Dimension of vector spaces #
In this file we provide results about Module.rank
and FiniteDimensional.finrank
of vector spaces
over division rings.
Main statements #
For vector spaces (i.e. modules over a field), we have
rank_quotient_add_rank
: ifV₁
is a submodule ofV
, thenModule.rank (V/V₁) + Module.rank V₁ = Module.rank V
.rank_range_add_rank_ker
: the rank-nullity theorem.rank_dual_eq_card_dual_of_aleph0_le_rank
: The Erdős-Kaplansky Theorem which says that the dimension of an infinite-dimensional dual space over a division ring has dimension equal to its cardinality.
If a vector space has a finite dimension, the index set of Basis.ofVectorSpace
is finite.
The rank-nullity theorem
Given a family of n
linearly independent vectors in a space of dimension > n
, one may extend
the family by another vector while retaining linear independence.
Given a family of n
linearly independent vectors in a space of dimension > n
, one may extend
the family by another vector while retaining linear independence.
Given a nonzero vector in a space of dimension > 1
, one may find another vector linearly
independent of the first one.
This is mostly an auxiliary lemma for Submodule.rank_sup_add_rank_inf_eq
.
The ι
indexed basis on V
, where ι
is an empty type and V
is zero-dimensional.
See also FiniteDimensional.finBasis
.
Equations
Instances For
A vector space has dimension at most 1
if and only if there is a
single vector of which all vectors are multiples.
A submodule has dimension at most 1
if and only if there is a
single vector in the submodule such that the submodule is contained in
its span.
A submodule has dimension at most 1
if and only if there is a
single vector, not necessarily in the submodule, such that the
submodule is contained in its span.
Given a family of n
linearly independent vectors in a finite-dimensional space of
dimension > n
, one may extend the family by another vector while retaining linear independence.
Given a family of n
linearly independent vectors in a finite-dimensional space of
dimension > n
, one may extend the family by another vector while retaining linear independence.
Given a nonzero vector in a finite-dimensional space of dimension > 1
, one may find another
vector linearly independent of the first one.
A finite family of vectors is linearly independent if and only if its cardinality equals the dimension of its span.
A family of finrank K V
vectors forms a basis if they span the whole space.
Equations
- basisOfTopLeSpanOfCardEqFinrank b le_span card_eq = Basis.mk ⋯ le_span
Instances For
A finset of finrank K V
vectors forms a basis if they span the whole space.
Equations
- finsetBasisOfTopLeSpanOfCardEqFinrank le_span card_eq = basisOfTopLeSpanOfCardEqFinrank Subtype.val ⋯ ⋯
Instances For
A set of finrank K V
vectors forms a basis if they span the whole space.
Equations
- setBasisOfTopLeSpanOfCardEqFinrank le_span card_eq = basisOfTopLeSpanOfCardEqFinrank Subtype.val ⋯ ⋯
Instances For
Key lemma towards the Erdős-Kaplansky theorem from https://mathoverflow.net/a/168624
The Erdős-Kaplansky Theorem: the dual of an infinite-dimensional vector space over a division ring has dimension equal to its cardinality.
The Erdős-Kaplansky Theorem over a field.