Recalling Exercises 38, prove that every basis for R n contains exactly n vectors . 38. Show that any spanning set of R n must containat least n vectors. Proceed by showing that if { u 1 , u 2 , ... u p } are vectors in R n and if p < n , then there is a non-zero vector v in R n such that v T u i = 0 , 1 ≤ i ≤ p [ Hint: Write the constraints as and ( p × n ) v as above can v be a linear combination of u 1 , u 2 , ... u p ?
Recalling Exercises 38, prove that every basis for R n contains exactly n vectors . 38. Show that any spanning set of R n must containat least n vectors. Proceed by showing that if { u 1 , u 2 , ... u p } are vectors in R n and if p < n , then there is a non-zero vector v in R n such that v T u i = 0 , 1 ≤ i ≤ p [ Hint: Write the constraints as and ( p × n ) v as above can v be a linear combination of u 1 , u 2 , ... u p ?
Solution Summary: The author proves that the every basis of Rn contains n vectors.
Recalling Exercises 38, prove that every basis for
R
n
contains exactly
n
vectors.
38. Show that any spanning set of
R
n
must containat least
n
vectors. Proceed by showing that if
{
u
1
,
u
2
,
...
u
p
}
are vectors in
R
n
and if
p
<
n
, then there is a non-zero vector v in
R
n
such that
v
T
u
i
=
0
,
1
≤
i
≤
p
[Hint: Write the constraints as and
(
p
×
n
)
v as above can v be a linear combination of
u
1
,
u
2
,
...
u
p
?
Quantities that have magnitude and direction but not position. Some examples of vectors are velocity, displacement, acceleration, and force. They are sometimes called Euclidean or spatial vectors.
Consider the vectors u1 = (1, 2, 1), u2 = (0, 3, 2) and u3 = (−2, 1, 1). Determine whether the set S = {u1, u2, u3} is a spanning set for R^3
S is a linearly independent set,then for any x element of , x is not the linear combination of other vectors in S. Is this true or false?
Let S be a spanning set for a finite dimensional vector space V. Prove that there exists a subset S′ of S that forms a basis for V.Getting Started: S is a spanning set, but it may not be a basis because it may be linearly dependent. You need to remove extra vectors so that a subset S′ is a spanning set and is also linearly independent.(i) If S is a linearly independent set, then you are done. If not, remove some vector v from S that is a linear combination of the other vectors in S. Call this set S1.(ii) If S1 is a linearly independent set, then you are done. If not, then continue to remove dependent vectors until you produce a linearly independent subset S′.(iii) Conclude that this subset is the minimal spanning set S′.
Chapter 3 Solutions
Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)
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