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Logseq/Major Study/assets/3장_4가지_기본_부분공간_1785222523974_0.edn

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:content {:text "벡터공간 (vector space)'"},
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:content {:text "Ar= 0 을 만족하는 벡터들만 모아서 벡터 공간"},
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:content {:text "기저(basis)"},
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:content {:text "공간 을 완벽하게 설명하는 벡터 집합"},
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:content {:text "벡터공간 R\"에서 선형결합 cut dw를 만들 수 있다"},
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:content {:text "3.1 절 연습 문제"},
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:content {:text "모든 선형 결합 cu+dw 가 벡터 공간에 속해야 한다"},
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:content {:text "영 벡터 0= ( 0, 0 ,・, 0 )만 포함된 공간 2 도 부분공간이다 ."},
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:content {:text "Z 는 가장 작은 벡터공간이다 "},
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:content {:text "가역행렬의 영 공간"},
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:content {:text "Az =0 의 유일한 해 가 영 벡터 z= 0이라면, 행렬 A 의 열 들은 선형 독립이고 , 행렬 A 의영 공간은 Z 이다"},
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:content {:text "행렬 로 이루어진 벡터공간"},
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:content {:text "함수 로 이루어진 벡터공간"},
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:content {:text "공간' 이라는 단어 는 벡터 나 행렬 또는 함수의 모든 선형결합이 공간 안에서 닫혀있음 을 의미 한다"},
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:content {:text "행렬 과 함수를 벡터 로 생각할 필요가 있다. 그러나 대부분 우리가 가정하는 벡터는 열벡터 이다.이들은 \" 개의 성분 을 가진 벡터이다(그러나 \"개의 성분 을 가진 모든 벡터가 아닐 수도 있다). R\" 의 내부에는 중요한 벡터 공간이 있고 , 이들은 R\" 의 부분공간"},
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:content {:text "3 차원 공간의 원점을 지나는 평면 은 R ' 처럼 보이 더라도 R' 과 다르다. "},
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:content {:text "정의 부분공간 판정법"},
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:content {:text "o+w가 부분공간 안에 있다 ."},
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:content {:text "cu가 부분 공간 안에 있다."},
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:content {:text "부분 집합"},
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:content {:text "부분공간"},
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:content {:text "모든 부분공간 이 영벡터를 포함 한다"},
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:content {:text "R의 평면이 부분공간이 되기 위해서는 원점 ( 0, 0, 0 )을 지나야 한다."},
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:content {:text "원점을 지나는 직선도 부분공간이다 "},
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:content {:text "항등 행렬 / 자체가 부분공간 이 될까 ? 물론 아니다 "},
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:content {:text "A와의 곱으로 나타낼 수 있는 벡터 b 는 행렬 A의 ' 열공간'을 형성"},
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:content {:text "Ar는 행렬 A의 열 들의 선형 결합"},
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:content {:text "이 선형 결합은행렬 A의 열공간을 생성 한다. 이는 열 벡터로 구성된 벡터 공간이다"},
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:content {:text "열공간 은 모든 열의 선형결합 으로 구성 된다"},
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:content {:text "가능한 모든 벡터 Az"},
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:content {:text "Ar 들이 열공간 C( A) 를 채운다."},
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:content {:text "방정식 Az=b 의 해 가 존재할 필요충분조건은벡터 6 가 행렬 A의 열공간 안에 있는 것이다."},
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:content {:text "벡터 6가 열공간 C (A ) 안에 있을 때, b 는 행렬 A의 열들의 선형 결합 이다 . 이때 선형 결합의계수는 방정식 Azr=b 의 해가 된다"},
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:content {:text "행렬 A 의 열만 모으면 이는 부분공간을 형성 하지 않는다는 것이다. 가역 행렬은 부분공간을 형성하지 않는다. 특이행렬 도 부분공간 을 형성하지 않는다 . 모든 선형결합 을 포함 해야한다. "},
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:content {:text "행렬 A의 행은 전치 행렬 A의 열이다"},
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:content {:text "A의 행공간 은 전치행렬 A 의 열 공간 CAT ) 이다."},
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:content {:text "생성 "},
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:content {:text "S가 N개의 벡터 만 포함한다면, S는 확실히 부분공간 이 아니다 . 그러나 S의 모든 벡터 의 선형결합을 포함한다면 벡터공간 V 가 만들어진다. 이 경우 집합 S 는 벡터 공간 V를 생성한다 ."},
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:content {:text "Q. 언제 10개의 벡터 가 R' 을 생성 하는가? 이는 가능하다 . 그러나 아무렇게나 고른 10 개의 벡터로는 R ' 을 생성 할 수 없다 "},
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:content {:text "정사각행렬 A 가 가역행렬 (랭크는 = 2) 일 때, 유일한 해는 0이다 . 그러면 행렬 A 의 영공간은 영벡터 만 포함한다"},
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:content {:text "특별해 S1 S2가 영공간 의 기저"},
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:content {:text "사다리꼴'이라는 용어는 1의 ' 계단 '을 의미한다 . 모든 영행은항상 RO 의 마지막 행에 위치 한다"},
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:content {:text "임의의 X\" 행렬 A로 시작 하여 소거법을 적용한다. 이에 따라행렬 A를 기약행 사다리꼴 Ro = rref ( A)로 바꾼다"},
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:content {:text "R의 모든 영행 을 제거하면 R 이 된다."},
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:content {:text "새로운 열 은 이전 열들과 선형종인가"},
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:content {:text "(A 의 선형 독립인 열) = (A의 선형종속인 열 ) X F"},
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:content {:text " A= CR과 영공간"},
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:content {:text "RO 안에있는 는 행렬 A에서 선형 독립인 열들 로 이루어진 행렬 C의 위치를 알려준다."},
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:content {:text "영행을 제거하면 A = CR 에서의 행 행렬 R"},
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:content {:text "원점만 있는 영 공간 Z는 굉장히 중요 하다. 이 경우 행렬 A의 열은 선형 독립이다 . 열들의 어떠한결합도 영벡터 를 만들지 않는다( 단, 0이 아닐 때 ). \"> 인 경우 에는 이런 상황이 발생하지않는다 . 다시 말해 R\" 에서 \"개의 선형 독립인 열을 가질 수 없다"},
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:content {:text ">m인 경우이므로 적어도 하나의 자유 변수가 있다"},
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:content {:text "방정식보다 미지수의 개수 가 더 많다고 가정 하자"},
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:content {:text "( n-2) 개의 자유변수가 있어야 한다 . Ar= 0 은 N (A ) 안에서 0이 아닌 해 를 가져야 한다."},
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:content {:text "블록 소거법 ( block elimination)을 설명 할 수 있다 . 이때 행렬 A의 랭크 는 ' 로 가정 한다"},
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:content {:text "PC를 이용해 행렬 A 의 열의 위치를 바꾼다"},
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:content {:text "W- 1B = [ IW-H ][IF]"},
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:content {:text "소거 를 통해 Az =0 에 대한 문제는 Ro =0 으로 변환되었다 . 자유변수 에는 특별한 값 (1 과 0) 이 주어졌다 . 그런 다음 역대 입법을 통해 피벗 변수를구했다. 우변 b 는 0이었으므로 주의 를 기울이지 않았다"},
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:content {:text "Ar = b는 동일한 해 ( 해가존재할 경우 ) 를 가진 더 간단한 연립방정식 Ror= d 로 축소된다. 이제 벡터 6 를 행렬 A의 열로 추가하자 "},
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:content {:text "가장 간단한 해 는 자유 변수를 0으로 선택한 경우이다"},
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:content {:text "자유변수 는 0이고, 피벗변수는 d로부터 구한다."},
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:content {:text "해가 존재 하려면 R의 영향이 있는 위치와 동일한 d 의 위치에도 0 이 있어야 한다"},
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:content {:text "자유변수에 0 을 할당 한 다음, 피벗 변수를 푸는 방법을 사용한다는 점에 주목하자"},
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:content {:text "자유변수가 없다 . 따라서 Ar=b 의 0 이 아닌 특별 해가 없다"},
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:content {:text "행렬 A 의 영 공간 안에는 = 0 밖에 없다"},
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:content {:text "Az = b와 Rod 의 특수 해는 최종 열 d 의 맨 위에 있다"},
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:content {:text "b3+6, +b2가 0 이 아니라면, Ax=b 의 해는 존재 하지 않는다"},
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:content {:text "r =m 인 경우 행렬 A는 최대 행 랭크 를 갖는다 . 이때 ' 행은 선형 독립이다 . 모든 행 에는 피벗 이 있다."},
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:content {:text "특수해 는 이 직선 위 의 한 점이 될 것이다. 영 공간벡터 1을 더하면 [ 그림 3.1] 과 같이 직선 을 따라 이동할 수 있다"},
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:content {:text "랜덤 선형 대수학(randomized linear algebra) 이다"},
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:content {:text "벡터공간의모든 벡터는 기저 벡터 들의 유일한 선형 결합 이다 ."},
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:content {:text "R'에서 세 벡터 는 선형 독립 일 수 없다. "},
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:content {:text "기저벡터 는 선형 독립 이며, 벡터 공간 전체를 생성 한다."},
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:content {:text "벡터 \"를 기저벡터의 선형 결합 으로 나타내는 방법 은 유일하다."},
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:content {:text "선형독립 인 벡터 의 모든 집합은 기저 로 확장될 수 있다.생성 하는 벡터 의 모든 집합은 기저 로 축소될 수 있다."},
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:content {:text "모든 기저에 있는 벡터의 개수를 벡터공간의 ' 차원 ' 이라 한다 "},
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:content {:text "01,... , Um 과 201, ... , W , 이 동일한 벡터 공간의 기저 벡터 일 때 , m = \" 이다 "},
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:content {:text "행렬 공간과 함수 공간 의 기저"},
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