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


6.2-6:
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 the eigenvectors.
Please use full frame sequences, because the shortcut fails on problems
other than 2x2.

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6.2-8:
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.

=====================================================================
6.2-18:
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);

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6.2-20:
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).


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6.2-26:
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).

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6.2-28:
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:
A1:=<1,1,0,1|0,1,1,1|0,0,2,1|0,0,0,2>^+;
linalg[eigenvects](A1);
                 
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