Project 1.sagews

# 1. Consider the following set vectors in R^5. Is this set of vectors lineraly independent or linearly dependent
A = matrix(QQ, [[9,13,5,6,-1],[14,15,-7,-6,4],[-8,-9,12,-5,-9],[-5,-6,-8,9,8],[13,14,15,2,11]])
A

[ 9 13 5 6 -1] [14 15 -7 -6 4] [-8 -9 12 -5 -9] [-5 -6 -8 9 8] [13 14 15 2 11]

A.rref()

[ 1 0 0 0 5] [ 0 1 0 0 -4] [ 0 0 1 0 0] [ 0 0 0 1 1] [ 0 0 0 0 0]

# Linearly dependent because x_5 is a free variable so there is more than one solution
## x_5 is a free variable for what system? Explain what you're doing. 4/5

#2 Span

C = matrix(QQ, [[9,13,5,6,-1,41],[14,15,-7,-6,4,12],[-8,-9,12,-5,-9,36],[-5,-6,-8,9,8,-63],[13,14,15,2,11,25]])

C

[ 9 13 5 6 -1 41] [ 14 15 -7 -6 4 12] [ -8 -9 12 -5 -9 36] [ -5 -6 -8 9 8 -63] [ 13 14 15 2 11 25]

C.rref()

[ 1 0 0 0 5 -17] [ 0 1 0 0 -4 16] [ 0 0 1 0 0 2] [ 0 0 0 1 1 -4] [ 0 0 0 0 0 0]

# Yes it is in the span because x_5 is free so there is at least one solution, so C [41,12,36,-63,25] does span A
## You mean to say that the given vector is in the span of the columns of A. 4/5

#3 span R^5?

# No becuase it does not have a unique solution because x_5 is a free variable. There is not a leading one in the last row. 
## This has nothing to do with free variables, but rather there being some vector for which the system is inconsistent. Can you demonstrate that? 3/5


#4 Matrix Transformation 
A = matrix(QQ,[[12,10,-6,-3,7,-10,0],[-7,-6,4,7,-9,5,0],[9,9,-9,-5,5,-1,0],[-4,-3,1,6,-8,9,0],[8,7,-5,-9,11,-8,0]])
A

[ 12 10 -6 -3 7 -10 0] [ -7 -6 4 7 -9 5 0] [ 9 9 -9 -5 5 -1 0] [ -4 -3 1 6 -8 9 0] [ 8 7 -5 -9 11 -8 0]

A.rref()

[ 1 0 2 0 2 0 0] [ 0 1 -3 0 -2 0 0] [ 0 0 0 1 -1 0 0] [ 0 0 0 0 0 1 0] [ 0 0 0 0 0 0 0]

# No T is not one-to-one because there is more than one solution because x_6 is free. In order for t(x)=0 to be one-to-one there has to be only the trivial solution.
## Right. 5/5

#5 No the linear transformation is #4 is not onto because not everything in R^6 is in the range
## How do you see that not everything is in the range? Note that if it were onto the range would be R^5, not R^6. 3/5


A = matrix(QQ, [[1,3,0,2],[-5,-5,7,4],[3,5,2,1],[1,-1,2,-3]])
A

[ 1 3 0 2] [-5 -5 7 4] [ 3 5 2 1] [ 1 -1 2 -3]