純粋・応用数学・数学隣接分野（含むガロア理論）19

[過去ﾛｸﾞ] 純粋・応用数学・数学隣接分野（含むガロア理論）19 (1002ﾚｽ)
上下前次1-新
通常表示 512ﾊﾞｲﾄ分割ﾚｽ栞
抽出解除ﾚｽ栞

このｽﾚｯﾄﾞは過去ﾛｸﾞ倉庫に格納されています｡
次ｽﾚ検索歴削→次ｽﾚ栞削→次ｽﾚ過去ﾛｸﾞﾒﾆｭｰ

ﾘﾛｰﾄﾞ規制です｡10分ほどで解除するので､他のﾌﾞﾗｳｻﾞへ避難してください｡

739: 現代数学の系譜 雑談 ◆yH25M02vWFhP  [] 2025/04/04(金) 10:26:20.63 ID:nFnX0O4C

>>734 補足
>4 Horn, Roger A.; Johnson, Charles R. (2013). Matrix Analysis, second edition. Cambridge University Press. ISBN 9780521839402.

これの海賊版PDFが見つかった
P62
Theorem 1.3.12. Let A, B ∈ Mn be diagonalizable. Then A and B commute if and only if they are simultaneously diagonalizable.
Proof.
Assume that A and B commute, perform a similarity transformation on both A and B that diagonalizes A (but not necessarily B) and groups together any repeated eigenvalues of A. Ifμ1,...,μd are the distinct eigenvalues of A and n1,...,nd are their respective multiplicities, then we may assume that
略す

P63
Observation 1.3.18. Suppose that n ≥ 2. A given A ∈ Mn is similar to a block triangular matrix of the form (1.3.17) if and only if some nontrivial subspace of Cn is A-invariant. Moreover, if W ⊆ Cn isanonzero A-invariantsubspace,thensomevector in W is an eigenvector of A. A given family F ⊆ Mn is reducible if and only if there is some k ∈{2,...,n −1} and a nonsingular S ∈ Mn such that S−1AS has the form (1.3.17) for every A ∈ F.

The following lemma is at the heart of many subsequent results.

Lemma1.3.19. Let F ⊂ Mn beacommutingfamily. Then some nonzero vector in Cn is an eigenvector of every A ∈ F.
Proof. 略す

P64
Lemma 1.3.19 concerns commuting families of arbitrary nonzero cardinality. Our next result shows that Theorem 1.3.12 can be extended to arbitrary commuting families of diagonalizable matrices.

Definition 1.3.20. A family F ⊂ Mn is said to be simultaneously diagonalizable if there is a single nonsingular S ∈ Mn such that S−1AS is diagonal for every A ∈ F.

Theorem 1.3.21. Let F ⊂ Mn be a family of diagonalizable matrices. Then F is a commuting family if and only if it is a simultaneously diagonalizable family. Moreover, for any given A0 ∈ F and for any given ordering λ1,...,λn of the eigenvalues of A0, there is a nonsingular S ∈ Mn such that S−1A0S = diag(λ1,...,λn) and S−1BS is diagonal for every B ∈ F.

つづく

http://rio2016.5ch.net/test/read.cgi/math/1725190538/739

上下前次1-新書関写板覧索設栞歴

ｽﾚ情報赤ﾚｽ抽出画像ﾚｽ抽出歴の未読ｽﾚ AAｻﾑﾈｲﾙ

ぬこの手ぬこTOP 0.041s