Download T. Penttila 1) Which of the following sets are spanning sets for R3?

Mathematics 369
Homework (due Feb 7)
T. Penttila
1) Which of the following sets are spanning sets for R3 ? In each case show that the set spans if true; respectively
give an example of a vector in R3 that does not lie in the span if false.
a) {(1, 2, 0)T , (0, 1, 0)T , (0, 2, 1)T } spans R3 , as a(1, 2, 0) + (−2a + b − 2c)(0, 1, 0) + c(0, 2, 1) = (a, b, c).
Since R3 has dimension 3, it follows that it is linearly independent.
b) {(1, 2, 0)T , (0, 1, 0)T , (0, 2, 1)T , (1, 2, 3)T } spans R3 , by (a).
It is NOT linearly independent, as (1, 2, 3)T = (1, 2, 0)T − 6(0, 1, 0)T + 3(0, 2, 1)T
c) {(2, 1, −2)T , (3, 2, −2)T , (2, 2, 0)T , (4, 3, −2)T } does not span R3 , (1, 0, 0) is not in the span.
It is NOT linearly independent, as (2, 2, 0)T = −2(2, 1, −2)T + 2(3, 2, −2)T . d) {(2, 1, −2)T , (3, 2, −2)T , (2, 2, 0)T }
does not span R3 , (1, 0, 0) is not in the span.
It is NOT linearly independent, as (2, 2, 0)T = −2(2, 1, −2)T + 2(3, 2, −2)T . e) {(1, 1, 1)T } does not span R3 , (1, 0, 0)
is not in the span.
It is linearly independent, as is any single non-zero vector.
2) Which of the following sets are spanning sets for P3 (the set of polynomials of degree up to 2)? Show that the set
spans, or give an example of a polynomial that does not lie in the span.
a) {x − 1, x2 − 1, 3x2 + 2x − 5} does not span P3 ,1 is not in the span
It is NOT linearly independent, as 3x2 + 2x − 5 = 2(x − 1) + 3(x2 − 1).
2c
2
b) {x2 + 1, 2x + 3, x} spans P3 , as a(x2 + 1) + (− a3 + 3c )(2x + 3) + ( 2a
3 + b − 3 )x = ax + bx + c.
It is linearly independent, as P3 has dimension 3.
c) {1 + x, x + x2 , 1 + x2 , 1 − x2 } spans P3 , as (− a2 + b2 + 2c )(1 + x) + ( 2a + b2 − 2c )(x + x2 ) + ( a2 − 2b + 2c )(1 + x2 ) =
ax2 + bx + c.
It is NOT linearly independent, as 1 − x2 = 1 + x − (x + x2 ).
3) Which of the sets in problem 1 and 2 are linearly independent? (Prove that they are independent, or show a
nontrivial linear relation among the vectors.)See above.


8 −1 0
23
2 1 −44 . Determine a spanning set for the nullspace of A.
4) Let A =  −16
9 −1 0
26
{(−3, −1, −2, 1)} is spanning set for the nullspace of A.
5) Show: Let V be a vector space, X ⊂ V and y ∈ V . We form a new set Z = X ∪ {y} by adding the vector y to X.
a) Show that if y ∈ Span(X) then Z is linearly dependent.
Let X = {x1 , . . . , xn }. If y ∈ Span(X), then y = a1 x1 + . . . + an xn . Hence a1 x1 + . . . + an xn − y = 0, so Z is linearly
dependent.
b) Assume that X is linearly independent. Show that if Z is linearly dependent, then y ∈ Span(X).
Suppose a1 x1 + . . . + an xn + an+1 y = 0, with not all ai equal to 0. If an+1 = 0, this gives a non-trivial linear dependence
a1
on X, contrary to X being linearly independent. So an+1 is not 0. Hence y = (− an+1
)x1 + . . . + (− f racan an+1 )xn , is in
Span(X).
6) a) Let V be a vector space and S, T ≤ V be two subspaces. Show that the intersection S ∩ T is a subspace of V .
S ∩ T is non-empty, so 0 is in both S and T , and hence in S ∩ T .
Let x, y ∈ S ∩ T . Then x, y ∈ S, so x + y ∈ S, since S is a subspace. Similarly, x, y ∈ T , so x + y ∈ T , since T is a
subspace. Thus x + y ∈ S ∩ T . So S ∩ T is closed under addition.
Let x ∈ S ∩ T , and c be a scalar. Then x ∈ S, so cx ∈ S,since S is a subspace. Similarly, x ∈ T , so cx ∈ T , since T is a
subspace. Thus cx ∈ S ∩ T . So S ∩ T is closed under scalar multiplication.
b) Give an example of a vector space V and subspaces S, T ≤ V such that the union S ∪ T is not a subspace of V .
V = R2 , S = Span((1, 0)t ),T = Span((0, 1)T ).