- This exercise is recommended for all readers.
- Problem 1
Which of these subsets of the vector space of matrices
are subspaces under the inherited operations?
For each one that is a subspace, parametrize its description.
For each that is not, give a condition that fails.
-
-
-
-
- Answer
By Lemma 2.9, to see if each
subset of is a subspace, we need only
check if it is nonempty and closed.
- Yes, it is easily checked to be nonempty and closed.
This is a parametrization.
By the way, the parametrization also shows that it is a subspace,
it is given as the span of the two-matrix set,
and any span is a subspace.
- Yes; it is easily checked to be nonempty and closed.
Alternatively, as mentioned in the prior answer, the existence
of a parametrization shows that it is a subspace.
For the parametrization,
the condition can be rewritten as .
Then we have this.
- No.
It is not closed under addition.
For instance,
is not in the set.
(This set is also not closed under scalar multiplication,
for instance, it does not contain the zero matrix.)
- Yes.
- This exercise is recommended for all readers.
- Problem 2
Is this a subspace of :
?
If it is then parametrize its description.
- Answer
No, it is not closed. In particular, it is not closed under scalar multiplication because it does not contain the zero polynomial.
- This exercise is recommended for all readers.
- Problem 3
Decide if the vector lies in the span of the set, inside of the
space.
- ,
,
in
- ,
,
in
- ,
,
in
- Answer
- Yes, solving the linear system arising from
and .
- Yes; the linear system arising from
gives that .
- No; any combination of the two given matrices has a zero
in the upper right.
- This exercise is recommended for all readers.
- Problem 5
Which of these sets spans ?
That is, which of these sets has the property that any three-tall
vector can be expressed as a suitable linear combination of the
set's elements?
-
-
-
-
-
- Answer
- Yes, for any this equation
has the solution , , and
.
- Yes, the equation
gives rise to this
so that, given any , , and , we can compute that
, , and
.
- No.
In particular, the vector
cannot be gotten as a linear combination since the two given
vectors both have a third component of zero.
- Yes.
The equation
leads to this reduction.
We have infinitely many solutions.
We can, for example, set to be zero and solve for
, , and in terms of , , and by the usual
methods of back-substitution.
- No.
The equation
leads to this reduction.
This shows that not every three-tall vector can be so expressed.
Only the vectors satisfying the restriction that
are in the span.
(To see that any such vector is indeed expressible,
take and
to be zero and solve for and in terms of , , and
by back-substitution.)
- This exercise is recommended for all readers.
- Problem 6
Parametrize each subspace's description.
Then express each subspace as a span.
- The subset
of the three-wide row vectors
- This subset of
- This subset of
- The subset of
- The subset of of quadratic polynomials
such that
- Answer
-
The obvious choice for the set that spans is
.
-
One set that spans this space consists of those three matrices.
- The system
gives and .
So one description is this.
That shows that a set spanning this subspace consists of those
two matrices.
- The gives
.
So the subspace is the span of the set .
- The set
parametrized as
has the spanning set .
- This exercise is recommended for all readers.
- Problem 7
Find a set to span the given subspace of the given space.
(Hint. Parametrize each.)
- the -plane in
-
in
-
in
-
in
- The set in the space
- in
- Answer
Each answer given is only one out of many possible.
- We can parametrize in this way
giving this for a spanning set.
- As a way to solve it, is to express as to get this parametrization.
and get this as a spanning set.
- Here as a way to go, is to notice that and depend on and , but not each other, suggesting to choose and as free variables. Then re-expressing,
and to obtain this parametrization.
and for a possible spanning set we have.
- Again, as a way to do it, re-express and to get parametrization.
which gives this for a spanning set.
- As there are no restrictions, we can get a spaning set right away, skipping parametrization.
- For same reason as the previous subproblem, we can skip parametrization and go to a spaning set immediatedly.
- This exercise is recommended for all readers.
- Problem 9
Decide if each is a subspace of the vector space of real-valued
functions of one real variable.
- The
even functions
.
For example, two members of this set are
and .
- The odd
functions
.
Two members are and .
- Answer
Of course, the addition and scalar multiplication operations are the
ones inherited from the enclosing space.
- This is a subspace.
It is not empty as it contains at least the two example functions
given.
It is closed because if are even and
are scalars then we have this.
- This is also a subspace; the check is similar to
the prior one.
- Problem 11
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 where
the system has variables.
What about a non-homogeneous linear system; do its solutions form a
subspace (under the inherited operations)?
- Answer
No, such a set is not closed. For one thing, it does not contain the zero vector.
- Problem 13
Finish the proof of Lemma 2.9.
- Answer
Item (1) is checked in the text.
Item (2) has five conditions.
First, for closure, if and then
as
.
Second, because the operations in are inherited from ,
for and , the scalar product
in equals the product
in , and that equals
in , which equals
in .
The check for the third, fourth, and fifth conditions are similar to the second conditions's check just given.
- Problem 14
Show that each vector space has only one trivial subspace.
- Answer
An exercise in the prior subsection shows that every vector space has only one zero vector (that is, there is only one vector that is the additive identity element of the space). But a trivial space has only one element and that element must be this (unique) zero vector.
- This exercise is recommended for all readers.
- Problem 15
Show that for any subset of a vector space,
the span of the span equals the span
.
(Hint.
Members of are linear combinations of members of .
Members of are linear combinations of
linear combinations of members of .)
- Answer
As the hint suggests, the basic reason is the Linear Combination Lemma
from the first chapter.
For the full proof, we will show mutual containment between the two sets.
The first containment
is an instance of the more general, and obvious, fact that for any
subset of a vector space, .
For the other containment,
that ,
take vectors from , namely
, ...,
,
and note that any linear combination of those
is a linear combination of elements of
and so is in . That is, simply recall that a linear combination of linear combinations (of members of ) is a linear combination (again of members of ).
- Problem 16
All of the subspaces that we've seen use zero in their
description in some way.
For example, the subspace in Example 2.3 consists of
all the vectors from with a second component of zero.
In contrast,
the collection of vectors from 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 were that the three components add to one then it would
not be a subspace (again, it would fail to be closed).
This exercise shows that a reliance on zero is not strictly necessary.
Consider the set
under these operations.
- Show that it is not a subspace of .
(Hint. See Example 2.5).
- Show that it is a vector space.
Note that by the prior item,
Lemma 2.9 can not apply.
- Show that any subspace of must pass through the origin,
and so any subspace of must involve zero in its description.
Does the converse hold?
Does any subset of that contains the origin become a
subspace when given the inherited operations?
- Answer
- It is not a subspace because these are not the inherited
operations.
For one thing, in this space,
while this does not, of course, hold in .
- We can combine the argument showing closure under
addition with the argument showing closure under
scalar multiplication into one single argument
showing closure under linear combinations of two vectors.
If are in then
(note that the definition of addition in this space is that
the first
components combine as ,
so the first component of the last vector does not say
"").
Adding the three components of the last vector gives
.
Most of the other checks of the conditions are easy (although the
oddness of the operations keeps them from being routine).
Commutativity of addition goes like this.
Associativity of addition has
while
and they are equal.
The identity element with respect to this addition operation
works this way
and the additive inverse is similar.
The conditions on scalar multiplication are also easy.
For the first condition,
while
and the two are equal.
The second condition compares
with
and they are equal.
For the third condition,
while
and the two are equal.
For scalar multiplication by we have this.
Thus all the conditions on a vector space are met by these two
operations.
Remark.
A way to understand this vector space is to think of it as
the plane in
displaced away from the origin by along the -axis.
Then addition becomes: to add two members of this space,
(such that and )
move them back by to place them in and
add as usual,
and then move the result back out by along the -axis.
Scalar multiplication is similar.
- For the subspace to be closed under the inherited scalar
multiplication, where is a member of that subspace,
must also be a member.
The converse does not hold.
Here is a subset of that contains the origin
(this subset has only two elements) but is not a subspace.
- Problem 18
Is a space determined by its subspaces?
That is, if two vector spaces have the same subspaces, must the
two be equal?
- Answer
Yes; any space is a subspace of itself, so each space contains the other.
- Problem 19
- Give a set that is closed under scalar multiplication
but not addition.
- Give a set closed under addition but not scalar
multiplication.
- Give a set closed under neither.
- Answer
- The union of the -axis and the -axis
in is one.
- The set of integers, as a subset of , is one.
- The subset of is one,
where is any nonzero vector.
- Problem 20
Show that the span of a set of vectors does not depend on the order in
which the vectors are listed in that set.
- Answer
Because vector space addition is commutative, a reordering of summands leaves a linear combination unchanged.
- Problem 21
Which trivial subspace is the span of the empty set?
Is it
or some other subspace?
- Answer
We always consider that span in the context of an enclosing space.
- Problem 22
Show that if a vector is in the span of a set then adding that
vector to the set won't make the span any bigger.
Is that also "only if"?
- Answer
It is both "if" and "only if".
For "if",
let be a subset of a vector space and assume
satisfies
where
are scalars and
.
We must show that .
Containment one way,
is obvious.
For the other direction,
, note that if a
vector is in the set on the left then it has the form
where the 's are
scalars and the 's are in .
Rewrite that as
and note that
the result is a member of the span of .
The "only if" is clearly true— adding enlarges the span to include at least .
- This exercise is recommended for all readers.
- Problem 23
Subspaces are subsets and so we naturally consider how "is a subspace of"
interacts with the usual set operations.
- If are subspaces of a vector space, must
be a subspace?
Always? Sometimes? Never?
- Must be a subspace?
- If is a subspace, must
its complement be a subspace?
(Hint. Try some test subspaces from Example 2.19.)
- Answer
- Always.
Assume that are subspaces of .
Note that
their intersection is not empty as both contain the zero vector.
If and are
scalars then because
each vector is in and so a linear combination is in ,
and for the same reason.
Thus the intersection is closed.
Now Lemma 2.19 applies.
- Sometimes (more precisely, only if or
).
To see the answer is not "always", take to be ,
take to be the -axis, and to be the
-axis.
Note that
as the sum is in neither nor .
The answer is not "never" because if or
then clearly is a subspace.
To show that is a subspace only if one
subspace contains the other, we assume that
and and prove that
the union is not a subspace.
The assumption that is not a subset of means that
there is an with .
The other assumption gives a with
.
Consider .
Note that sum is not an element of or else
would be in , which it is not.
Similarly the sum is not an element of .
Hence the sum is not an element of , and so the union
is not a subspace.
- Never.
As is a subspace, it contains the zero vector, and therefore
the set that is 's complement does not.
Without the zero vector, the complement cannot be a vector space.
- This exercise is recommended for all readers.
- This exercise is recommended for all readers.
- Problem 26
Because "span of" is an operation on sets we naturally consider
how it interacts with the usual set operations.
- If are subsets of a vector space, is
?
Always? Sometimes? Never?
- If are subsets of a vector space, is
?
- If are subsets of a vector space, is
?
- Is the span of the complement equal to the complement of
the span?
- Answer
- Always;
if then a linear combination of elements of
is also a linear combination of elements of .
- Sometimes (more precisely, if and only if
or ).
The answer is not "always" as is shown by this example from
because of this.
The answer is not "never" because if either set contains the other
then equality is clear.
We can
characterize equality as happening only when either set contains
the other by assuming (implying the
existence of a vector with
)
and (giving a with
), noting
,
and showing that
.
- Sometimes.
Clearly
because any linear combination of vectors from
is a combination of vectors from and also a combination of
vectors from .
Containment the other way does not always hold.
For instance, in , take
so that is the -axis
but is the trivial subspace.
Characterizing exactly when equality holds is tough.
Clearly equality holds if either set contains the other, but that is
not "only if" by this example in .
- Never, as the span of the complement is a subspace, while
the complement of the span is not (it does not contain the zero
vector).
- Problem 28
Find a structure that is closed under linear combinations, and yet is not a vector space.
(Remark. This is a bit of a trick question.)
- Answer
For this to happen, one of the conditions giving the sensibleness of the
addition and scalar multiplication operations must be violated.
Consider with these operations.
The set is closed under these operations.
But it is not a vector space.