Writing the Solution as the sum of a vector in the row space and a vector homogeous solution. 

Let us return to the example  

>	with(linalg):

 

>	with(LinearAlgebra):

 

>	A:=<<2,2,1>|<2,-2,-1>|<4,6,3>>;

 

>	X:=<<x,y,z>>;

 

>	b:=<<3,4,2>>;

 

>	xs:=BackwardSubstitute(GaussianElimination(<A|b>));

 

>	xp:=subs(_t[1]=0,xs);

 

>	xh:=xp-(subs(_t[1]=1,xs));

 

Let us look at how xh is related to the null space. 

>	RS:=RowSpace(A);

 

>	NullSpace(A);

 

Which is the the same generator as xh execpt that it is 2*xh 

(1.1)

 

>	x1:=RS[1]^%T;

 

>	x2:=RS[2]^%T;

 

The basis vectors in row space x1 and x2 are are always linearly independent.  Also since xh is perpendicular to x1 and x2 then xh, x1,x2 form a linearly indepentdent  set of vectors.  Because they are linearly independent a number three then they form  a basis for all of This means that every vector can be written as a linear combination of the three vectors.   We will thus write xp as a linear combination of the three linearly indpendent vectors.  We are not really interested in finding the the coefficient of xh since that would be redundant.   We will find the coefficients of the above set of basis vectors.  This vector will be called xr in the following set of Maple commands.   

>	Basis1:=<x1|x2|xh>;

 

>	det(Basis1);

 

>	Basis([x1,x2,xh]);

 

(1.2)

 

The Basis1 matrix must always be a square matrix.  The Basis command is a with(LinearAlgebra) and will well us the linearly independent vectors. 

>	c:=BackwardSubstitute(GaussianElimination(<Basis1|xp>));

 

We need only use the first two entries from c because the 3/5  would be taken times xh and we already know a basis for the homogenous solution.   Remember that we did need xh though to get three linearly independent vectors.  The three linearly independent vectors will be a basis for Which gives the unique solution for the c's above.   

>	xr:=c[1]*x1+c[2]*x2;

 

>

 

>	(xr)^%T.xh;

 

>

A.xr;

 

>	xa:=xr+alpha*xh;

 

>

A.xa;

 

This has been a rather long process but let us try to summerize what we have just done.   Since the row space vectors are orthogonal to the null space vectors the basis for the row space and the null space are linearly independent set of vectors.  When combined they will become a basis for the dimension of the matrix A.  This means that the particular solution must be in the span of the combination of the null space vectors and the row space vectors.  We set up a system of equation to solve to see what linear combination of these vectors will give the particular soltuion.  The only ones we are interested in is the coefficients of the row space vectors.  Why?  The coefficient for the homogenous vector is all ready in the arbitrary paramenter for the homogenous vector.   

Exercises   

Exercise 2.3.3 (a), (b) to write the solution to as xr + xh where xr is the solution in the row space.  You will need to use only one of the set of basis vectors.  The best set of basis vectors  is the one obtained from using Gauss Jordan on the coefficient matrix.  You can use the gaussjord command.  Next, compare this with the set of vector obtained by using the Maple command Rowspace.   Exercise 2.3.6   

Assignment is due on October 20, 2008.   

>