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Mathematical Expression Editor
Additional Exercises for Chapter 10: Abstract Vector Spaces
Suppose you have and the operation is as
follows:
Scalar multiplication is defined in the usual way. Is this a vector space? Explain why
or why not.
Suppose you have and the operation is defined as follows.
Scalar multiplication is defined in the usual way. Is this a vector space? Explain why
or why not.
Suppose you have and scalar multiplication is defined as while vector
addition is defined as usual. Is this a vector space? Explain why or why
not.
Suppose you have and the operation is defined as follows.
Scalar multiplication is same as usual. Is this a vector space? Explain why or why
not.
Consider all the functions defined on a non empty set which have values in . Is this a vector
space? Explain. The operations are defined as follows. Here signify functions and is a scalar.
Consider the set of symmetric matrices. That is, In other words, . Show that this
set of symmetric matrices is a vector space and a subspace of the vector space of
matrices.
Consider the set of all vectors in such that Let the vector space operations be the
usual ones. Is this a vector space? Is it a subspace of ?
Consider the vectors in such that . Is this a subspace of Is it a vector space? The
addition and scalar multiplication are the usual operations.
Define the operation of vector addition on by Let scalar multiplication be the usual
operation. Is this a vector space with these operations? Explain.
Let the vectors be real numbers. Define vector space operations in the usual way.
That is means to add the two numbers and means to multiply them. Is with these
operations a vector space? Explain.
Let the scalars be the rational numbers and let the vectors be real numbers which
are the form for rational numbers. Show that with the usual operations, this is a
vector space.
Let be the set of all polynomials of degree 2 or less. That is, these are of the form . Addition is
defined as
and scalar multiplication is defined as
Show that, with this definition of the vector space operations that is a vector
space. Now let denote those polynomials such that . Is a subspace of
Explain.
Let be subspaces of a vector space and consider defined as the set of all where
and . Show that is a subspace of .
Let be subspaces of a vector space . Then consists of all vectors which are in both
and . Show that is a subspace of .
Let be subspaces of a vector space Then consists of all vectors which are in either
or . Show that is not necessarily a subspace of by giving an example where fails to
be a subspace.
Let consist of the real valued functions which are defined on an interval For is the
name of the function which satisfies . For a real number, . Show this is
a vector space.
The axioms of a vector space all hold because they hold
for a vector space. The only thing left to verify is the assertions about the
things which are supposed to exist. would be the zero function which sends
everything to . This is an additive identity. Now if is a function, . Then
Hence . For each let and if Then these vectors are obviously linearly
independent.
Let the vectors be polynomials of degree no more than 3. Show that with the usual
definitions of scalar multiplication and addition wherein, for a polynomial, and for
polynomials this is a vector space.
This is just a subspace of the vector space of
functions because it is closed with respect to vector addition and scalar
multiplication. Hence this is a vector space.
Let . Is in
Let . Is in ?
Let . Is in
Show that the spanning set in Question prb:10.26 is a spanning set for , the vector space of all
matrices.
Consider the vector space of polynomials of degree at most . Determine whether the
following is a basis for .
Hint: There is a isomorphism from to . It is defined as
follows:
Then extend linearly. Thus
It follows that if
is a basis for then the polynomials will be a basis for because they will be
independent. Recall that an isomorphism takes a linearly independent set to a
linearly independent set. Also, since is an isomorphism, it preserves all linear
relations.
Find a basis in for the subspace
If the above three vectors do not yield a basis, exhibit one of them as a linear
combination of the others.
This is the situation in which you have a spanning set
and you want to cut it down to form a linearly independent set which is also a
spanning set. Use the same isomorphism above. Since is an isomorphism, it preserves
all linear relations so if such can be found in the same linear relations will be present
in .
Find a basis in for the subspace
If the above three vectors do not yield a basis, exhibit one of them as a linear
combination of the others.
Find a basis in for the subspace
If the above three vectors do not yield a basis, exhibit one of them as a linear
combination of the others.
Find a basis in for the subspace
If the above three vectors do not yield a basis, exhibit one of them as a linear
combination of the others.
Find a basis in for the subspace
If the above three vectors do not yield a basis, exhibit one of them as a linear
combination of the others.
Find a basis in for the subspace
If the above three vectors do not yield a basis, exhibit one of them as a linear
combination of the others.
Find a basis in for the subspace
If the above three vectors do not yield a basis, exhibit one of them as a linear
combination of the others.
Here are some vectors.
If these are linearly independent, extend to a basis for all of .
Here are some vectors.
If these are linearly independent, extend to a basis for all of .
Here are some vectors.
If these are linearly independent, extend to a basis for all of .
Here are some vectors.
If these are linearly independent, extend to a basis for all of .
Determine if the following set is linearly independent. If it is linearly dependent,
write one vector as a linear combination of the other vectors in the set.
Determine if the following set is linearly independent. If it is linearly dependent,
write one vector as a linear combination of the other vectors in the set.
Determine if the following set is linearly independent. If it is linearly dependent,
write one vector as a linear combination of the other vectors in the set.
Determine if the following set is linearly independent. If it is linearly dependent,
write one vector as a linear combination of the other vectors in the set.
If you have vectors in and the vectors are linearly independent, can it always be
concluded they span
Yes. If not, there would exist a vector not in the span. But
then you could add in this vector and obtain a linearly independent set of vectors
with more vectors than a basis.
If you have vectors in is it possible they are linearly independent? Explain.
No.
They can’t be.
Let be the polynomials of degree no more than 3. Determine which of the following
are bases for this vector space.
(a)
(b)
(a)
(b)
Suppose Then combine the terms according to power of Is there a non
zero solution to the system
Solution is: Therefore, these are linearly independent.
Define by .
Find .
Answer:
Is a linear transformation? If so, prove it. If not, give a counterexample.
Let
Let and .
Suppose is a linear transformation such that
Verify that vectors and are in by expressing each as a linear combination of and
.
Show that is in by expressing it as a linear combination of and .
Find and express it as a linear combination of and .
In the context of the above problem, consider polynomials
Show that this collection of polynomials is linearly independent on an interval if and
only if
is an invertible matrix.
Let denote the of these polynomials. Suppose
Then collecting terms according to the exponent of you need to have
The matrix of coefficients is just the transpose of the above matrix. There exists
a non trivial solution if and only if the determinant of this matrix equals
0.
Let Is a subspace of ?
This is not a subspace. is in it, but is not.
Let Is a subspace of ?
This is not a subspace.
Let be a subset of given by In words, is the set of all symmetric matrices. Is a
subspace of ?
Let be a subset of given by Is a subspace of ?
Let be a subset of given by Is a subspace of ?
Let be a subset of given by Is a subspace of ?
Let be a linear transformation such that Find .
By linearity we have , , and
.
Thus.
Consider the following functions Explain why each of these functions is not
linear.
(a)
(b)
(c)
(d)
Suppose is a linear transformation such that
Find the matrix of . That is find such that .
Suppose is a linear transformation such that
Find the matrix of . That is find such that .
Consider the following functions . Show that each is a linear transformation and
determine for each the matrix such that .
(a)
(b)
(c)
(d)
Let and be subspaces of and respectively and let be a linear transformation.
Suppose that is linearly independent. Show that it must be the case that is also
linearly independent.
If , then using linearity properties of we get Since we assume
that is linearly independent, we must have all , and therefore we conclude that is
also linearly independent.
Let
Let where is the matrix
Give a basis for .
Let
Let where is the matrix
Find a basis for . In this case, the original vectors do not form an independent
set.
Since the third vector is a linear combinations of the first two, then the image of the
third vector will also be a linear combinations of the image of the first two. However
the image of the first two vectors are linearly independent (check!), and hence form a
basis of the image.
Thus a basis for is:
If is linearly independent and is a one to one linear transformation, show that is
also linearly independent. Give an example which shows that if is only linear, it can
happen that, although is linearly independent, is not. In fact, show that it can
happen that each of the equals 0.
Let and be subspaces of and respectively and let be a linear transformation.
Show that if is onto and if is a basis for then .
Define as follows.
Find a basis for . Also find a basis for
Define as follows.
where on the right, it is just matrix multiplication of the vector which is meant.
Explain why is an isomorphism of to .
Suppose is a linear transformation given by
where is a matrix. Show that is an isomorphism if and only if is invertible.
Suppose is a linear transformation given by
where is an matrix. Show that is never an isomorphism if . In particular, show that
if cannot be onto and if then cannot be one to one.
Define as follows.
where on the right, it is just matrix multiplication of the vector which is meant.
Show that is one to one. Next let Show that is an isomorphism of and
.
In the above problem, find a matrix such that the restriction of to gives the same
result as on . Hint: You might let be such that
now find another vector such that
is a basis. You could pick
for example. Explain why this one works or one of your choice works. Then you could
define to equal some vector in Explain why there will be more than one such matrix
which will deliver the inverse isomorphism on .
Now let equal and let be a linear transformation where
and
Explain why is an isomorphism. Determine a matrix which, when multiplied on the
left gives the same result as on and a matrix which delivers on .
You need to have
Now enlarge to obtain a basis for . You could add in for example, and then pick
another vector in and let equal this other vector. Then you would have
This would involve picking for the new vector in the vector Then you could find .
You can do something similar to find a matrix for denoted as .
Let and let
Find a basis of consisting of vectors in .
In this case and a basis for consisting of vectors in can be obtained by taking any
(nonzero) vector from .
Let be a linear transformation given by Find a basis for and .
A basis for is and a basis for is . There are many other possibilities for the specific bases, but in this case and
.
Let be a linear transformation given by Find a basis for and .
In this case and (pick any basis of ).
Let and let
Extend this basis of to a basis of .
There are many possible such extensions, one is (how do we know?):
Let be a linear transformation given by What is ?
We can easily see that , and thus .
Consider the following functions which map to .
(a)
multiplies the component of by a nonzero number
(b)
replaces the component of with times the component added to the
component.
(c)
switches the and components.
Show these functions are linear transformations and describe their matrices such
that .
(a)
The matrix of is the elementary matrix which multiplies the diagonal
entry of the identity matrix by .
(b)
The matrix of is the elementary matrix which takes times the row and
adds to the row.
(c)
The matrix of is the elementary matrix which switches the and the rows
where the two components are in the and positions.
Suppose is a linear transformation such that
Find the matrix of . That is find such that .
Suppose is a linear transformation such that
Find the matrix of . That is find such that .
Suppose is a linear transformation such that
Find the matrix of . That is find such that .
Suppose is a linear transformation such that
Find the matrix of . That is find such that .
Suppose is a linear transformation such that
Find the matrix of . That is find such that .
Consider the following functions . Show that each is a linear transformation and
determine for each the matrix such that .
(a)
(b)
(c)
(d)
Consider the following functions Explain why each of these functions is not
linear.
(a)
(b)
(c)
(d)
Find the matrix for where
Recall that and so the desired matrix has column
equal to Therefore, the matrix desired is
Find the matrix for where
Find the matrix for where
Let be a basis of and let be a vector in . Find .
Let be a basis of and let be a vector in . Find .
.
Let be a linear transformation defined by .
Consider the two bases and
Find the matrix of with respect to the ordered bases and .
Challenge Exercises
Denote by the set of real valued sequences. For two of these, define their sum to be
given by
and define scalar multiplication by
Is this a special case of Problem prb:10.5? Is this a vector space?
Let be the set of ordered pairs of complex numbers. Define addition and scalar multiplication in
the usual way.
Here the scalars are from . Show this is a vector space.
Let be the set of functions defined on a nonempty set which have values in a vector
space Is this a vector space? Explain.
Consider functions defined on having values in . Explain how, if is the set of all
such functions, can be considered as .
Let be the component of a vector . Thus a
typical element in is .
Let the field of scalars be , the rational numbers and let the vectors be of the form
where are rational numbers. Show that this collection of vectors is a vector space
with field of scalars and give a basis for this vector space.
When you add two of
these you get one and when you multiply one of these by a scalar, you get another
one. A basis is . By definition, the span of these gives the collection of vectors. Are
they independent? Say where are rational numbers. If then which can’t happen
since is rational. If then which again can’t happen because on the left is a
rational number and on the right is an irrational. Hence both and so this is a
basis.
Suppose is a finite dimensional vector space. Based on the exchange
theorem above, it was shown that any two bases have the same number
of vectors in them. Give a different proof of this fact using the earlier
material in the book.
Suppose and are two bases with Then define
by
Consider the linear transformation, Argue it is a one to one and onto mapping from
to Now consider a matrix of this linear transformation and its reduced row echelon
form.
This is obvious because when you add two of these you get one and when you
multiply one of these by a scalar, you get another one. A basis is . By definition, the
span of these gives the collection of vectors. Are they independent? Say where are
rational numbers. If then which can’t happen since is rational. If then which again
can’t happen because on the left is a rational number and on the right is an
irrational. Hence both and so this is a basis.
If and are polynomials in , define
Show that this is an inner product on .
Let denote the space of all functions from the set to with pointwise addition and
scalar multiplication. Show that is an inner product on if .