Linear Algebra, 2020a

Section I. Definition of Vector Space 105
̌ 2.30 Decide if each is a subspace of the vector space of real-valued functions of one
real variable.
(a) The even functions {f: R → R | f(−x) =f(x) for all x}. For example, two members
of this set are f 1 (x) =x 2 and f 2 (x) =cos(x).
(b) The odd functions {f: R → R | f(−x) =−f(x) for all x}. Two members are
f 3 (x) =x 3 and f 4 (x) =sin(x).
2.31 Example 2.16 says that for any vector ⃗v that is an element of a vector space V,
the set {r · ⃗v | r ∈ R} is a subspace of V. (This is simply the span of the singleton
set {⃗v}.) Must any such subspace be a proper subspace?
2.32 An example following the definition of a vector space shows that the solution
set of a homogeneous linear system is a vector space. In the terminology of this
subsection, it is a subspace of R n where the system has n variables. What about
a non-homogeneous linear system; do its solutions form a subspace (under the
inherited operations)?
2.33 [Cleary] Give an example of each or explain why it would be impossible to do
so.
(a) A nonempty subset of M 2×2 that is not a subspace.
(b) A set of two vectors in R 2 that does not span the space.
2.34 Example 2.19 shows that R 3 has infinitely many subspaces. Does every nontrivial
space have infinitely many subspaces?
2.35 Finish the proof of Lemma 2.9.
2.36 Show that each vector space has only one trivial subspace.
2.37 Show that for any subset S of a vector space, the span of the span equals the
span [[S]] = [S]. (Hint. Members of [S] are linear combinations of members of S.
Members of [[S]] are linear combinations of linear combinations of members of S.)
2.38 All of the subspaces that we’ve seen in some way use zero in their description.
For example, the subspace in Example 2.3 consists of all the vectors from R 2 with
a second component of zero. In contrast, the collection of vectors from R 2 with a
second component of one does not form a subspace (it is not closed under scalar
multiplication). Another example is Example 2.2, where the condition on the
vectors is that the three components add to zero. If the condition there were that
the three components add to one then it would not be a subspace (again, it would
fail to be closed). However, a reliance on zero is not strictly necessary. Consider
the set
⎛ ⎞
x
{ ⎝y⎠ | x + y + z = 1}
z
under these operations.
⎛ ⎛ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
x 1 + x 2 − 1 x rx − r + 1
⎞ ⎞
x 1 x 2
⎝y 1
⎠ + ⎝y 2
⎠ = ⎝
z 1 z 2
y 1 + y 2
z 1 + z 2
⎠
r ⎝ ⎠ = ⎝
(a) Show that it is not a subspace of R 3 .(Hint. See Example 2.7).
(b) Show that it is a vector space. Note that by the prior item, Lemma 2.9 can
not apply.
y
z
ry
rz
⎠