Span, Linear Independence, and Basis
Problem 1:
Look at Vector([2,-1,1]) and let
L=Span of (2,-1,1) = {t (2,-1,1) where t is real}. Notice L is a line
through the origin in R3 and we can graph the vector
in Maple (without the arrowhead) as:
with(LinearAlgebra): with(plots):
a:=spacecurve({[2*t,-1*t,1*t,t=0..1]},color=red):
display(a);
Part 1:
Find a vector w1 so that
{Vector([2,-1,1]), w1}
is a basis for some plane P1.
Part 2:
Find a vector w2
not on the same line through the origin as w1
so that {Vector([2,-1,1]), w2} is also a basis for the same
plane P1
Part 3: In Maple, use spacecurve commands and display
as above to show that all three vectors lie in the same plane but
no 2 are on the same line
[use different colors like black, blue, green..., and one display
command like display(a,b,c);]
Part 4:
Describe all the vectors w for which
{Vector([2,-1,1]), w} is a basis for the same plane
P1 [Hint - linear combinations are in the same geometric space
so think about what linear combinations
a Vector([2,-1,1]) + b w1
you can use that will give a basis
ie what a's and b's you can use].
Part 5:
Find a vector u
so that {Vector([2,-1,1]), u}
is a basis for a different plane P2 through the origin.
Part 6:
Add u to your graph from Part 3 to show it lies outside the
plane.
Problem 2: Let W be the subspace of R4
spanned by the vectors u1:=(1,2,3,4),
u2:=(4,2,1,5), and
u3:=(3,5,1,7).
Determine if the vectors v:=(8,9,5,16) and
w:=(7,2,1,3) are in W by setting up the augmented matrices and solving
using
M:=Matrix([[
ReducedRowEchelonForm(M).
Next, determine whether
{u1,
u2,
u3, v}
and
{u1,
u2,
u3, w}
are linearly independent sets by setting up the homogeneous
equations and solving.
Does the set of all five vectors span R4?
If you finish before we come back together to go over these,
write down your answers for your notes..
Definitions Review Recall that
linear combinations are in the same geometric space.
A collection of vectors is a
basis if it both spans and is linearly
independent. For a plane, 2 linearly independent vectors (ie not on the
same line) will span and 2 vectors that span are also linearly independent.
The vectors
v1, v2, ..., v
n
span
S if each element of S can be written as a linear combination
of these vectors. In other words, every s in S can be written as
-
s =
c1v1 + c2
v2 + ... + cnvn
for some constants ci.
The vectors v1, v2, ..., vn are linearly independent, if and only if the following condition is satisfied:
Whenever c1, c2, ..., cn are elements of R such that:
- c1v1 + c2v2 + ... + cnvn = 0
then ci = 0 for i = 1, 2, ..., n.