# Linear Algebra/Rangespace and Nullspace/Solutions

## Solutions[edit]

*This exercise is recommended for all readers.*

- Problem 1

Let be given by . Which of these are in the nullspace? Which are in the rangespace?

- Answer

First, to answer whether a polynomial is in the nullspace, we have to consider it as a member of the domain . To answer whether it is in the rangespace, we consider it as a member of the codomain . That is, for , the question of whether it is in the rangespace is sensible but the question of whether it is in the nullspace is not because it is not even in the domain.

- The polynomial is not in the nullspace because is not the zero polynomial in . The polynomial is in the rangespace because is mapped by to .
- The answer to both questions is, "Yes, because ." The polynomial is in the nullspace because it is mapped by to the zero polynomial in . The polynomial is in the rangespace because it is the image, under , of .
- The polynomial is not in the nullspace because is not the zero polynomial in . The polynomial is not in the rangespace because there is no member of the domain that when multiplied by gives the constant polynomial .
- The polynomial is not in the nullspace because . The polynomial is in the rangespace because it is the image of .
- The polynomial is not in the nullspace because . The polynomial is not in the rangespace because of the constant term.

*This exercise is recommended for all readers.*

- Problem 2

Find the nullspace, nullity, rangespace, and rank of each map.

- given by
- given by
- given by
- the zero map

- Answer

- The nullspace is
- The nullspace is this.
- The nullspace is
- The nullspace is all of so the nullity is three. The rangespace is the trivial subspace of so the rank is zero.

*This exercise is recommended for all readers.*

- Problem 3

Find the nullity of each map.

- of rank five
- of rank one
- , an onto map
- , onto

- Answer

For each, use the result that the rank plus the nullity equals the dimension of the domain.

*This exercise is recommended for all readers.*

- Problem 4

What is the nullspace of the differentiation transformation ? What is the nullspace of the second derivative, as a transformation of ? The -th derivative?

- Answer

Because

we have this.

In the same way,

for .

- Problem 5

Example 2.7 restates the first condition in the definition of homomorphism as "the shadow of a sum is the sum of the shadows". Restate the second condition in the same style.

- Answer

The shadow of a scalar multiple is the scalar multiple of the shadow.

- Problem 6

For the homomorphism given by find these.

- Answer

- Setting gives and and , so the nullspace is .
- Setting gives that , and , and . Taking as a parameter, and renaming it gives this set description .
- This set is empty because the range of includes only those polynomials with a term.

*This exercise is recommended for all readers.*

- Problem 7

For the map given by

sketch these inverse image sets: , , and .

- Answer

All inverse images are lines with slope .

*This exercise is recommended for all readers.*

- Problem 8

Each of these transformations of is nonsingular. Find the inverse function of each.

- Answer

These are the inverses.

For instance, for the second one, the map given in the question sends and then the inverse above sends . So this map is actually self-inverse.

- Problem 9

Describe the nullspace and rangespace of a transformation given by .

- Answer

For any vector space , the nullspace

is trivial, while the rangespace

is all of , because every vector is twice some other vector, specifically, it is twice . (Thus, this transformation is actually an automorphism.)

- Problem 10

List all pairs that are possible for linear maps from to .

- Answer

Because the rank plus the nullity equals the dimension of the domain (here, five), and the rank is at most three, the possible pairs are: , , , and . Coming up with linear maps that show that each pair is indeed possible is easy.

- Problem 11

Does the differentiation map have an inverse?

- Answer

No (unless is trivial), because the two polynomials and have the same derivative; a map must be one-to-one to have an inverse.

*This exercise is recommended for all readers.*

- Problem 12

Find the nullity of the map given by

- Answer

The nullspace is this.

Thus the nullity is .

- Problem 13

- Prove that a homomorphism is onto if and only if its rank equals the dimension of its codomain.
- Conclude that a homomorphism between vector spaces with the same dimension is one-to-one if and only if it is onto.

- Answer

- One direction is obvious: if the homomorphism is onto then its range is the codomain and so its rank equals the dimension of its codomain. For the other direction assume that the map's rank equals the dimension of the codomain. Then the map's range is a subspace of the codomain, and has dimension equal to the dimension of the codomain. Therefore, the map's range must equal the codomain, and the map is onto. (The "therefore" is because there is a linearly independent subset of the range that is of size equal to the dimension of the codomain, but any such linearly independent subset of the codomain must be a basis for the codomain, and so the range equals the codomain.)
- By Theorem 2.21, a homomorphism is one-to-one if and only if its nullity is zero. Because rank plus nullity equals the dimension of the domain, it follows that a homomorphism is one-to-one if and only if its rank equals the dimension of its domain. But this domain and codomain have the same dimension, so the map is one-to-one if and only if it is onto.

- Problem 14

Show that a linear map is nonsingular if and only if it preserves linear independence.

- Answer

We are proving that is nonsingular if and only if for every linearly independent subset of the subset of is linearly independent.

One half is easy— by Theorem 2.21, if is singular then its nullspace is nontrivial (contains more than just the zero vector). So, where is in that nullspace, the singleton set is independent while its image is not.

For the other half, assume that is nonsingular and so by Theorem 2.21 has a trivial nullspace. Then for any , the relation

implies the relation . Hence, if a subset of is independent then so is its image in .

*Remark.* The statement is that a linear map is nonsingular if and only if it preserves independence for *all* sets (that is, if a set is independent then its image is also independent). A singular map may well preserve some independent sets. An example is this singular map from to .

Linear independence is preserved for this set

and (in a somewhat more tricky example) also for this set

(recall that in a set, repeated elements do not appear twice). However, there are sets whose independence is not preserved under this map;

and so not all sets have independence preserved.

- Problem 15

Corollary 2.17 says that for there to be an onto homomorphism from a vector space to a vector space , it is necessary that the dimension of be less than or equal to the dimension of . Prove that this condition is also sufficient; use Theorem 1.9 to show that if the dimension of is less than or equal to the dimension of , then there is a homomorphism from to that is onto.

- Answer

(We use the notation from Theorem 1.9.) Fix a basis for and a basis for . If the dimension of is less than or equal to the dimension of then the theorem gives a linear map from to determined in this way.

We need only to verify that this map is onto.

Any member of can be written as a linear combination of basis elements . This vector is the image, under the map described above, of . Thus the map is onto.

- Problem 16

Let be a homomorphism, but not the zero homomorphism. Prove that if is a basis for the nullspace and if is not in the nullspace then is a basis for the entire domain .

- Answer

By assumption, is not the zero map and so a vector exists that is not in the nullspace. Note that is a basis for , because it is a size one linearly independent subset of . Consequently is onto, as for any we have for some scalar , and so .

Thus the rank of is one. Because the nullity is given as , the dimension of the domain of (the vector space ) is . We can finish by showing is linearly independent, as it is a size subset of a dimension space. Because is linearly independent we need only show that is not a linear combination of the other vectors. But would give and applying to both sides would give a contradiction.

*This exercise is recommended for all readers.*

- Problem 17

Recall that the nullspace is a subset of the domain and the rangespace is a subset of the codomain. Are they necessarily distinct? Is there a homomorphism that has a nontrivial intersection of its nullspace and its rangespace?

- Answer

Yes. For the transformation of given by

we have this.

*Remark.* We will see more of this in the fifth chapter.

- Problem 18

Prove that the image of a span equals the span of the images. That is, where is linear, prove that if is a subset of then equals . This generalizes Lemma 2.1 since it shows that if is any subspace of then its image is a subspace of , because the span of the set is .

- Answer

This is a simple calculation.

*This exercise is recommended for all readers.*

- Problem 19

- Prove that for any linear map and any , the set has the form
**coset**of and is denoted . - Consider the map given by
- Conclude from the prior two items that for any linear system of the form
- Show that this map is linear
- Show that the -th derivative map is a linear transformation of
for each . Prove that this map is a linear transformation of that space

- Answer

- We will show that the two sets are equal by mutual inclusion. For the direction, just note that equals , and so any member of the first set is a member of the second. For the direction, consider . Because is linear, implies that . We can write as , and then we have that , as desired, because .
- This check is routine.
- This is immediate.
- For the linearity check, briefly, where are scalars and have components and , we have this.
- Each power of the derivative is
linear because of the rules

- Problem 20

Prove that for any transformation that is rank one, the map given by composing the operator with itself satisfies for some real number .

- Answer

Because the rank of is one, the rangespace of is a one-dimensional set. Taking as a basis (for some appropriate ), we have that for every , the image is a multiple of this basis vector— associated with each there is a scalar such that . Apply to both sides of that equation and take to be

to get the desired conclusion.

- Problem 21

Show that for any space of dimension , the **dual space**

is isomorphic to . It is often denoted . Conclude that .

- Answer

Fix a basis for . We shall prove that this map

is an isomorphism from to .

To see that is one-to-one, assume that and are members of such that . Then

and consequently, , etc. But a homomorphism is determined by its action on a basis, so , and therefore is one-to-one.

To see that is onto, consider

for . This function from to

is easily seen to be linear, and to be mapped by to the given vector in , so is onto.

The map also preserves structure: where

we have

so .

- Problem 22

Show that any linear map is the sum of maps of rank one.

- Answer

Let be linear and fix a basis for . Consider these maps from to

for any . Clearly is the sum of the 's. We need only check that each is linear: where we have .

- Problem 23

Is "is homomorphic to" an equivalence relation? (*Hint:* the difficulty is to decide on an appropriate meaning for the quoted phrase.)

- Answer

Either yes (trivially) or no (nearly trivially).

If "is homomorphic to" is taken to mean there is a homomorphism from into (but not necessarily onto) , then every space is homomorphic to every other space as a zero map always exists.

If "is homomorphic to" is taken to mean there is an onto homomorphism from to then the relation is not an equivalence. For instance, there is an onto homomorphism from to (projection is one) but no homomorphism from onto by Corollary 2.17, so the relation is not reflexive.^{[1]}

- Problem 24

Show that the rangespaces and nullspaces of powers of linear maps form descending

and ascending

chains. Also show that if is such that then all following rangespaces are equal: . Similarly, if then .

- Answer

That they form the chains is obvious. For the rest, we show here that implies that . Induction then applies.

Assume that . Then is the same map, with the same domain, as . Thus it has the same range: .

- ↑ More information on equivalence relations is in the appendix.