c994919546
- 7단계 재구성: 정의(QᵀQ vs QQᵀ 함정) → 3대장(반사행렬 상세: I−2uuᵀ = 수직 성분 부호 반전) → 보존 → Q 사영(역행렬 소멸, 굿노트 유도 반영) → b 분해(EX4 질문 답변) → 그람-슈미트 → A=QR·최소제곱·mn² 유도 - 하이라이트 53개 연결 (텍스트 44 + 영역 9), OCR 교정 43건 - 확인문제 4.4A(아다마르)·4.4B 정리, p.263 A vs Q 요약표 임베드 - 연습문제 2·4·12·15·18·32 선정·해설 (표준 구조) Co-Authored-By: Claude Fable 5 <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_01Ufr6wUhJAzBaqcaBiRJ1xk
2098 lines
117 KiB
Clojure
Executable File
2098 lines
117 KiB
Clojure
Executable File
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:content {:text "두 벡터 가 직교한다는 것은 두 벡터 의 내적 이 0임을 의미한다 ."},
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:content {:text "4가지 기본 부분공간 은 서로 직교하는 관계이다 "},
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:content {:text "식 Az= 0은 2가 행렬 A의 각 행 과 직교 함을 의미하며 , 식 Ay= 0 은 y 가 행렬 A 의 각 열과직교 함을 의미한다"},
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:content {:text "직선 위의 모든 벡터와 평면 위에 모든 벡터 가 서로 직교 하면, 그 직선( 1 차원 부분 공간)과 평면 ( 2차원 부분 공간 ) 은 서로 직교 한다."},
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:content {:text "행렬 A의 영공간 은 A의 행공간 과 직교 한다."},
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:content {:text " 의 열공간 C( A) 와 AT 의 영공간 N (AT ) 가 직교한다 "},
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:content {:text "전체 공간 을 구성하는 직교하는 두 부분 공간의 쌍을 특별히 직교여 공간이라 한다"},
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:content {:text "V의 직교 여공간 V- 은V 와 직교 하는 모든 벡터 를 포함 한다."},
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:content {:text "직교 하는 두 부분공간 에 속하는 유일한 벡터는 영벡터이다"},
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:content {:text "해 의 유일성은 해 가 존재 함을 의미 하고, 해 의 존재성은 해가 유일 함을 의미 한다"},
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:content {:text "행렬 A는행공간 에서 열 공간 으로 의 가역변환이다"},
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:content {:text "행 공간의 기저벡터 개와 영공간 의 기저벡터 ( 2-1)개 를 더하면 + ( nr ) = \"개의 벡터가 된다 . 이 \" 개의 벡터는 선형독립 이다 . 따라서 R \"을 생성한다"},
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:content {:text "벡터 b 는 S 안에서 가장 가까운 점 p 로 사영된다 . 그러면 오차 벡터 e =b-p 는 부분 공간 S와 직교 한다. "},
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:content {:text "항등행렬 / 에 대해 행렬 - P 도 사영행렬 이어야 한다"},
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:content {:text "P 가 하나의 부분공간 위로 사영 할 때 , I -P 는 그 부분 공간과 직교 하는 부분공간 위로 사영 한다."},
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:content {:text "p = Ai( 사)"},
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:content {:text "위 의 ' ( hat) 표기는 열 공간 안에서 가장 가까운 벡터 를 제공하는 최적 의 선택 임을 의미 한다 . \" = 1인 경우 , 그 최적 의 선택은 호 - 이다"},
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:y1 905.3059539794922,
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:content {:text "A 는 직사각 행렬 이다. 즉 A의 역행렬은 없다. > 개인 경우에는 A'가 존재하지 않으므로( ATA )' 를 A- '와 ( AT)' 로 나눌 수 없다 "},
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:content {:text "Az= b의 해 가 존재하지 않으면, 이 식의 양변에 AT 를 곱하고 AT Af= ATb를 푼다."},
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:content {:text "임의 의 에 대한 제곱 오차 ||| Az- 6 ||=|| Azp ||+ ||e || 2"},
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:content {:text "식 (4 )의 두 방정식은 ATA=ATb이다."},
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:content {:text "영공간 N (A ) 가 얼마나 작은지에 주목하라 . N( A) 는 단 하나 의 점만 포함 한다. 선형독립 인 열을이용 하면 Az= 0의 유일한 해 는 z = 0임을 알 수 있다. 그러면 ATA 는 가역행렬이다. 방정식 ATA = ATb는 최적의 벡터 눈을 완전히 결정 한다. 오차e= b-p는 ATe= 0을 만족 한다 "},
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:content {:text "직선 맞추기 (fitting a line) 는 최소 제곱 법에 대한 가장 명확한 응용 이다"},
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:content {:text "기저는 벡터공간을 생성 하는 선형독립인 벡터들 로 구성 된다."},
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:content {:text "내적 9. 9가 모두 0일 때 91 ... 9 개 은 서로 직교한다 . 좀 더 정확히 말하면 ≠ 일 때만 qq=0 이다. 한발 더 나아가 각 벡터 91 ... 9 들을 자기 자신 의 크기 로 나누면 정규직교벡터 가 된다. 이를 통해 모든 벡터 의 크기 는 1 이 된다"},
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:content {:text "일반적으로 QQT≠ 이다"},
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:content {:text "Q가 정사각행렬 일 때 , QQ=1이면 QT= Q-' 이다. 즉 ( 전치행렬)= (역행렬 ) 이다"},
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:content {:text "열 들이 정규성 은 만족 하지 않고 직교성 만 만족하는 경우, 내적은 여전히 대각 행렬( 항등 행렬은 아님) 을 만든다 "},
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:content {:text "Q 가 직사각 행렬이라고 해도 QQ=I는 성립 한다"},
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:content {:text "만약 Q 가 정사각행렬이면 QQT =/ 도 성립"},
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:content {:text "Q 의 역행렬 은 전치행렬 QT"},
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:content {:text "이와 같은 경우 정사각 행렬 Q를 직교행렬 이라 한다."},
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:content {:text "직교행렬인지 를 빠르게 판정 하는 방법은 QTQ=1 인지 확인하는 것이다"},
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:content {:text "모든 치환행렬 은 직교행렬이다"},
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:content {:text "회전 은 모든 벡터 의 크기를 유지 한다. 반사와 치환도 마찬가지 이다. 그러므로 행렬 Q를 곱한다고해도 벡터의 크기와 벡터 간의 사잇각 은 변하지 않는다 "},
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:content {:text "(Qz) ( Qz) = QQ = TI=IT] 이므로, Qz ' 은 || ||' 과 같음"},
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:content {:text "모든 벡터 2에 대해 ||Q || =|||||"},
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:content {:text "(Qx) (Qy) = xQTQy=xTy"},
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:content {:text "Qz=b 의 최소 제곱 해는 i =Q6 이다. 그리고 사영행렬은 QQ \" 이다"},
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:content {:text "ATA는 이제 QQ =I "},
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:content {:text "사영은 p = Qi=QQTb"},
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:content {:text "연결하는 세 번째 행렬 이 존재한다 . 이 세 번째 행렬이 A= QR에 있는 삼각행렬 R 이다 (이는1장 에서의 R 이 아니다)"},
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:content {:text "R= QTA는 상삼각행렬"},
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:content {:text "선형독립 인 열을 갖는 모든 행렬 A 는 A= QR 분해를 갖는다 . mx \" 행렬 Q 는 정규직 교하 는열 을 갖는 행렬이고 , 정사각행렬 R은 양 의 주대각성분 을 갖는 상삼각행렬 이다."},
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:content {:text "ATA = ( QR) TQR =RTQTQR =RTR"},
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:content {:text "최소 제곱 문제 RTRE =RTQTb 또는 R² = QTb 또는 i = R -QT6 "},
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:content {:text "필요한 계산은 그림- 슈미트 과정에 있는 '번의 곱셈이다."},
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