# Linear Algebra/Any Matrix Represents a Linear Map/Solutions

## SolutionsEdit

*This exercise is recommended for all readers.*

- Problem 1

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

- ,
- ,
- ,

- Answer

- Yes;
we are asking if there are scalars and such that
- No;
we are asking if there are scalars and
such that
- 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

*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

*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
- The second member of the basis is mapped
- 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

- 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
- 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.

- 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 .

- 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

- Answer

- This is immediate from Theorem 2.3.
- 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

- 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
- 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 .

- Answer

See the following section.