Linear Algebra Exercises-n-Answers.pdf

40 Linear Algebra, by Hefferon
outermost ‘+’ occurs right before ⃗v k+1 .
= (· · · ⃗v 1 · · · ⃗v m · · · ) + ((· · · (⃗v m+1 + ⃗v m+2 ) + · · · + ⃗v k ) + ⃗v k+1 )
Apply the associativity of the sum of three things
= (( · · · ⃗v 1 · · · ⃗v m · · · ) + ( · · · (⃗v m+1 + ⃗v m+2 ) + · · · ⃗v k )) + ⃗v k+1
and finish by applying the inductive hypothesis inside these outermost parenthesis.
Two.I.1.43 (a) We outline the check of the conditions from Definition 1.1.
Item (1) has five conditions. First, additive closure holds because if a 0 + a 1 + a 2 = 0 and
b 0 + b 1 + b 2 = 0 then
(a 0 + a 1 x + a 2 x 2 ) + (b 0 + b 1 x + b 2 x 2 ) = (a 0 + b 0 ) + (a 1 + b 1 )x + (a 2 + b 2 )x 2
is in the set since (a 0 + b 0 ) + (a 1 + b 1 ) + (a 2 + b 2 ) = (a 0 + a 1 + a 2 ) + (b 0 + b 1 + b 2 ) is zero. The
second through fifth conditions are easy.
Item (2) also has five conditions. First, closure under scalar multiplication holds because if
a 0 + a 1 + a 2 = 0 then
r · (a 0 + a 1 x + a 2 x 2 ) = (ra 0 ) + (ra 1 )x + (ra 2 )x 2
is in the set as ra 0 + ra 1 + ra 2 = r(a 0 + a 1 + a 2 ) is zero. The second through fifth conditions here
are also easy.
(b) This is similar to the prior answer.
(c) Call the vector space V . We have two implications: left to right, if S is a subspace then it is closed
under linear combinations of pairs of vectors and, right to left, if a nonempty subset is closed under
linear combinations of pairs of vectors then it is a subspace. The left to right implication is easy;
we here sketch the other one by assuming S is nonempty and closed, and checking the conditions of
Definition 1.1.
Item (1) has five conditions. First, to show closure under addition, if ⃗s 1 , ⃗s 2 ∈ S then ⃗s 1 + ⃗s 2 ∈ S
as ⃗s 1 + ⃗s 2 = 1 · ⃗s 1 + 1 · ⃗s 2 . Second, for any ⃗s 1 , ⃗s 2 ∈ S, because addition is inherited from V , the
sum ⃗s 1 + ⃗s 2 in S equals the sum ⃗s 1 + ⃗s 2 in V and that equals the sum ⃗s 2 + ⃗s 1 in V and that in
turn equals the sum ⃗s 2 + ⃗s 1 in S. The argument for the third condition is similar to that for the
second. For the fourth, suppose that ⃗s is in the nonempty set S and note that 0 ·⃗s = ⃗0 ∈ S; showing
that the ⃗0 of V acts under the inherited operations as the additive identity of S is easy. The fifth
condition is satisfied because for any ⃗s ∈ S closure under linear combinations shows that the vector
0 · ⃗0 + (−1) · ⃗s is in S; showing that it is the additive inverse of ⃗s under the inherited operations is
routine.
The proofs for item (2) are similar.
Subsection Two.I.2: Subspaces and Spanning Sets
Two.I.2.20 By Lemma 2.9, to see if each subset of M 2×2 is a subspace, we need only check if it is
nonempty and closed.
(a) Yes, it is easily checked to be nonempty and closed. This is a paramatrization.
( ) ( ) 1 0 0 0 ∣∣
{a + b a, b ∈ R}
0 0 0 1
By the way, the paramatrization also shows that it is a subspace, it is given as the span of the
two-matrix set, and any span is a subspace.
(b) Yes; it is easily checked to be nonempty and closed. Alternatively, as mentioned in the prior
answer, the existence of a paramatrization shows that it is a subspace. For the paramatrization, the
condition a + b = 0 can be rewritten
( )
as a = −b. Then
(
we have
)
this.
−b 0 ∣∣ −1 0 ∣∣
{
b ∈ R} = {b
b ∈ R}
0 b
0 1
(c) No. It is not closed under addition. For instance,
( ) ( ) ( )
5 0 5 0 10 0
+ =
0 0 0 0 0 0
is not in the set. (This set is also not closed under scalar multiplication, for instance, it does not
contain the zero matrix.)