Linear Algebra/Any Matrix Represents a Linear Map/Solutions

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Solutions[edit]

This exercise is recommended for all readers.
Problem 1

Decide if the vector is in the column space of the matrix.

  1. ,
  2. ,
  3. ,
Answer
  1. 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.
  2. 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.
  3. 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.

  1. ,
  2. ,
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.

  1. 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.
  2. 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.

  1. To what vector in the codomain is the first member of mapped?
  2. The second member?
  3. 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
  1. The first member of the basis
    is mapped to
    which is this member of the codomain.
  2. The second member of the basis is mapped
    to this member of the codomain.
  3. 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.

  1. Find the set of column vectors representing the members of the nullspace of any map represented by this matrix.
  2. Find the nullity of any such map.
  3. Find the set of column vectors representing the members of the rangespace of any map represented by this matrix.
  4. Find the rank of any such map.
  5. Check that rank plus nullity equals the dimension of the domain.
Answer

Let the matrix be , and suppose that it rperesents 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.

  1. The only representation of the zero vector in the codomain is
    and so the set of representations of members of the nullspace is this.
  2. 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.
  3. The set of representations of members of the rangespace is this.
  4. 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.
  5. 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.

Answer

This is immediate from Corollary 2.6.

Problem 11

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 .

  1. 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.
  2. Are there bases for such that this matrix
    represents a map from to whose range is the plane subspace of ?
Answer
  1. This is immediate from Theorem 2.3.
  2. 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.

  1. Show that this map is linear.
  2. Show that for any linear map there is an such that .
  3. 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
  1. 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.
  2. 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.
  3. 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 .

  1. 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 .
  2. Suppose that are represented with respect to by and . Give the matrix representing with respect to by expressing it in terms of and .
  3. 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 .
Answer

See the following section.