えーと、おサルさん>>7-10 いきなり 難しい定理のサイトに飛んで 消化不良ですよ まず 順番として 下記 高校数学の美しい物語 次元定理の意味，具体例，証明 さらに 数学の風景 線形写像の次元定理dim V = rank f + dim ker fの証明 を見なさい。後者は、図解が美しいよ。
その上で 英 wikipedia ”等しい有限次元のベクトル空間の線型変換の場合、単射性または全射性のいずれかが全単射性を意味することになります。 （原文 It follows that for linear transformations of vector spaces of equal finite dimension, either injectivity or surjectivity implies bijectivity.）” が、キモです。百回音読しましょうねｗ ；ｐ）
(参考) https://manabitimes.jp/math/1077 高校数学の美しい物語 次元定理の意味，具体例，証明 2021/03/07 行列における次元定理 A を m×n 実行列とするとき， rankA+dim(KerA)=n 目次 次元定理について 具体例 次元定理のイメージ 次元定理の証明 次元定理について rankA は A のランク（階数）です。→行列のランクの意味（８通りの同値な定義） dim は次元， KerA は A のカーネル（核）です。→行列のカーネル（核）の性質と求め方
英 wikipedia https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem Rank–nullity theorem (google訳) ランク-ヌル定理（階数零定理） 階数零定理は線型代数学の定理であり、次のことを主張します。 略す したがって、等しい有限次元のベクトル空間の線型変換の場合、単射性または全射性のいずれかが全単射性を意味することになります。 （原文 It follows that for linear transformations of vector spaces of equal finite dimension, either injectivity or surjectivity implies bijectivity.）
A third fundamental subspace When T:V→W is a linear transformation between two finite-dimensional subspaces, with n=dim(V) and m=dim (W) (so can be represented by an m×n matrix M), the rank–nullity theorem asserts that if T has rank r, then n－r is the dimension of the null space of M, which represents the kernel of T. In some texts, a third fundamental subspace associated to T is considered alongside its image and kernel: the cokernel of T is the quotient space W/Im(T), and its dimension is m－r. This dimension formula (which might also be rendered dim Im(T)+dimCoker(T)=dim(W) together with the rank–nullity theorem is sometimes called the fundamental theorem of linear algebra.[7][8]
再定式化と一般化 この定理は、ベクトル空間の場合の代数学の第一同型定理の記述であり、分割補題に一般化されます。 より現代的な言葉で言えば、この定理はベクトル空間の短完全列はそれぞれ分割される、と表現することもできる。 0→U→V→R→0 はベクトル空間の短完全列 であるので、 U⊕R≅Vしたがって dim（U）+ dim（R）=dim（V）. 略す We see that we can easily read off the index of the linear map T from the involved spaces, without any need to analyze T in detail. This effect also occurs in a much deeper result: the Atiyah–Singer index theorem states that the index of certain differential operators can be read off the geometry of the involved spaces.
ついでに 独 wikipedia https://de.wikipedia.org/wiki/Rangsatz Rangsatz Der Rangsatz oder Dimensionssatz ist ein Satz aus dem mathematischen Teilgebiet der linearen Algebra. Er zeigt einen Zusammenhang zwischen den Dimensionen der Definitionsmenge, des Kerns und des Bildes einer linearen Abbildung zwischen zwei Vektorräumen auf. （google 英訳） Table of contents 1 Sentence 2 Proofs 2.1 Proof of the Hom
仏 wikipedia https://fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8me_du_rang Théorème du rang （google 英訳） Rank theorem In mathematics , and more precisely in linear algebra , the rank theorem links the rank of a linear application and the dimension of its kernel . It is a corollary of an isomorphism theorem . It can be interpreted by the notion of linear application index . In finite dimension, it allows in particular to characterize the invertibility of a linear application or of a matrix by its rank. (引用終り) 以上 []
有限次元の場合、dim V = dim im f だったら dim ker f=0 だから R^nの標準基底の像が線形独立なら 当然基底になる
し・か・し、無限次元ではそんなことは言えない というのは∞＝∞＋ｘのとき、ｘ＝０なんていえないから
>”等しい有限次元のベクトル空間の線型変換の場合、 >単射性または全射性のいずれかが全単射性を意味することになります。 >（It follows that for linear transformations of vector spaces of equal finite dimension, >either injectivity or surjectivity implies bijectivity.）” >が、キモです。百回音読しましょうね
ついでに、”Proof that every vector space has a basis”貼るよ ”This proof relies on Zorn's lemma, which is equivalent to the axiom of choice. Conversely, it has been proved that if every vector space has a basis, then the axiom of choice is true.[9]”
(参考) https://en.wikipedia.org/wiki/Basis_(linear_algebra) Basis (linear algebra) This article deals mainly with finite-dimensional vector spaces. However, many of the principles are also valid for infinite-dimensional vector spaces. Basis vectors find applications in the study of crystal structures and frames of reference.
Proof that every vector space has a basis Let V be any vector space over some field F. Let X be the set of all linearly independent subsets of V.
The set X is nonempty since the empty set is an independent subset of V, and it is partially ordered by inclusion, which is denoted, as usual, by ⊆.
Let Y be a subset of X that is totally ordered by ⊆, and let LY be the union of all the elements of Y (which are themselves certain subsets of V).
Since (Y, ⊆) is totally ordered, every finite subset of LY is a subset of an element of Y, which is a linearly independent subset of V, and hence LY is linearly independent. Thus LY is an element of X. Therefore, LY is an upper bound for Y in (X, ⊆): it is an element of X, that contains every element of Y.
As X is nonempty, and every totally ordered subset of (X, ⊆) has an upper bound in X, Zorn's lemma asserts that X has a maximal element. In other words, there exists some element
156 名前：Lmax of X satisfying the condition that whenever Lmax ⊆ L for some element L of X, then L = Lmax.
It remains to prove that Lmax is a basis of V. Since Lmax belongs to X, we already know that Lmax is a linearly independent subset of V.
If there were some vector w of V that is not in the span of Lmax, then w would not be an element of Lmax either. Let Lw = Lmax ∪ {w}. This set is an element of X, that is, it is a linearly independent subset of V (because w is not in the span of Lmax, and Lmax is independent). As Lmax ⊆ Lw, and Lmax ≠ Lw (because Lw contains the vector w that is not contained in Lmax), this contradicts the maximality of Lmax. Thus this shows that Lmax spans V.
Hence Lmax is linearly independent and spans V. It is thus a basis of V, and this proves that every vector space has a basis.
This proof relies on Zorn's lemma, which is equivalent to the axiom of choice. Conversely, it has been proved that if every vector space has a basis, then the axiom of choice is true.[9] Thus the two assertions are equivalent. (引用終り) 以上 []
・Hilbert spaceの Hilbert dimension は、下記 "As a consequence of Zorn's lemma, every Hilbert space admits an orthonormal basis; furthermore, any two orthonormal bases of the same space have the same cardinality, called the Hilbert dimension of the space.[94]" (which may be a finite integer, or a countable or uncountable cardinal number). ・”The Hilbert dimension is not greater than the Hamel dimension (the usual dimension of a vector space).” ”As a consequence of Parseval's identity,[95] 略 ” ・なお、>>146-147 "Proof that every vector space has a basis"では、有限和は 陽には使われていない なので ”The set X is nonempty since the empty set is an independent subset of V, and it is partially ordered by inclusion, which is denoted, as usual, by ⊆. Let Y be a subset of X that is totally ordered by ⊆, and let LY be the union of all the elements of Y (which are themselves certain subsets of V). Since (Y, ⊆) is totally ordered, every finite subset of LY is a subset of an element of Y, which is a linearly independent subset of V, and hence LY is linearly independent. Thus LY is an element of X. Therefore, LY is an upper bound for Y in (X, ⊆): it is an element of X, that contains every element of Y. As X is nonempty, and every totally ordered subset of (X, ⊆) has an upper bound in X, Zorn's lemma asserts that X has a maximal element. In other words, there exists some element Lmax of X satisfying the condition that whenever Lmax ⊆ L for some element L of X, then L = Lmax.” とやっているので、⊆ による順序は Hilbert space でも そのまま使える あとは、直交基底と 位相的な収束の話を 色付けすれば、よさそうだ
(参考) https://en.wikipedia.org/wiki/Hilbert_space Hilbert space
Hilbert dimension As a consequence of Zorn's lemma, every Hilbert space admits an orthonormal basis; furthermore, any two orthonormal bases of the same space have the same cardinality, called the Hilbert dimension of the space.[94] For instance, since l^2(B) has an orthonormal basis indexed by B, its Hilbert dimension is the cardinality of B (which may be a finite integer, or a countable or uncountable cardinal number).
The Hilbert dimension is not greater than the Hamel dimension (the usual dimension of a vector space).
As a consequence of Parseval's identity,[95] if {ek}k ∈ B is an orthonormal basis of H, then the map Φ : H → l^2(B) defined by Φ(x) = ⟨x, ek⟩k∈B is an isometric isomorphism of Hilbert spaces: it is a bijective linear mapping such that ⟨Φ(x),Φ(y)⟩l^2(B)=⟨x,y⟩H for all x, y ∈ H. The cardinal number of B is the Hilbert dimension of H. Thus every Hilbert space is isometrically isomorphic to a sequence space l^2(B) for some set B.
