Linear continuation – Serlo

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The principle of linear continuation states that every linear map is exactly determined by the images of the basis vectors. It provides an alternative way to characterize a linear map.


So far, we have mostly specified linear maps by saying where each vector of a vector space is mapped. Those are a lot of vectors, e.g. infinitely many for . Is there a way to specify the map with less vectors? Perhaps finitely many ones?

For every vector of our starting vector space we have to provide the information to which vector of the target vector space it should be mapped. Every such vector can be represented within a basis: If is a -vector space with basis and , then there are unique coefficients such that holds.

Now, consider a linear map into another -vector space . The basis vectors of then have images . Now, an important trick follows: we can use these images as building bricks to construct : by linearity (= additivity + homogeneity) of , we have that:

This is amazing: For any , the image can be reconstructed using . Than means the information how the (often infinitely) many are mapped by can be condensed in specifying only vectors! For a linear map , knowing three vectors already suffices to know the image of all infinitely many vectors.

The following theorem assures mathematically that this reconstruction works for any finite dimensional vector space:

Principle of linear continuation [Bearbeiten]

Theorem (Linear continuation)

Let be a field, and two -vector spaces and a basis of . Further, let be any vectors from . Then, there exists exactly one linear map with for all .

How to get to the proof?

First we have to find and define a suitable map . This map is basically given in the "motivation" section. But, is it really mathematically well-defined?

Once we have chosen a map, we should check that it is indeed linear and satisfies the requirement . Thus a suitable map exists.

Finally we have to show that the map with these properties is uniquely determined. To do this, we assume that there is another map with the same properties. Then we have to show that this map with is identical.


Let . Since form a basis of , there are clearly certain coefficients such that . Now we set

Because the coefficients are uniquely determined, the map is well-defined.

Further, it follows immediately that satisfies the requirement for every , because for every we have that:

Now we show that is linear. For this, let with and as well as . Then:

Aktuelles Ziel: additivity

Aktuelles Ziel: homogeneity

Finally we want to show that is uniquely determined by the properties of being linear and for every mapping the basis vector to . To do this, suppose there is a second map with exactly these two properties. We then have to show that . Let for this be arbitrarily. Then:

We have shown that and take the same value for every vector . So both maps are the same and we are done with the proof of uniqueness.


In the premise on the principle of linear continuation, a basis of occurs. That is, must be finite-dimensional. However, might be infinite-dimensional.

Actually, the statement also holds for being infinite-dimensional. The proof works similar to the one above.


Example 1[Bearbeiten]


We consider the -vector space with the basis where and . It can easily be seen that this is basis. (you may now think a moment about why) Let and be two vectors. By the theorem above, there hence exists a unique linear map given by and . What is the image of for a general vector ?

We proceed as in the theorem on the principle of linear continuation: let be a vector in . First, we represent as a linear combination of basis vectors . So we determine such that . They are given by:

So we need to solve the system of equations

for and . Subtracting the second equation from the first, we obtain . To get , we substitute this result into the second equation:

If we resolve for , we get . Consequently, the linear combination we are looking for is .

By the proof of the theorem above, we know how acts on :

So the has the general image

Example 2[Bearbeiten]


We consider the map with .

As basis of we choose . Then

So we could also specify the linear map by requiring that it maps to and to . This only requires fixing two vectors.

Example 3[Bearbeiten]


Is there a linear map with and ?

Assuming there is such a map, then we would have:

This is a contradiction. Hence such a linear map cannot exist.

Question: A linear map should be specified by exactly 2 vectors and we have 2 vectors. Then why is there a contradiction, anyway?

The vectors and are linearly dependent, but the function values we assigned to them are not multiples of each other. This is where the contradiction comes from. However, this does not contradict the theorem of linear continuation. Because there the function values are given for a basis.

Properties of the linear continuation[Bearbeiten]

In the following, and are two -vector spaces, is a basis of and are vectors in . Let be a linear map with for all . Because of the above theorem such a linear map exists and it is unique.

Theorem (Properties of the linear continuation)

In particular we have that is surjective if and only if is a generator of .

How to get to the proof?

We establish the first statement by showing equality of sets. That is, we prove that and hold.

For the first inclusion we consider an element . So there exists a such that holds. We can write this as a linear combination of the basic elements of . Together with the linearity of it can then be shown that we may also write as a linear combination of .

For the other inclusion "" we now consider a . Then we can write as a linear combination of . Since holds, is representable as a linear combination of . And since is linear, we can now show that lies in .

Thus we can easily prove that is surjective exactly if is a generator of using the following statements:

  • is surjective if and only if holds.
  • is a generator of if and only if holds.
  • (our already proved statement).



: Let . Then there is a with . Since is a basis of , there are coefficients such that . Now we have:

i.e., we managed to write as a linear combination of , such that .

: Let , then there are coefficients such that . By definition of we have:

In particular, this implies the second statement:

Beweisschritt: is surjective, if and only if is a generator of .

If is surjective, then:

(according to the statement above).

Therefore, is a generator of .

Conversely, if is a generator, then we have that , and is surjective.

Theorem (Injective maps send bases to linearly independent vectors)

is injective, if and only if is linearly independent.

How to get to the proof? (Injective maps send bases to linearly independent vectors)

For equivalence, we need to show two implications. In the proof of "" we want to show that the vectors are linearly independent if is injective. We assume that is injective and consider the zero vector as a linear combination of , i.e. with . We now want to prove that all coefficients vanish. If we replace in our linear combination with the respective and use the linearity of , we get


We know that because is linear. So


Using injectivity of , it follows that . Since the basis is linearly independent, we have for all .

In the proof of "", our goal is to show that is injective if are linearly independent. To do this, we consider two vectors with . We want to show that . Since forms a basis of , we can represent and as a linear combination of them:

and with

To prove , it is enough to show that for holds. With and the linearity of we get

Because of we get the representation

Because of the linear independence of their linear combinations are unique and one has for all .

Proof (Injective maps send bases to linearly independent vectors)

We need to establish two directions.

Proof step: If is injective, then the are linearly independent.

Let and let

For any linear mapping, it is also true that . Since is injective, we have

Further, since is a basis of :

Thus, the are linearly independent.

Proof step: If the are linearly independent, then is injective.

Let with . Then, there are some with and . We have that:

If are linearly independent, the representation is unique, so . Thus is injective.

Theorem (Bijective maps send bases to bases)

is bijective if and only if is a basis of .

How to get to the proof? (Bijective maps send bases to bases)

We simply combine the statements of the last two theorems.

Proof (Bijective maps send bases to bases)

Proof step: If is bijective, then is a basis of .

Since is bijective, also is injective and surjective. Therefore, according to the last two theorems, form a linearly independent generator. This generator is always a basis.

Proof step: If is a basis of , then is bijective.

Suppose is a basis - so in particular it is linearly independent and a generator. Then, we have by the last two theorems that is injective and surjective - so in particular bijective.


Exercise (Linear maps under some conditions)

Let and . Is there an -linear map that satisfies ?

How to get to the proof? (Linear maps under some conditions)

First you should check if the vectors are linearly independent. If this is the case, is a basis of because of . Using the principle of linear continuation, the existence of such a linear map would follow . Let thus :

But then also and so must be fulfilled. However, this equation has not only the "trivial" solution . In fact, the upper equation is satisfied for . Thus, one obtains

For such a map , the relation would then have to hold, which is a contradiction to

Solution (Linear maps under some conditions)

Let us first assume that such a linear map would exist. By the following calculation

we see that should hold. But this is a contradiction to the other conditions, because those would imply

So there is no such .