# MATH 2270
# PROJECT 4
# November 8, 1999
# 
# This project is about the fundamental subspaces associated with
# matrices and about coordinates with respect to different bases. 
# I remind that in each problem you should use Maple 
# to solve the problem, not pen-and-paper computations! 
#      A Maple text version of this project may be found at the web site
# http://www.math.utah.edu/~kapovich/teaching2.html/
#
# Go to that page and click on the "4-th maple assignment".  

# Use "split execution groups" command to edit this textfile.  

# First load our tools:
#  
> with(linalg):
# Here's the matrix we will work with today.  We will keep it moderately
# sized, but of course we could make it much bigger without causing
# Maple any problems. 
# 
> A:=matrix([[ 1,1,1,2,6 ],[ 2,3,-2,1,-3],[3,5,-5,1,-8],[4,3,8,2,3]]);

                         [1    1     1    2     6]
                         [                       ]
                         [2    3    -2    1    -3]
                    A := [                       ]
                         [3    5    -5    1    -8]
                         [                       ]
                         [4    3     8    2     3]

# We will think of this as the matrix of a linear transformation L: R^5-> R^4,
# L(x)=Ax. 

# Problem 1. Compute rank and nullity  of A using reduction 
# of A to row echelon form. 

# Problem 2.  A basis for the rowspace of A is staring at you in the rref(A) 
# matrix. What is this basis? 

# Problem 3. Find a basis for the null-space of A. To solve this problem you 
# can either use the "rref" command or "linsolve": 
> linsolve(A, b); 
# in this command A is the matrix, b is the vector 
# of the right-hand side (in the present problem you will use zero vector).
# If you do not want to use the symbols like _t_1 as your parameters 
# you can use the command:
> linsolve(A, b, 'r', s);
# Now the names of the parameters will be s_1, s_2,...
#Type 
>r;
#to see what the rank of the matrix is. 


# Problem 4. Use Maple to find a basis v1, v2, v3 for the column space of A 
# (i.e. of the image of  L).  
# Recall that one way to find this basis is to do
# column operations, putting your matrix into reduced column echelon
# form.  You can do this in Maple by transposing, computing the rref of
# the transpose, and transposing back:
> rcef:=B->transpose(rref(transpose(B)));
>     #a procedure for computing reduced
>     #column echelon form
> rcef(A);


# Problem 5.  Maple will compute bases for these spaces with  single commands. 
# Compare the answers you've gotten above with Maple's answers:  Since
# you know bases are not unique, you can pretty well guess that Maple is
# using methods close to the ones we used, since its answers should be
# quite similar to some of yours:
> rowspace(A); # Gives a basis for the space spanned by row vectors.
> nullspace(A);#nullspace basis
> colspace(A); #nice column space basis, i.e. a basis for the image of L. 
> v:=convert(colspace(A),list); 
#The above comman is making a list of vectors in the basis. 
#This keeps track of order in the set v of vectors 
#which form basis of the image of L.  
     
>     v1:=v[1]; v2:= v[2]; v3:= v[3] 
# These vectors form a basis of the image of L. 
# Compare these vectors 
# with vectors from the problem 4.  
> r1:=row(rref(A),1); #r_1,r_2,r_3 basis for rowspace(A)
> r2:=row(rref(A),2);
> r3:=row(rref(A),3);
> G:=convert(nullspace(A),list); #making a list
>     #keeps track of order in a set

> n1:=G[1];n2:=G[2]; #nullspace basis. 


# Problem 6. Use Maple to verify that the vectors r1, r2, r3, n1, n2 form 
# a basis of R^5. Hint: use reduced row echelon form or the determinant. Explain your solution: namely, how 
# the value of the determinant or shape of the reduced echelon form is related to the fact that we have a basis.  
# Check that each of the vectors r1, r2, r3 is orthogonal to each of the vectors n1,n2. I.e. the row-space is orthogonal 
# to the null-space of the matrix. 
 
# Problem 7.  Let us call E={e1,e2,e3,e4,e5}, i.e. the set of standard basis
# vectors in R^5.  Let us call our new basis  S={r1,r2,r3,n1,n2}. 
# Find the transition matrices P_{E <-S} and P_{S<-E}, which 
# convert S coordinates to E-coordinates and vice-versa. Recall that the matrix P_{E <-S} 
# is very easy to find and the matrix P_{S<-E} is easily obtained from the matrix P_{E <-S}.  

# Problem 8. Using transition matrix find the coordinates of the following vectors with respect to
# the S-basis:
# 8a)  a=(0,1,-4,0,-3)
# 8b)  b=(1,0,0,0,0)

# Problem 9. Note that v_1, v_2, v_3 are three linearly independent vectors in
# R^4. However to get a basis we need four vectors. Find a nonzero vector v_4 
# which is orthogonal to v_1, v_2 and v_3. (Hint: form a matrix with the rows v1, v2, v3 and then 
# find a basis of its null-space. Then check that the vector which 
# you got is indeed orthogonal to v1, v2, v3.) 
# Use Maple to check that v1, v2, v3, v4 are
# linearly independent. Why do they form a basis of R^4 ? 


# Problem 10.  Let F={e1,e2,e3,e4} be the standard basis in R^4.  Let
# T:={v1,v2,v3,v4}. Find the transition matrices P_{F <-T} and P_{T<-F}.
#