%% 2250-notes-S2013-6.2.txt
%% S2008: 6.2:  8, 20, 26. One stapled package.
%% S2012: 6.2:  6, 18, 28. 
%% S2013: 6.2:  6, 20, 26, 28. 

=====================================================================
6.2-6:  A:=Matrix([[10,-6],[12,-7]]); 
        Find invertible P and diagonal D such that AP=PD.

 The answer: P is the matrix of eigenvectors of A and D is the diagonal
 matrix of corresponding eigenvalues. The formula for the characteristic
 equation is

                 (-lambda)^2 + (10-7)(-lambda) + 2 = 0
                                   or
                              r^2-3r+2=0.

 The roots are the eigenvalues lambda = 1, 2. Supply frame sequence
 details to obtain the eigenvectors v1 and v2. Then D=diag(1,2),
 P=aug(v1,v2) are the eigenpair packages and AP=PD, Fourier's model
 holds and the matrix is diagonalizable.

 WARNING: Edwards-Penney examples use a shortcut to find eigenvectors.
 Please use full frame sequences, to practise the method, because the
 shortcut fails on problems other than 2x2.

 MAPLE ANSWER CHECK. 
     A:=Matrix([[10,-6],[12,-7]]); 
     linalg[eigenvects](A);
      # Learn how to decode the output, by maple help examples.

 SAMPLE HAND SOLUTION for lambda=2.

  Define C=A-(2)I, then
          C = A-(2)I
            = Matrix([[10,-6],[12,-7]])-2*Matrix([[1,0],[0,1]])
            = Matrix([[8,-6],[12,-9]])      
  Find the reduced row echelon form rref(C). This form gives the answers
  x for the homogeneous problem Cx=0:
          rref(C)=Matrix([[1,-3/4],[0,0]])      
          x=<x1,x2>, 
          x1=(3/4)t1, x2=t1 (by the last frame algorithm).         
  Take the partial(s) on the invented symbol(s), in this case symbol t1,
  to find a basis, which is the basis of eigenvectors for eigenvalue
  lambda (lambda=2 for this case). Verify this formula:
          partial of x on t1 = Vector([3/4,1])
  Then Vector([3/4,1]) is the eigenvector for lambda=2.
  
=====================================================================
6.2-8:  A:=Matrix([[11,-15],[6,-8]]); 
        Find invertible P and diagonal D such that AP=PD.

 The answer: P is the matrix of eigenvectors of A and D is the diagonal
 matrix of corresponding eigenvalues. The formula for the characteristic
 equation is

                 (-lambda)^2 + (11-8)(-lambda) + 2 = 0
                                   or
                              r^2-3r+2=0.

 The roots are the eigenvalues lambda = 1, 2. Supply frame sequence
 details to obtain the eigenvectors v1 and v2. Then D=diag(1,2),
 P=aug(v1,v2) are the eigenpair packages and AP=PD, Fourier's model
 holds and the matrix is diagonalizable.

 MAPLE ANSWER CHECK. 
     A:=Matrix([[11,-15],[6,-8]]); 
     linalg[eigenvects](A);
      # Learn how to decode the output, by maple help examples.

=====================================================================
6.2-18: A:=Matrix([[6,-5,2],[4,-3,2],[2,-2,3]]); 
        Find invertible P and diagonal D such that AP=PD.

 The answer: P is the matrix of eigenvectors of A and D is the diagonal
 matrix of corresponding eigenvalues. The 6.1 problems imply that the
 characteristic equation is [use r=lambda]

          (-r)^3 + trace(A)(-r)^2 + (m1+m2+m3)(-r) + det(A)=0

 where m1,m2,m3 are the minors of A along the diagonal and trace(A) is
 the sum of the diagonal elements of A.The result for this problem is 

          (-r)^3 + 6(-r)^2 + (-5+14+2)(-r) + 6 = 0

 This formula attacks the college algebra problem of finding roots, but
 the cubic is not factored. The rational root theorem can help, because
 it suggests to try 1, 2, 3, 6 and -1, -2, -3, -6 as possible rots. The
 answer: roots 1,2,3.

 There are three distinct eigenvalues 1,2,3 and therefore the matrix is
 diagonalizable. Find the eigenvectors v1,v2,v3 from three frame
 sequences B --> rref(B) followed by applying the last frame algorithm
 to find the eigenvector [3 times!].

 The book's answers are correct. D=diag(1,2,3), P=aug(v1,v2,v3).

 MAPLE ANSWER CHECK:
    A:=<6,-5,2|4,-3,2|2,-2,3>^+;
   linalg[eigenvects](A);

=====================================================================
6.2-20: A:=Matrix([[2,0,0],[-6,11,2],[6,-15,0]]); 
        Find invertible P and diagonal D such that AP=PD.

 The answer: P is the matrix of eigenvectors of A and D is the diagonal
 matrix of corresponding eigenvalues. The 6.1 problems imply that the
 characteristic equation is [use r=lambda]

          (-r)^3 + trace(A)(-r)^2 + (m1+m2+m3)(-r) + det(A)=0

 where m1,m2,m3 are the minors of A along the diagonal and trace(A) is
 the sum of the diagonal elements of A. This formula is perhaps the
 worst way to attack the college algebra problem because it is not
 factored.

 ALWAYS try to use cofactors to expand det(A-lambda I), in order to get
 free factorizations. In the present case, 2-lambda is a factor obtained
 from the cofactor rule applied to row 1 of det(A-lambda I).

 There are three distinct eigenvalues 2,5,6 and therefore the matrix is
 diagonalizable. Find the eigenvectors v1,v2,v3 from three frame
 sequences B --> rref(B) followed by applying the last frame algorithm
 to find the eigenvector [3 times!].

 The book's answers are correct. D=diag(2,5,6), P=aug(v1,v2,v3).

=====================================================================
6.2-26: A:=Matrix([[1,0,0,1],[0,1,0,1],[0,0,1,1],[0,0,0,2]]); 
        Find invertible P and diagonal D such that AP=PD.

 The answer: P is the matrix of eigenvectors of A and D is the diagonal
 matrix of corresponding eigenvalues. There is no easy way to find the
 characteristic equation, the only possibility is to expand the 4x4
 determinant relation det(A-lambda I)=0.

 The book's answer is correct. Beware: there are two 6.2-25 answers and
 no answer for 6.2-26. The second 6.2-25 answer is actually 6.2-26.

 Use the triangular rule to obtain the characteristic polynomial
 det(A-lambda I)=0 with roots lambda = 1,1,1,2. There are two frame
 sequences B --> rref(B) to calculate. For lambda=1 the last frame
 algorithm gives symbols t1,t2,t3 and you will take three partials to
 find v1,v2,v3. For lambda=2 the last frame algorithm gives one symbol
 t1 and the partial on t1 gives v4. Then

                  D=diag(1,1,1,2), P=aug(v1,v2,v3,v4).

 MAPLE ANSWER CHECK: 
   A:=<1,0,0,1|0,1,0,1|0,0,1,1|0,0,0,2>^+;
   linalg[eigenvects](A);
                 
=====================================================================
6.2-28: A:=Matrix([[1,1,0,1],[0,1,1,1],[0,0,2,1],[0,0,0,2]]); 
        Find invertible P and diagonal D such that AP=PD.

 The answer: P is the matrix of eigenvectors of A and D is the diagonal
 matrix of corresponding eigenvalues. There is no easy way to find the
 characteristic equation, the only possibility is to expand the 4x4
 determinant relation det(A-lambda I)=0. The book's answer is correct.

 Use the triangular rule to obtain the characteristic polynomial
 det(A-lambda I)=0 with roots lambda = 1,1,2,2. There are two frame
 sequences B --> rref(B) to calculate. For lambda=1 the last frame
 algorithm gives one symbols t1 and you will take a partial on t1, which
 finds v1. For lambda=2 the last frame algorithm gives one symbol t1 and
 the partial on t1 gives v2. Then A is not diagonalizable.

 MAPLE ANSWER CHECK: 
   A:=<1,1,0,1|0,1,1,1|0,0,2,1|0,0,0,2>^+;
   linalg[eigenvects](A);
                 
=====================================================================