linear_independence.sagews

The first two cells below show how to check whether a given set of vectors is linearly independent.

# Sec. 7.3, exercise 10: Check for linear in/dependence of the vectors
# x_1=(1, 2, -1, 0); x_2=(2, 3, 1, -1); x_3=(-1, 0, 2, 2); x_4=(3, -1, 1, 3)
#
# METHOD 1
# Check whether the matrix of column vectors (x_1, x_2, x_3, x_4) 
# is non-singular. Note: since we can only check non-singular 
# for square matrices, this method is limited to cases 
# where # of vectors = # of dimension of vector-space.

x1 = vector( [1, 2, -1, 0] )
x2 = vector( [2, 3, 1, -1] )
x3 = vector( [-1, 0, 2, 2] )
x4 = vector( [3, -1, 1, 3] )
A = matrix( [x1, x2, x3, x4])  # define matrix of column vectors
det(A)                         # compute its determinant

$-70$

# METHOD 2
# Create a "free module" of dimension 4 and define the vectors 
# in that vector space/free module.  Then use the builtin 
# attribute "are_linearly_dependent" to get true/false.
M = QQ^4                # Creates a "free module" of dimension 4 over the field of rationals
v1 = M( [1, 2, -1, 0] ) # Defines the vectors in the vector space/free module M
v2 = M( [2, 3, 1, -1] )
v3 = M( [-1, 0, 2, 2] )
v4 = M( [3, -1, 1, 3] )
M.are_linearly_dependent([v1, v2, v3, v4])  # Check whether the vectors are linearly dependent

$\mathrm{False}$

Let's modify the above example and address cases where the number of vectors does not equal the dimensions of the vector space.

We will consider both possibilities: (1) fewer vectors than dimensions; and (2) more vectors than dimensions.

# Case 1: Fewer vectors than dimensions.
# Sec. 7.3, modify exercise 10: Check for linear in/dependence of 
# the first 3 vectors x_1=(1, 2, -1, 0); x_2=(2, 3, 1, -1); x_3=(-1, 0, 2, 2).
# We know from the previous exercise that they must be 
# linearly independent.  
# Let's check for this by forming the matrix of column vectors 
# and then row-reducing it to reduced echelon form:
x1 = vector( [1, 2, -1, 0] )
x2 = vector( [2, 3, 1, -1] )
x3 = vector( [-1, 0, 2, 2] )
Ashort = transpose( matrix( [x1, x2, x3]) )  # define matrix of column vectors
Ashort.rref()                   # reduced echelon form

print "\n Since every row has diagonal entry 1, they are linearly independent"

$\left(\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right)$

Since every row has diagonal entry 1, they are linearly independent

# Case 2: More vectors than dimensions.
# Sec. 7.3, modify exercise 10: Check for linear in/dependence of 
# the 5 vectors x_1=(1, 2, -1, 0); x_2=(2, 3, 1, -1); x_3=(-1, 0, 2, 2); 
# x_4=(3, -1, 1, 3); x_5=(0, -1, -2, 3).
# A set of 5 vectors in 4 dimensions will always be 
# linearly dependent.  
x1 = vector( [1, 2, -1, 0] )
x2 = vector( [2, 3, 1, -1] )
x3 = vector( [-1, 0, 2, 2] )
x4 = vector( [3, -1, 1, 3] )
x5 = vector( [0, -1, -2, 3] )
Als = transpose( matrix( [x1, x2, x3, x4, x5]) )  # define matrix of column vectors
Als.rref()

[ 1 0 0 0 57/35] [ 0 1 0 0 -9/7] [ 0 0 1 0 9/35] [ 0 0 0 1 2/5]

# Verify
(57/35)*x1 - (9/7)*x2 + (9/35)*x3 + (2/5)*x4

(0, -1, -2, 3)

# Example showing how to row-reduce a matrix to echelon form.
# (There is also some miscellaneous other stuff mixed in below.)
# Note that when the matrix is singular, I don't get the general 
# parametric form of the solution of A*X = 0.  Instead it returns X=0.

A1 = matrix([ [3,-6], [-1, 2] ])
Y1 = vector([0, 0])
print A1, Y1
#X1 = A1\Y1
#print X1
#print det(A1)
#A1.echelon_form()      # plain echelon form
A1.rref()               # reduced echelon form

[ 3 -6] [-1 2] (0, 0) [ 1 -2] [ 0 0]

# Another example of row-reduction, this time with a rectangular matrix.

A2 = matrix([ [3,-6, 4], [-1, 2, 5]])
Y2 = vector([0, 0])
print A2, Y2
X2 = A2\Y2
print X2
#A2.echelon_form()
A2.rref()

[ 3 -6 4] [-1 2 5] (0, 0) (0, 0, 0) [ 1 -2 0] [ 0 0 1]

# Yet another example of row-reduction, again with a rectangular matrix.

A3 = matrix([ [1, 3, 5], [2, 4, 6]])
A3.rref()

[ 1 0 -1] [ 0 1 2]

# And some more...

A4 = matrix([ [-1, 2], [0, 0], [2, -4]])
A4.rref()

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

# Sec. 7.3, exercise 7

A5 = matrix([ [1, 0, 1], [1, 1, 0], [0, 1, 1]])
A5.rref()

[1 0 0] [0 1 0] [0 0 1]

A6 = matrix([ [1, 1, 3], [2, 1, 0] ])
A6.rref()

[ 1 0 -3] [ 0 1 6]