solns4a

solns4a - MATH 235/W08 Solutions for Assignment 4A...

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Unformatted text preview: MATH 235/W08 Solutions for Assignment 4A Eigenvalues, Eigenvectors, Diagonalization Questions 2,3,4,5,6,7,12 handed in Wed. Feb. 27/08. Though many solutions are done using MATLAB, the details for a solution by hand are included! Many extra details are also included. Please try working through the MATLAB solutions! Contents 1 Eigenpairs 2 2 Diagonalize 2 2.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 3 Linear Transformation 17 3.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 4 Similarity and Rank 18 4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 5 Eigenvalue Properties 18 5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 6 Diagonalizability 19 7 Real Eigenvalues 21 7.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 8 Evaluate A n 22 9 Simultaneous Diagonalization 22 10 Characteristic Polynomial 22 11 Initial Value Problem 22 12 MATLAB and Recurrences 23 12.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 1 1 Eigenpairs Find the eigenvalues and eigenvectors of A = 2 6 6 4 1 1 4 1 a 2 c b 2 3 7 7 5 . Is A diagonalizable? 2 Diagonalize Diagonalize, if possible, the following matrices, i.e. find the eigenvalues and then find the corresponding eigenspaces. A = 2 4 − 1 3 5 − 3 1 − 1 2 3 5 , B = 2 4 − 2 − 4 i 4 i − 4 i 2 − 4 i 2 3 5 , C = 2 6 6 6 6 4 4 4 2 3 − 2 1 − 2 − 2 2 6 12 11 2 − 4 9 20 10 10 − 6 15 28 14 5 − 3 3 7 7 7 7 5 . 2.1 Solution BEGIN SOLUTION: We now work on the three matrices: 1. The following MATLAB program yields the result for the matrix A . Note that the eigenvalues of A are: -1,1,2. And, distinct eigenvalues implies diagonalizable. %%MATLAB PROGRAM STARTS%%%%%%%%%%%%%%%%%%%%%%%%%%%% !rm temp.txt diary temp.txt clear all A = [ 0 -1 3 5 0 -3 1 -1 2 ]; [u,d]=eig(A); eigs=diag(d); disp([’eigenvalues of A are: ’,num2str((eigs’))]) disp(’The eigenvalues are distinct; so A is diagonalizable’) disp(’We can confirm that these are the eigenvalues using the determinant’) for i=1:size(A,1), detp=double(det(A-eigs(i)*eye(size(A,1)))); disp([’For eigenvalue =’,num2str(eigs(i)),’ det A-lambda I is ’,... num2str(detp)]); end disp(’ ’) disp(’We can now find a basis for the eigenspace for each eigenvalue’) 2 P=zeros(size(A,1)); disp(’ ’) for i=1:size(A,1), disp([’For eigenvalue =’,num2str(eigs(i)),’ basis for nullity is ’]) P(:,i)=null(A-eigs(i)*eye(size(A,1))); P(:,i) end disp(’The matrix P of eigenvectors is’) P disp(’ ’) disp(’now diagonalize A using inv(P)*A*P’) P\A*P diary off %%MATLAB PROGRAM ENDS%%%%%%%%%%%%%%%%%%%%%%%%%%%% and the output is: %%MATLAB OUTPUT STARTS%%%%%%%%%%%%%%%%%%%%%%%%%%%% eigenvalues of A are: -1 1 2 The eigenvalues are distinct; so A is diagonalizable We can confirm that these are the eigenvalues using the determinant For eigenvalue =-1 det A-lambda I is...
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solns4a - MATH 235/W08 Solutions for Assignment 4A...

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