Linear continuation – Serlo

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.

Motivation

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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

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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.

Proof

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.

Hint

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.

Examples

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Example 1

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Example

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

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Example

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

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Example

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

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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).

Proof

Beweisschritt:  

 : 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.

Exercises

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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  .