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S2013 notes on problems 3.4-20,30,34,40
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3.4-20
The system of equations is already a reduced echelon system, that is,
in augmented matrix form the first frame equals the last frame, which
is the RREF. This is a problem with three lead variables and two free variables. It is
the infinitely many solution case. The algorithm used is as follows:

Assign symbols t1, t2 to the free variables [book: symbols s,t]

Back-substitute the invented symbols into the equations of the last frame.
Then solve for the lead variables.

Report the variable list equalities, in variable list order, as the general scalar
soluton.

Write the scalar solution as a vector solution of the form u = t1 v1 + t2 v2,
where u, v1, v2 are column vectors. Argue that v1 = partial of u w/r/t t1
and v2 = partial of u w/r/t t2.

maple.
A:=matrix([[1,-3,0,0,7],[0,0,1,0,-5],[0,0,0,1,-10]]);
linsolve(A,vector([0,0,0]));

3.4-22
The homogeneous system is in reduced echelon form Ax=0 with

  A=[[1,-1,0,7,3],[0,0,1,-1,-2]],  x=[x1,x2,x3,x4,x5]

The last frame algorithm is applied to the scalar equations

   x1 -x2 +     7x4 + 3x5 = 0,
            x3 - x4 - 2x5 = 0.

to obtain lead variables x1,x3 and free variables x2,x4,x5,
then assign invented symbols t1, t2, t3 to the free variables 
and write the answer in variable list order:

   x1 = t1 -7t2 -3t3,
   x2 = t1,
   x3 = t2 + 2t3,
   x4 = t2,
   x5 = t3.

3.4-29
This result is called the Cayley-Hamilton theorem. It is the basis for
solving system of differential equations of the form x'=Ax, where A is
a square matrix. The general theorem reads 

     "A square matrix A satifies its own characteristic equation."

For a 2x2 matrix A, the result is given in 3.4-29. 

    (-r)^2 + trace(A)(-r) + det(A)  = 0    Characteristic equation
    (-A)^2 + trace(A)(-A) + det(A)I = 0    Cayley-Hamilton identity

The characteristic equation is the formal expansion det(A-rI)=0. In
college algebra, 2x2 and 3x3 determinants were studied. The topic of
determinant theory appears in 3.6 of the textbook.


3.4-30
See 3.4-29 for the Cayley-Hamilton formula for a general 2x2 matrix A.
For the given matrix A, this formula is

                r^2 -4r + 3 = 0   Characteristic equation

                A^2 -4A + 3I = O  Cayley-Hamilton identity

Then
                A^2 = 4A - 3I,
                A^3 = A(A^2)
                    = A(4A - 3I)
                    = 4A^2-3AI
                    = 4A^2 - 3A         because AI=A
                    = 4(4A-3I)-3A
                    = 13A - 12I

Because of these formulas, the powers A^2, A^3, A^4, A^5 can be computed
from linear combinations of A and I. No matrix multiply is needed.

Ans Check: Use maple as follows

                with(linalg):
                A:=matrix([[2,1],[1,2]]);
                evalm(A^5);

Then maple will print A^5, which if it matches your A^5 done the other way,
proves that your computation is error free.

3.4-34
Use 3.4-29, the Cayley Hamilton theorem as follows. The equation A^2 = O
is the same as A^2 -(a+d)A+(ad-bc)=O where the components a,b,c,d of A
have been chosen so that a+d=0 and ad-bc=0. This means the trace of A is
zero and the determinant of A is zero. There are lots of matrices A with
these conditions satisfied: pick one and show how you got it.

== more details on 3.4-34 ==
The book has no answer. Suggested is to use the Cayley-Hamilton identity
A^2-(a+d)A+(ad-bc)I=0 where A=[[a,b],[c,d]], I=[1,0],[0,1]]=2x2 identity
matrix and 0=[0,0,],[0,0]]=2x2 zero matrix.

Initially, a^2=b^2=c^2=d^2=1 because each of a,b,c,d is either +1 or -1.

Using Cayley-Hamilton (Exercise 3.4-29) and A^2=0 implies the identity

		(a+d)A=(ad-bc)I
 which is equivalent to
                    (a+d)[[a,b],[c,d]]=(ad-bc)[[1,0],[0,1]]

One way to make this identity true is to let

		a+d=0,
                    ad-bc=0.

Combine this with a^2=b^2=c^2=d^2=1 to find possible values for a,b,c,d.
For instance, a=1 and then d=-1 (a+d=0 true) and then ad-bc=0 becomes
-1-bc=0, which is true for b=1,c=-1.

There are a number of possible solutions. Can you display all answers for
matrix A, using these assumptions?

3.4-36
The book has no answer. This is a variant of 3.4-34, where it is suggested 
to use the Cayley-Hamilton identity A^2-(a+d)A+(ad-bc)I=0 where 
A=[[a,b],[c,d]], I=[1,0],[0,1]]=2x2 identity matrix and 
0=[0,0,],[0,0]]=2x2 zero matrix.

Initially, a=d=0 because the main diagonal of A is assumed all zeros.

Using Cayley-Hamilton (Exercise 3.4-29) and A^2=I implies the identity

		(a+d)A=(ad-bc+1)I
 which is equivalent to (using a=d=0)
                    (0+0)[[0,b],[c,0]]=(0-bc+1)[[1,0],[0,1]]

One way to make this identity true is to let

                    -bc+1=0.

Combine this with a=d=0 to find possible values for a,b,c,d.
For instance, b=c=1 and then -bc+1=0. For this case,

       A=[[0,1],[1,0]].

There are a number of possible solutions. Can you display all answers for
matrix A, using these assumptions?


3.4-40
(a) Given A x_0 = 0 and A x_1 = b, then define x = x_0 + x_1.
We show that A x = b, as follows:

  (LHS of eq Ax=b)  =  Ax
                    =  A(x_0 + x_1)             substitute
                    =  Ax_0 + Ax_1              matrix multiply rules
                    =  0    + b                 use given equations
                    =  b                        vector toolkit property
                    =  (RHS of eq Ax=b)
The proof is complete.

(b) Give A x_1 = b  and A x_2 = b, then define x = x_1 - x_2.
Show that A x = 0.

The details, left to you, parallel the matrix algebra details of (a).
In particular, do not break A and x into components; use only formal
matrix algebra rules, as in (a) above.