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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