Decide if the vector is in the column space of the matrix.
,
,
,
Answer
Yes;
we are asking if there are scalars and such that
which gives rise to a linear system
and Gauss' method produces and .
That is, there is indeed such a pair of scalars and so the vector
is indeed in the column space of the matrix.
No;
we are asking if there are scalars and
such that
and one way to proceed is to consider the resulting linear system
that is easily seen to have no solution.
Another way to proceed is to note
that any linear combination of the columns on the left
has a second component half as big as its first component,
but the vector on the right does not meet that criterion.
Yes; we can simply observe that the vector
is the first column minus the second.
Or, failing that, setting up the relationship among the columns
and considering the resulting linear system
gives the additional information (beyond that there is at least one
solution) that there are infinitely many solutions.
Parametizing gives and , and so taking to
be zero gives a particular solution of , , and
(which is, of course, the observation made at the start).
This exercise is recommended for all readers.
Problem 2
Decide if each vector lies in the range of the map from
to represented with respect to the standard bases by the matrix.
,
,
Answer
As described in the subsection, with respect to the standard bases,
representations are transparent,
and so, for instance, the first matrix describes this map.
So, for this first one, we are asking whether there are scalars such that
that is, whether the vector is in the column space of the matrix.
Yes.
We can get this conclusion by setting up the resulting linear system
and applying Gauss' method, as usual.
Another way to get it is to note by inspection of the equation of
columns that taking , and , and will do.
Still a third way to get this conclusion is to note that the rank
of the matrix is two, which equals the dimension of the
codomain, and so the map is onto— the range is all of and
in particular includes the given vector.
No; note that all of the columns in the matrix have a second
component that is twice the first, while the vector does not.
Alternatively, the column space of the matrix is
(which is the fact already noted, but was arrived at by calculation
rather than inspiration), and the given vector is not in this set.
This exercise is recommended for all readers.
Problem 3
Consider this matrix, representing a transformation of ,
and these bases for that space.
To what vector in the codomain
is the first member of mapped?
The second member?
Where is a general vector from the domain (a vector with
components and ) mapped?
That is, what transformation of is represented with
respect to by this matrix?
Answer
The first member of the basis
is mapped to
which is this member of the codomain.
The second member of the basis is mapped
to this member of the codomain.
Because the map that the matrix represents is the identity
map on the basis, it must be the identity on all members of the
domain.
We can come to the same conclusion in another way by considering
which is mapped to
which represents this member of .
Problem 4
What transformation of
is represented with respect to
and
by this matrix?
Answer
A general member of the domain, represented with respect to the
domain's basis as
is mapped to
and so the linear map represented by the matrix with respect to these
bases
is projection onto the first component.
This exercise is recommended for all readers.
Problem 5
Decide if is in the range of the map from to
represented with respect to and
by this matrix.
Answer
Denote the given basis of
by .
Then application of the linear map is represented by matrix-vector
addition.
Thus, the first vector in is mapped to the element
of represented with respect to by
and that element is .
The other two images of basis vectors are calculated similarly.
We can thus decide if is in the range of the map by
looking for scalars , , and such that
and obviously , , and suffice.
Thus it is in the range, and in fact it is the image of
this vector.
Problem 6
Example 2.8 gives a matrix that is
nonsingular, and is therefore associated with maps that are nonsingular.
Find the set of column vectors representing the members of
the nullspace of any map represented by this matrix.
Find the nullity of any such map.
Find the set of column vectors representing the members of
the rangespace of any map represented by this matrix.
Find the rank of any such map.
Check that rank plus nullity equals the dimension of the
domain.
Answer
Let the matrix be , and suppose that it represents
with respect to bases and .
Because has two columns, is two-dimensional.
Because has two rows, is two-dimensional.
The action of on a general member of the domain is this.
The only representation of the zero vector in the codomain
is
and so the set of representations of members of the nullspace is
this.
The representation map
and its inverse
are isomorphisms, and so preserve the dimension of subspaces.
The subspace of that is in the prior item is
one-dimensional.
Therefore, the image of that subspace under the inverse of the
representation map— the nullspace of ,
is also one-dimensional.
The set of representations of members of the rangespace is
this.
Of course, Theorem 2.3 gives that
the rank of the map equals the rank of the matrix, which is one.
Alternatively, the same argument that was used above for the
nullspace gives here that the dimension of the rangespace is one.
One plus one equals two.
This exercise is recommended for all readers.
Problem 7
Because
the rank of a matrix equals the rank of any map it represents, if
one matrix represents two different maps
(where )
then the dimension of the rangespace of
equals the dimension of the rangespace of .
Must these equal-dimensioned rangespaces actually be the same?
Answer
No, the rangespaces may differ. Example 2.2 shows this.
This exercise is recommended for all readers.
Problem 8
Let be an -dimensional space with bases and
.
Consider a map that sends, for ,
the column vector representing with
respect to to the column vector representing with
respect to .
Show that is a linear transformation of .
Answer
Recall that the represention map
is an isomorphism.
Thus, its inverse map
is also an isomorphism.
The desired transformation of is then this composition.
Because a composition of isomorphisms is also an isomorphism,
this map
is an isomorphism.
Problem 9
Example 2.2 shows that changing the pair of
bases can change the map that a matrix
represents, even though the domain and codomain remain the same.
Could the map ever not change?
Is there a matrix , vector spaces and , and
associated pairs of bases and (with
or or both)
such that the map represented
by with respect to equals the map represented
by with respect to ?
Answer
Yes. Consider
representing a map from to .
With respect to the standard bases
this matrix
represents the identity map.
With respect to
this matrix again represents the identity. In fact, as long as the starting and ending bases are equal— as long as — then the map represented by is the identity.
This exercise is recommended for all readers.
Problem 10
A square matrix is a diagonal matrix if it is all zeroes
except possibly for the entries on its upper-left to lower-right
diagonal— its entry, its entry, etc.
Show that a linear map is an isomorphism if there are bases such that,
with respect to those bases, the map is represented by a diagonal matrix
with no zeroes on the diagonal.
Describe geometrically the action on of
the map represented with respect to the standard
bases by this matrix.
Do the same for these.
Answer
The first map
stretches vectors by a factor of three in the
direction and by a factor of two in the direction.
The second map
projects vectors onto the axis.
The third
interchanges first and second components
(that is, it is a reflection about the line ).
The last
stretches vectors parallel to the axis, by an amount equal to three times their distance from that axis (this is a skew.)
Problem 12
The fact that for any linear map the rank plus the nullity
equals the dimension of the domain shows that a necessary
condition for the existence of a homomorphism between two spaces, onto
the second space, is that there be no gain in dimension.
That is, where is onto, the dimension of must
be less than or equal to the dimension of .
Show that this (strong) converse holds:
no gain in dimension implies that
there is a homomorphism and, further,
any matrix with the correct size and correct rank
represents such a map.
Are there bases for such that
this matrix
represents a map from to whose range is
the plane subspace of ?
Yes.
This is immediate from the prior item.
To give a specific example, we can
start with as the basis for the domain, and then
we require a basis for the codomain .
The matrix gives the action of the map as this
and there is no harm in finding a basis so that
that is, so that the map represented by with respect to
is projection down onto the plane.
The second condition gives that the third member of
is .
The first condition gives that the first member of plus twice
the second equals , and so this basis will do.
Problem 13
Let be an -dimensional space and suppose
that .
Fix a basis for and consider the map
given
by the dot product.
Show that this map is linear.
Show that for any linear map there is
an such that .
In the prior item we fixed the basis and varied the
to get all possible linear maps.
Can we get all possible linear maps by fixing an and
varying the basis?
Answer
Recall that the representation map
is linear (it is actually
an isomorphism, but we do not need that it is one-to-one or onto
here).
Considering the column vector to be a matrix
gives that the map from to that takes a column vector
to its dot product with is linear (this is a matrix-vector
product and so Theorem 2.1 applies).
Thus the map under consideration is linear because
it is the composistion of two linear maps.
Any linear map is represented by some
matrix
(the matrix has columns because is -dimensional and it
has only one row because is one-dimensional).
Then taking to be the column vector that is the transpose
of this matrix
has the desired action.
No.
If has any nonzero entries then
cannot be the zero map (and if is the zero vector
then can only be the zero map).
Problem 14
Let be vector spaces with bases .
Suppose that
is represented with respect to by the matrix .
Give the matrix representing the scalar multiple
(where ) with
respect to by expressing it in terms of .
Suppose that are represented with
respect to by and .
Give the matrix representing with
respect to by expressing it in terms of and .
Suppose that is represented
with respect to by and
is represented with respect to
by .
Give the matrix representing with
respect to by expressing it in terms of and .