transformations
Exercise 6.5. (Compositions of Linear Transformations)
Consider successive rotations of \(\mathbb{R}^{3}\) by \(30\,^{\circ}\) about the \(z\)-axis, then by \(60\,^{\circ}\) about the \(x\)-axis, and then by \(45\,^{\circ}\) about the \(y\)-axis. If it is desired to execute the three rotations by a single rotation about an appropriate axis, what axis and angle should be used?
Solution.
% Angle as a degree.
ang1=30; ang2=60; ang3=45;
% Convert angles from degrees to radians.
theta1=((ang1)*(pi/180));
theta2=((ang2)*(pi/180));
theta3=((ang3)*(pi/180));
format short;
% Rotation about z-axis with the angle 30.
Rz = [cos(theta1) -sin(theta1) 0; sin(theta1) cos(theta1) 0; 0 0 1];
% Rotation about x-axis with the angle 60.
Rx = [1 0 0; 0 cos(theta2) -sin(theta2); 0 sin(theta2) cos(theta2)];
% Rotation about y-axis with the angle 45.
Ry = [cos(theta3) 0 sin(theta3); 0 1 0; -sin(theta3) 0 cos(theta3)];
% Composition of the three rotation matrices R = Ry*Rx*Rz.
R = Ry*Rx*Rz;
% Find the axis of rotation of R,
% by finding the eigenvector of R corresponding to the eigenvalue lambda=1.
% Find a nonzero vector satisfying Rx = x.
x = null(eye(3) - R);
% Find the angle of rotation of R.
% Note that w=(-x2,x1,0) is orthogonal to the axis of rotation x=(x1,x2,x3).
w = [-x(2) x(1) 0]';
theta = acos((dot(w, R*w))/((norm(w)*norm(R*w))));
% Convert the angle from radians to degrees.
ang = ((theta)*(180/pi));
disp('The angle of rotation in degrees is'); disp(ang);
disp('The axis of rotation is'); disp(x');
MATLAB results.
The angle of rotation in degrees is
69.3559
The axis of rotation is
-0.9350 -0.3525 -0.0391
Chapter 7. Dimension and Structure
7.1 Basis and Dimension
Exercise 7.1. (Linear Combination and Independence)
Are any of the vectors in the set
\[S = \{ (2,6,3,4,2), \hspace{1.5mm}(3,1,5,8,3), \hspace{1.5mm}(5,1,2,6,7), \hspace{1.5mm}(8,4,3,2,6), \hspace{1.5mm}(5,5,6,3,4) \}\]linear combinations of predecessors? Justify your answer.
Solution. One strategy is to form a matrix \(V\) of the column vectors \(\mathbf{v}_{k}\) mentioned above and decide whether the system \(V \mathbf{x} = \mathbf{0}\) has nontrivial solutions. If so, then at least one column is a linear combination of previous ones. Otherwise, the columns are linearly independent.
v1 = [2 6 3 4 2]'; v2 = [3 1 5 8 3]'; v3 = [5 1 2 6 7]';
v4 = [8 4 3 2 6]'; v5 = [5 5 6 3 4]';
% Construct V of the column vectors v1,v2,v3,v4 and v5.
V = [v1 v2 v3 v4 v5];
format short;
% Find the reduced row echelon form of V.
rref_V = rref(V);
disp('The reduced row echelon form of A is'); disp(rref_V);
MATLAB results.
The reduced row echelon form of A is
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
Since the reduced row echelon form of \(V\) has \(5\) pivots, the columns of \(V\) are linearly independent. Hence, no column of \(V\) can be a linear combination of any other columns.
7.2 Properties of Bases
Exercise 7.2.
In this problem, we make a function file CheckBasis.m to check that the vectors \(\mathbf{v}_{1}\), \(\mathbf{v}_{2}\), \(\mathbf{v}_{3}\) and \(\mathbf{v}_{4}\) form a basis of \(\mathbb{R}^4\) using the equivalent statements (a), (g), (h), and (o) of Theorem 7.2.7 in the textbook.
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Complete the shadow part (/////) of the m-file given below referring to the comments and the execution results.
%--- your function file ---% function [Result]=CheckBasis(v1, v2, v3, v4, case_num) % if case_num=1, check the statement (a), % if case_num=2, check the statement (g), % if case_num=3, check the statement (h). % Construct the matrix V. //////////////////////// % Use the switch statement to check % whether one of the statements (a), (g), and (h) holds. switch case_num case 1 fprintf('* You enter %d: statement (a) *\n', case_num); //////////////////////// if //////////////////////// disp('Given vectors form a basis of 4 dimensional space.'); else disp('Given vectors do not form a basis of 4 dimensional space.'); end case 2 fprintf('* You enter %d: statement (g) *\n', case_num); Result=det(V); if Result~=0 disp('Given vectors form a basis of 4 dimensional space.'); else disp('Given vectors do not form a basis of 4 dimensional space.'); end /////////////////////////// % check statement (h) //////////////////////// //////////////////////// //////////////////////// //////////////////////// //////////////////////// //////////////////////// //////////////////////// end endThe execution results will be as follows in the Command Window:
>> % Let vectors v1, v2 ,v3 ,v4, v5 be: v1=[1 0 0 0]'; v2=[0 2 0 0]'; v3=[0 0 4 5]'; v4=[0 0 0 -1]'; v5=[0 0 0 1]'; >> C = CheckBasis(v1, v2, v3, v4, 3) * You entered 3: statement (h) * Given vectors form a basis of 4 dimensional space C = -8 >> CheckBasis(v1, v2, v4, v5, 1); * You entered 1: statement (a) * Given vectors do not form a basis of 4 dimensional space. >> determinant=CheckBasis(v1, v2, v3, v5, 2) * You entered 2: statement (g) * Given vectors form a basis of 4 dimensional space. determinant = 8 -
Using
CheckBasis.mfrom 1, check whether-
\(\mathbf{v}_{1}=(-1, 0, 1, 0)^{T}\), \(\mathbf{v}_{2}=(2, 3, -2, 6)^{T}\), \(\mathbf{v}_{3}=(0, -1, 2, 0)^{T}\) and \(\mathbf{v}_{4}=(0, 0, 1, 5)^{T}\) form a basis of \(\mathbb{R}^4\).
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\(\mathbf{v}_{1}=(a, b, c, d)^{T}\), \(\mathbf{v}_{2}=(-b, a, d, -c)^{T}\), \(\mathbf{v}_{3}=(-c, -d, a, b)^{T}\) and \(\mathbf{v}_{4}=(-d, c, -b, a)^{T}\) form a basis of \(\mathbb{R}^4\).
(You may need to use the Symbolic toolbox of MATLAB)
-
Solution.
CheckBasis.m% ----- your function file ----- %`\ function [Result]=CheckBasis(v1, v2, v3, v4, case_num) % if case_num=1, check the statement (a), % if case_num=2, check the statement (g), % if case_num=3, check the statement (h). % Construct the matrix V. V=[v1 v2 v3 v4]; switch case_num case 1 fprintf('* You entered %d: statement (a) *\n', case_num); Result=rref(V); if det(Result)~=0 disp(' Given vectors f orm a basis of 4 dimensional space.'); else disp('Given vectors do not form a basis of 4 dimensional space.'); end case 2 fprintf('* You entered %d: statement (g) *\n', case_num); Result=det(V); if Result~=0 disp(' Given vectors form a basis of 4 dimensional space.'); else disp(' Given vectors do not form a basis of 4 dimensional space.'); end case 3 fprintf('* You entered %d: statement (h) *\n', case_num); [Q, D]=eig(V); % D: the eigenvalue matrix of V such that VQ = QD. Result=prod(diag(D)); if Result~=0 disp(' Given vectors form a basis of 4 dimensional space'); else disp(' Given vectors do not form a basis of 4 dimensional space.'); end end end-
2-1.
>> v1=[-1 0 1 0]'; v2=[2 3 -2 6]'; v3=[0 -1 2 0]'; v4 = [0 0 1 5]'; CheckBasis(v1, v2, v3, v4, 1); CheckBasis(v1, v2, v3, v4, 2); CheckBasis(v1, v2, v3, v4, 3);MATLAB results.
* You entered 1: statement (a) * Given vectors f orm a basis of 4 dimensional space. * You entered 2: statement (g) * Given vectors form a basis of 4 dimensional space. * You entered 3: statement (h) * Given vectors form a basis of 4 dimensional space2-2.
>> syms a b c d; >> v1=[a;b;c;d]; v2=[-b;a;d;-c]; v3=[-c;-d;a;b]; v4 = [-d;c;-b;a]; >> CheckBasis(v1, v2, v3, v4, 1); >> CheckBasis(v1, v2, v3, v4, 2);MATLAB results.
* You enter 1: statement (a) * Given vectors form a basis of 4 dimensional space. * You enter 2: statement (g) * Given vectors form a basis of 4 dimensional space.
7.3 The Fundamental Spaces of a Matrix
Exercise 7.3. (Need the Symbolic toolbox)
In this problem, we make a function file getFSinfo.m to get the dimension and basis of the fundamental spaces of a given matrix. For example, we execute the followings:
>> A=[1 0 0 0 2; -2 1 -3 -2 -4; 0 5 -14 -9 0; 2 10 -28 -18 4];
>> getFSinfo(A);
Then, the Command Window displays the results as follows:
Given matrix is:
1 0 0 0 2
-2 1 -3 -2 -4
0 5 -14 -9 0
2 10 -28 -18 4
== Dimension of the fundamental spaces of a given matrix ==
dim(row(A))=dim(col(A)): 3, dim(null(A)): 2, dim(null(A_trans)): 1
== Basis of the fundamental spaces of a given matrix (in row vectors) ==
row(A)
1 0 0 0 2
0 1 0 1 0
0 0 1 1 0
col(A)
1 0 0 2
0 1 0 0
0 0 1 2
null(A)
0 -1 -1 1 0
-2 0 0 0 1
null(A_trans)
-2 0 -2 1
*****************************************************
-
Complete the missing parts of the m-file
getFSinfogiven as follows:%--- function file 'getFSinfo.m' ---% function [info]=getFSinfo(A) % row(A): basis and dimension /////// missing part /////// % col(A): basis and dimension /////// missing part /////// % null(A): basis and dimension /////// missing part /////// % null(A'): basis and dimension /////// missing part /////// disp('Given matrix is:'); disp(A); fprintf('== Dimension of the fundamental spaces of given matrix == \n'); fprintf('dim(row(A))=dim(col(A)): %d,', rank_A); fprintf('\t dim(null(A)): %d,\t dim(null(A_trans)): %d \n\n', nullity, nullity_T); fprintf('== Basis of the fundamental spaces of given matrix (in row vectors) == \n'); disp(' row(A)'); disp(double(rowA_basis)); disp(' col(A)'); disp(double(colA_basis)); disp(' null(A)'); disp(nullA_basis); disp(' null(A_trans)'); disp(nullAtrans_basis); fprintf('\n*****************************************************\n'); endYou may use the MATLAB commands rank, colspace, rref, null and so on.
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Using your function file
getFSinfo.m, find the dimension and basis of the fundamental spaces of\(A = \begin{bmatrix} 3 & 2 & 1 & 3 & 5 \\ 6 & 4 & 3 & 5 & 7 \\ 9 & 6 & 5 & 7 & 9\\ 3 & 2 & 0 & 4 & 8 \end{bmatrix}, B = \begin{bmatrix} 3 & -1 & 3 & 2 & 5\\ 5 & -3 & 2 & 3 & 4\\ 1 & -3 & -5 & 0 & -7\\ 7 & -5 & 1 & 4 & 1 \end{bmatrix}, C = \begin{bmatrix} 1 & 3 & 2 & 1\\ -2 & -6 & 0 & -6\\ 3 & 9 & 1 & 8\\ -1 & -3 & -3 & -6\\ 1 & 3 & 2 & 1\\ 4 & 12 & 1 & 11 \end{bmatrix}\).
Solutions.
getFSinfo.m% ----- function file 'getFSinfo.m' ----- %` function info = getFSinfo(A) [m,n]=size(A); % size of matrix A % row(A): basis and dimension rank_A = rank(A); % rank of A rowA = colspace(sym(A')); % way1: MATLAB command colspace % rowA2=rref(A); % way2: reduced row echelon form rowA_basis=rowA(:, 1:rank_A)'; % basis of row(A) % col(A): basis and dimension colA=colspace(sym(A)); colA_basis=colA(:, 1:rank_A)'; % basis of col(A) % null(A): basis and dimension nullA=null(A, 'r'); nullity=n-rank_A; % way1. Dimension theorem nullA_basis=nullA(:, 1:nullity)'; % nullity=size(nullA,2); % way2. return the corresponding value % null(A'): basis and dimension nullAtrans=null(A', 'r'); % nullAtrans(:, 1:size(nullAtrans,2))'; basis of null(A') nullity_T=m-rank_A; % way1. using the relation nullAtrans_basis=nullAtrans(:,1:nullity_T)'; % nullity=size(nullAtrans,2); % way2. return the corresponding value % fprintf('****************************************************************\n'); disp('Given matrix is:'); disp(A); fprintf('== Dimension of the fundamental spaces of given matrix == \n'); fprintf('dim(row(A))=dim(col(A)): %d,', rank_A); fprintf('\t dim(null(A)): %d,\t dim(null(A_trans)): %d \n\n', nullity, nullity_T); fprintf('== Basis of the fundamental spaces of given matrix (in row vectors) == \n'); disp(' row(A)'); disp(double(rowA_basis)); disp(' col(A)'); disp(double(colA_basis)); disp(' null(A)'); disp(nullA_basis); disp(' null(A_trans)'); disp(nullAtrans_basis); fprintf('\n****************************************************************\n'); end-
*Commands.
A=[3 2 1 3 5; 6 4 3 5 7; 9 6 5 7 9; 3 2 0 4 8]; B=[3 -1 3 2 5; 5 -3 2 3 4; 1 -3 -5 0 -7; 7 -5 1 4 1]; C=[1 3 2 1; -2 -6 0 -6 ;3 9 1 8; -1 -3 -3 -6; 1 3 2 1; 4 12 1 11]; getFSinfo(A); getFSinfo(B); getFSinfo(C);MATLAB results.
Given matrix is: 3 2 1 3 5 6 4 3 5 7 9 6 5 7 9 3 2 0 4 8 == Dimension of the fundamental spaces of given matrix == dim(row(A))=dim(col(A)): 2, dim(null(A)): 3, dim(null(A_trans)): 2 == Basis of the fundamental spaces of given matrix (in row vectors) == row(A) 1.0000 0.6667 0 1.3333 2.6667 0 0 1.0000 -1.0000 -3.0000 col(A) 1 0 -1 3 0 1 2 -1 null(A) -0.6667 1.0000 0 0 0 -1.3333 0 1.0000 1.0000 0 -2.6667 0 3.0000 0 1.0000 null(A_trans) 1 -2 1 0 -3 1 0 1 ***************************************************** Given matrix is: 3 -1 3 2 5 5 -3 2 3 4 1 -3 -5 0 -7 7 -5 1 4 1 == Dimension of the fundamental spaces of given matrix == dim(row(A))=dim(col(A)): 3, dim(null(A)): 2, dim(null(A_trans)): 1 == Basis of the fundamental spaces of given matrix (in row vectors) == row(A) 1.0000 0 1.7500 0.7500 0 0 1.0000 2.2500 0.2500 0 0 0 0 0 1.0000 col(A) 1 0 -3 0 0 1 2 0 0 0 0 1 null(A) -1.7500 -2.2500 1.0000 0 0 -0.7500 -0.2500 0 1.0000 0 null(A_trans) 3 -2 1 0 ***************************************************** Given matrix is: 1 3 2 1 -2 -6 0 -6 3 9 1 8 -1 -3 -3 -6 1 3 2 1 4 12 1 11 == Dimension of the fundamental spaces of given matrix == dim(row(A))=dim(col(A)): 3, dim(null(A)): 1, dim(null(A_trans)): 3 == Basis of the fundamental spaces of given matrix (in row vectors) == row(A) 1 3 0 0 0 0 1 0 0 0 0 1 col(A) 1.0000 0 0.5000 0 1.0000 0.5000 0 1.0000 -1.2500 0 0 -1.7500 0 0 0 1.0000 0 0 null(A) -3 1 0 0 null(A_trans) -0.5000 1.2500 1.0000 0 0 0 -1.0000 0 0 0 1.0000 0 -0.5000 1.7500 0 0 0 1.0000 *****************************************************
Exercise 7.4. (Bases for the Fundamental Spaces, Need the Symbolic toolbox)
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Use the MATLAB commands
\[A = \left[\begin{array}{rrrr} 2& \hspace{2mm} -1& \hspace{5mm} 3&\hspace{4mm} 5\\ 4 & -3 & 1 & 3 \\ 3 & -2 & 3 & 4 \\ 4 & -1 & 15 & 17 \\ 7 & -6 & -7 & 0 \end{array} \right].\]symand colspace to find a basis for the column space of the matrix -
Use the same MATLAB commands in 1 to find a basis for the row space of \(A\).
-
Confirm that the basis obtained in 2 is consistent with the basis obtained from the reduced row echelon form of \(A\).
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Tell what happens if you use the MATLAB command
orth?
Solutions.
-
% Set a matrix A whose entries are symbolic objects. A = sym([2 -1 3 5; 4 -3 1 3; 3 -2 3 4; 4 -1 15 17; 7 -6 -7 0]); % Find a basis for the column space of A. col_basis = colspace(A); disp('A basis for the column space of A is'); disp(col_basis(:,1)'); disp(col_basis(:,2)'); disp(col_basis(:,3)');MATLAB results.
A basis for the column space of A is [ 1, 0, 0, 2, 1] [ 0, 1, 0, -3, 5] [ 0, 0, 1, 4, -5] -
% Set a matrix A_transpose whose entries are symbolic objects. A_transpose = sym([2 -1 3 5; 4 -3 1 3; 3 -2 3 4; 4 -1 15 17; 7 -6 -7 0]'); % Finding a basis for the row space of A is equivalent to % finding a basis for the column space of A_transpose. rowbasis = colspace(A_transpose); disp('A basis for the row space of A is'); disp(rowbasis(:,1)'); disp(rowbasis(:,2)'); disp(rowbasis(:,3)');MATLAB results.
A basis for the row space of A is [ 1, 0, 0, 6] [ 0, 1, 0, 7] [ 0, 0, 1, 0] -
% Set a matrix A. A = [2 -1 3 5; 4 -3 1 3; 3 -2 3 4; 4 -1 15 17; 7 -6 -7 0]; % Find the reduced row echelon form of A. rref_A = rref(A); % The nonzero rows of the reduced row echelon form of A % form a basis for the row space of A. disp('A basis for the row space of A is'); disp(rref_A(1,:)); disp(rref_A(2,:)); disp(rref_A(3,:));MATLAB results.
A basis for the row space of A is 1 0 0 6 0 1 0 7 0 0 1 0 -
% Set A. A = [2 -1 3 5; 4 -3 1 3; 3 -2 3 4; 4 -1 15 17; 7 -6 -7 0]; % The command orth gives an orthonormal basis for the column space of A. B = orth(A); disp('An orthonormal basis for the column space of A is'); disp('q1='); disp(B(:,1)'); disp('q2='); disp(B(:,2)'); disp('q3='); disp(B(:,3)');MATLAB results.
An orthonormal basis for the column space of A is q1= -0.2427 -0.1508 -0.2229 -0.9246 0.1177 q2= -0.1189 -0.3624 -0.2060 0.0253 -0.9008 q3= 0.3760 -0.6016 -0.5930 0.1848 0.3331
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