Exercises 8.1 - 8.2
Chapter 8. Diagonalization
8.1 Matrix Representations of Linear Transformations
Exercise 8.1.
Let \(T : \mathbb{R}^{5} \rightarrow \mathbb{R}^{3}\) be the linear operator given by the formula
\[\small{T(x_{1}, x_{2}, x_{3}, \hspace{1mm} x_{4}, x_{5}) = (7x_{1}+12x_{2}-5x_{3}, \hspace{1mm} 3x_{1}+10x_{2}+13x_{4}+x_{5}, \hspace{1mm} -9x_{1}-x_{3}-3x_{5})}\]and let \(B = \{\mathbf{v}_{1}, \hspace{1mm} \mathbf{v}_{2}, \hspace{1mm} \mathbf{v}_{3}, \hspace{1mm} \mathbf{v}_{4}, \hspace{1mm} \mathbf{v}_{5}\}\) and \(B' = \{\mathbf{v}'_{1}, \hspace{1mm} \mathbf{v}'_{2}, \hspace{1mm} \mathbf{v}'_{3}\}\) be the bases for \(\mathbb{R}^{5}\) and \(\mathbb{R}^{3}\), respectively, in which \(\mathbf{v}_{1} = (1, \hspace{1mm} 1, \hspace{1mm} 0, \hspace{1mm} 0, \hspace{1mm} 0)\), \(\mathbf{v}_{2} = (0, \hspace{1mm} 1, \hspace{1mm} 1, \hspace{1mm} 0, \hspace{1mm} 0)\), \(\mathbf{v}_{3} = (0, \hspace{1mm} 0, \hspace{1mm} 1, \hspace{1mm} 1, \hspace{1mm} 0)\), \(\mathbf{v}_{4} = (0, \hspace{1mm} 0, \hspace{1mm} 0, \hspace{1mm} 1, \hspace{1mm} 1)\), \(\mathbf{v}_{5} = (1, \hspace{1mm} 0, \hspace{1mm} 0, \hspace{1mm} 0, \hspace{1mm} 1)\), \(\mathbf{v}'_{1} = (1, \hspace{1mm} 2, \hspace{1mm} -1)\), \(\mathbf{v}'_{2} = (2, \hspace{1mm} 1, \hspace{1mm} 3)\), and \(\mathbf{v}'_{3} = (1, \hspace{1mm} 1, \hspace{1mm} 1)\).
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Find the matrix \([T]_{B', B}\).
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For the vector \(\mathbf{x} = (3, \hspace{1mm} 7, \hspace{1mm} -4, \hspace{1mm} 5, \hspace{1mm} 1)\), find \([\mathbf{x}]_{\tiny{B}}\) and use the matrix obtained in part 1. to compute \([T(\mathbf{x})]_{B'}\).
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Find the factorization of \([T]\) which is the standard matrix for the linear transformation \(T\) using Formula (28) in Section 8.1.
Solution.
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v1 = [1 1 0 0 0]'; v2 = [0 1 1 0 0]'; v3 = [0 0 1 1 0]'; v4 = [0 0 0 1 1]'; v5 = [1 0 0 0 1]'; nv1 = [1 2 -1]'; nv2 = [2 1 3]'; nv3 = [1 1 1]'; T = [7 12 -5 0 0; 3 10 0 13 1; -9 0 -1 0 -3]; B1 = [v1 v2 v3 v4 v5]; B2 = [nv1 nv2 nv3]; format short; % Find the matrix representation with respect to the bases B1 and B2. TB = T*B1; TB1B2 = B2\TB; disp('The matrix representation of T with respect to the basis B1 and B2 is'); disp(TB1B2);MATLAB results.
The matrix representation of T with respect to the basis B1 and B2 is 34.0000 5.0000 -22.0000 -11.0000 22.0000 40.0000 2.0000 -40.0000 -25.0000 25.0000 -95.0000 -2.0000 97.0000 61.0000 -65.0000 -
% Find the coordinate vector of x with respect to the basis B1. x = [3 7 -4 5 1]'; x_B1 = B1\x; disp('The coordinate vector of x with respect to the basis B is'); disp(x_B1'); % Find the coordinate vector of T(x) with respect to the basis B2. Tx_B2 = TB1B2 * x_B1; disp('The coordinate vector of T(x) with respect to the basis B'' is'); disp(Tx_B2');MATLAB results.
The coordinate vector of x with respect to the basis B is 9 -2 -2 7 -6 The coordinate vector of T(x) with respect to the basis B' is 131.0000 111.0000 -228.0000 -
% Transition matrix from B to the standard basis for R^n. U=B1; % Transition matrix from B' to the standard basis for R^m. V=B2; T=[7 12 -5 0 0 ; 3 10 0 13 1; -9 0 -1 0 -3]; disp('V'); disp(V); disp('TB1B2'); disp(TB1B2); disp('inv(U)'); disp(inv(U)); disp('V*TB1B2*inv(U)');disp(V*TB1B2*inv(U)); disp('T'); disp(T);MATLAB results.
V 1 2 1 2 1 1 -1 3 1 TB1B2 34.0000 5.0000 -22.0000 -11.0000 22.0000 40.0000 2.0000 -40.0000 -25.0000 25.0000 -95.0000 -2.0000 97.0000 61.0000 -65.0000 inv(U) 0.5000 0.5000 -0.5000 0.5000 -0.5000 -0.5000 0.5000 0.5000 -0.5000 0.5000 0.5000 -0.5000 0.5000 0.5000 -0.5000 -0.5000 0.5000 -0.5000 0.5000 0.5000 0.5000 -0.5000 0.5000 -0.5000 0.5000 V*TB1B2*inv(U) 7.0000 12.0000 -5.0000 0 0 3.0000 10.0000 0 13.0000 1.0000 -9.0000 -0.0000 -1.0000 -0.0000 -3.0000 T 7 12 -5 0 0 3 10 0 13 1 -9 0 -1 0 -3
8.2 Similarity and Diagonalizability
Exercise 8.2.
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Show that the matrix
\[A = \begin{bmatrix}-13&\hspace{1mm} -60&\hspace{1mm} -60\\ 10 & 42 & 40\\ -5 & -20 & -18 \end{bmatrix}\]is diagonalizable by finding the nullity of \(\lambda I - A\) for each eigenvalue \(\lambda\) with the use of Theorem 8.2.11 in the Section 8.2.
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Find a basis for \(\mathbb{R}^{3}\) consisting of eigenvectors of \(A\).
Solution.
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% For the exact computation of the eigenvalues, % we use symbolic computation. % Set A as a symbolic matrix. A = sym([-13 -60 -60; 10 42 40; -5 -20 -18]); n = length(A); % Find the eigenvalues of A by using the command eig. eigenvalues = eig(A); for j = 1 : n fprintf('The eigenvalue lambda is '); disp(eigenvalues(j)); % nullity(lambda*I - A) = n - rank(lambda*I - A); nullity = n - rank((eigenvalues(j) * eye(n)) - A); fprintf('The nullity of (lambda*I - A) is '); disp(nullity); end % Since the geometric multiplicity of each eigenvalue of A % is the same as the algebraic multiplicity, % by the Theorem 8.2.11, A is diagonalizable.MATLAB results.
The eigenvalue lambda is 2 The nullity of (lambda*I - A) is 2 The eigenvalue lambda is 2 The nullity of (lambda*I - A) is 2 The eigenvalue lambda is 7 The nullity of (lambda*I - A) is 1 -
% Since the eigenvalue = 2 of A has the multiplicity = 2, % find two linearly independent eigenvectors of A corresponding to lambda = 2. %Find a basis for the null space of (2*I-A). eigvec12=null((2 * eye(n)) - A); % Since the eigenvalue = 7 of A has the multiplicity = 1, % find an eigenvector of A corresponding to lambda = 7. %Find a basis for the null space of (7*I-A). eigvec3=null((7 * eye(n)) - A); p1 = eigvec12(:, 1); p2 = eigvec12(:, 2); p3 = eigvec3(:, 1); % By the Theorem 8.2.7, since the eigenvectors corresponding to % distinct eigenvalues are linearly independent, % the three obtained eigenvectors {p1, p2, p3} form a basis for R^{3}. disp('A basis {p1, p2, p3} for R^{3} consisting of the eigenvectors of A is'); fprintf('p1 ='); disp(p1'); fprintf('p2 ='); disp(p2'); fprintf('p3 ='); disp(p3');MATLAB results.
A basis {p1, p2, p3} for R^{3} consisting of the eigenvectors of A is p1 =[ -4, 1, 0] p2 =[ -4, 0, 1] p3 =[ 3, -2, 1]
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