Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 15
- Linear subspaces and dual space
Linear subspaces of some -vector space are in direct relation with linear subspaces of the dual space .
These orthogonal spaces are again linear subspaces, see Exercise 15.4 . Whether a linear form belongs to can be checked on a generating system of , see Exercise 15.5 . In the second semester, when we have inner products at hand, there will also be an orthogonal space for in itself.
We consider the linear subspace
The orthogonal space of consists of all linear forms
with and . Because a linear form is described by a row matrix with respect to the standard basis, we are dealing with the solution set of the linear system
The solution space is
Let be a finite-dimensional -vector space with a basis , , and the corresponding dual basis , . Let
for a subset . Then
Let be a -vector space and be a linear subspace of the dual space of . Then
is called the orthogonal space
of .Let a homogeneous linear system
over a field be given. We consider the -th equation as a kernel condition for the linear form
Let
denote the linear subspace of the dual space generated by these linear forms. Then is the solution space of this linear system.
In general, we have the relation
In particular,
Let be a -vector space with dual space
. Then the following statements hold.- For linear subspaces
,
we have
- For linear subspaces
,
we have
- Let be
finite-dimensional.
Then
and
- Let be
finite-dimensional.
Then
and
(1) and (2) are clear. (3). The inclusion
is also clear. Let , . Then we can choose a basis of and extend it to a basis of . The linear form vanishes on , therefore, it belongs to . Because of
we have .
(4). Let be a basis of , and let
denote the mapping where these linear forms are the components. Here, we have
Assume that the mapping is not surjective. Then is a strict linear subspace of and its dimension is at most . Let be a -dimensional linear subspace with
Due to Lemma 14.5 , there is a linear form
whose kernel is exactly . Write . Then
contradicting the linear independence of the . Moreover, is surjective and the statement follows from Theorem 11.5 .
Let be a finite-dimensional -vector space and a linear subspace. Then there exist linear forms on such that
is the kernel of a linear mapping and every linear subspace of is the solution space of a
linear system.Proof
- The dual mapping
Let denote a field, let and denote -vector spaces, and let
denote a -linear mapping. Then the mapping
This assignment arises from just considering the composition
The dual mapping is a special case of the situation described in Lemma 13.8 (1). In particular, the dual mapping is again linear.
Let denote vector spaces over a field and let
and
be
linear mappings. Then the following hold.- For the
dual mapping,
we have
- For the identity on , we have
- If is surjective then is injective.
- If is injective then is surjective.
(1). For , we have
(2) follows directly from .
(3). Let and
Because of the surjectivity of , there exist for every a such that . Therefore
and is itself the zero mapping. Due to Lemma 11.4 , injective.
(4). The condition means that we may consider as a linear subspace. Because of Lemma 9.12 , we can write
with another -linear subspace . A linear form
can always be extended to a linear form
for example, by defining on to be the zero form. This means the surjectivity.
Let be a field and let and be vector spaces over , where is finite-dimensional. Let
denote a linear mapping. Then there exist vectors and linear forms on such that[1]
Let be a basis of and the corresponding dual basis. We set
Then, for every vector , we have
where the last equation rests on Lemma 14.12 .
Let be a field, let denote an -dimensional -vector space with a basis , and let be an -dimensional vector space with a basis . Let and be the corresponding dual bases. Let
be a linear mapping, and suppose that it is described by the -matrix
with respect to the given bases. Then the dual mapping
is described by the transposed matrix with respect to the dual bases of
and .The claim means the equality[2]
in . This can be checked on the basis , . On one hand, we have
on the other hand, we have
- The bidual
Let be a field, and let be a -vector space. Then the dual space of the dual space , that is,
Let be a field, and let be a -vector space. Then there exists a natural injective linear mapping
If has finite dimension, then is an
isomorphism.Let be fixed. First of all, we show that is a linear form on the dual space . Obviously, is a mapping from to . The additivity follows from
where we have used the definition of the addition on the dual space. The compatibility with the scalar multiplication follows similarly from
In order to prove the additivity of , let be given. We have to show the equality
This is an equality inside of , in particular, it is an equality of mappings. So let be given. Then, the additivity follows from
The scalar compatibility follows from
In order to prove injectivity, let with be given. this means that for all linear forms , we have . But then, due to Fact *****, we have
By the criterion for injectiviy, is injective.
In the finite-dimensional case, the bijectivity follows from injectivity and from Corollary 13.12 .
Thus, the mapping sends a vector to the evaluation
(or evaluation mapping)
which evaluates a linear form an the point .
- Footnotes
- ↑ Here, is to be understood in the sense of Remark 14.4 .
- ↑ In , the relation hold. Note that here, the running index is the first index; in the equation claimed, the running index is the second index, corresponding to transposing.
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