Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 69
Two.III.4.39 It happens when at least one of W 1 , W 2 is trivial. But that is the only way it can happen.
To prove this, assume that both are non-trivial, select nonzero vectors ⃗w 1 , ⃗w 2 from each, and
consider ⃗w 1 + ⃗w 2 . This sum is not in W 1 because ⃗w 1 + ⃗w 2 = ⃗v ∈ W 1 would imply that ⃗w 2 = ⃗v − ⃗w 1 is
in W 1 , which violates the assumption of the independence of the subspaces. Similarly, ⃗w 1 + ⃗w 2 is not
in W 2 . Thus there is an element of V that is not in W 1 ∪ W 2 .
Two.III.4.40
(a) The set
( ) ( ) ( )
v1 ∣∣ v1 x
{
= 0 for all x ∈ R}
v 2 v 2 0
is easily seen to be the y-axis.
(b) The yz-plane.
(c) The z-axis.
(d) Assume that U is a subspace of some R n . Because U ⊥ contains the zero vector, since that vector
is perpendicular to everything, we need only show that the orthocomplement is closed under linear
combinations of two elements. If ⃗w 1 , ⃗w 2 ∈ U ⊥ then ⃗w 1 ⃗u = 0 and ⃗w 2 ⃗u = 0 for all ⃗u ∈ U. Thus
(c 1 ⃗w 1 + c 2 ⃗w 2 ) ⃗u = c 1 ( ⃗w 1 ⃗u) + c 2 ( ⃗w 2 ⃗u) = 0 for all ⃗u ∈ U and so U ⊥ is closed under linear
combinations.
(e) The only vector orthogonal to itself is the zero vector.
(f) This is immediate.
(g) To prove that the dimensions add, it suffices by Corollary 4.13 and Lemma 4.15 to show that
U ∩ U ⊥ is the trivial subspace {⃗0}. But this is one of the prior items in this problem.
Two.III.4.41 Yes. The left-to-right implication is Corollary 4.13. For the other direction, assume that
dim(V ) = dim(W 1 ) + · · · + dim(W k ). Let B 1 , . . . , B k be bases for W 1 , . . . , W k . As V is the sum of the
subspaces, any ⃗v ∈ V can be written ⃗v = ⃗w 1 + · · · + ⃗w k and expressing each ⃗w i as a combination of
⌢
vectors from the associated basis B i shows that the concatenation B 1 · · ·⌢
Bk spans V . Now, that
concatenation has dim(W 1 ) + · · · + dim(W k ) members, and so it is a spanning set of size dim(V ). The
concatenation is therefore a basis for V . Thus V is the direct sum.
Two.III.4.42 No. The standard basis for R 2 does not split into bases for the complementary subspaces
the line x = y and the line x = −y.
Two.III.4.43 (a) Yes, W 1 + W 2 = W 2 + W 1 for all subspaces W 1 , W 2 because each side is the span
of W 1 ∪ W 2 = W 2 ∪ W 1 .
(b) This one is similar to the prior one — each side of that equation is the span of (W 1 ∪ W 2 ) ∪ W 3 =
W 1 ∪ (W 2 ∪ W 3 ).
(c) Because this is an equality between sets, we can show that it holds by mutual inclusion. Clearly
W ⊆ W + W . For W + W ⊆ W just recall that every subset is closed under addition so any sum of
the form ⃗w 1 + ⃗w 2 is in W .
(d) In each vector space, the identity element with respect to subspace addition is the trivial subspace.
(e) Neither of left or right cancelation needs to hold. For an example, in R 3 take W 1 to be the
xy-plane, take W 2 to be the x-axis, and take W 3 to be the y-axis.
Two.III.4.44 (a) They are equal because for each, V is the direct sum if and only if each ⃗v ∈ V can
be written in a unique way as a sum ⃗v = ⃗w 1 + ⃗w 2 and ⃗v = ⃗w 2 + ⃗w 1 .
(b) They are equal because for each, V is the direct sum if and only if each ⃗v ∈ V can be written in
a unique way as a sum of a vector from each ⃗v = ( ⃗w 1 + ⃗w 2 ) + ⃗w 3 and ⃗v = ⃗w 1 + ( ⃗w 2 + ⃗w 3 ).
(c) Any vector in R 3 can be decomposed uniquely into the sum of a vector from each axis.
(d) No. For an example, in R 2 take W 1 to be the x-axis, take W 2 to be the y-axis, and take W 3 to
be the line y = x.
(e) In any vector space the trivial subspace acts as the identity element with respect to direct sum.
(f) In any vector space, only the trivial subspace has a direct-sum inverse (namely, itself). One way
to see this is that dimensions add, and so increase.
Topic: Fields
1 These checks are all routine; most consist only of remarking that property is so familiar that it does
not need to be proved.