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100755 index 0000000..71fd7fa Binary files /dev/null and b/assets/2장_선형방정식_Ax_=_b_풀기_1784805780272_0/6_6a636514-b25c-4535-a0de-ec772ee116c6_1784898835828.png differ diff --git a/pages/2.1 소거법과 역대입법.md b/pages/2.1 소거법과 역대입법.md index 923c6de..be23016 100755 --- a/pages/2.1 소거법과 역대입법.md +++ b/pages/2.1 소거법과 역대입법.md @@ -39,6 +39,18 @@ deck:: Logseq/전공공부/선형대수학 - **$A$ 에서 $U$ 로, $b$ 에서 $c$ 로 — 소거의 실제 과정** (p.76 예제, 식 (2)~(5)) - **소거행렬** $E_{ij}$: $i$ 행에서 $j$ 행에 $\ell_{ij}$ 를 곱한 것을 빼서 $(i,j)$ 위치에 0을 만드는 행렬. $E_{21}, E_{31}, E_{32}$ 순서로 $A$ 의 왼쪽에 곱한다 (정확한 형태는 [[2.2 소거행렬과 역행렬]]에서 상세히). - 📌 원문 — 소거행렬 (p.76): ((6a635932-a78c-4087-a764-e7eaabae9f9f)) + - 예제 (p.76, 식 (2)~(5)) — 시작 행렬 $A$ 와 우변 $b$, 첫 번째 피벗은 2: + - 📌 원문 — 식 (2) 시작 행렬 (p.76): [🔗 PDF 위치](((6a636504-582f-4014-a85a-ec2681812c56))) + - ![소거 예제 식 (2) — 시작 행렬 A와 b (p.76)](../assets/2장_p76_소거예제_식2_시작행렬.png) + - ① $E_{21}$ — 2행에서 1행×2를 빼서 피벗 2 아래에 0 생성: + - 📌 원문 — 식 (3) (p.76): [🔗 PDF 위치](((6a636509-c1bc-472d-9c7e-3bda5d7ab152))) + - ![소거 예제 식 (3) — E21 적용 (p.76)](../assets/2장_p76_소거예제_식3_E21.png) + - ② $E_{31}$ — 3행에서 1행을 빼서 1열 소거 완료: + - 📌 원문 — 식 (4) (p.76): [🔗 PDF 위치](((6a63650f-a68d-4a28-9ff8-f3c6e31db572))) + - ![소거 예제 식 (4) — E31 적용 (p.76)](../assets/2장_p76_소거예제_식4_E31.png) + - ③ $E_{32}$ — 3행에서 (새로운) 2행을 빼서 $U$ 완성. $E_{32}E_{31}E_{21}A = U$ 는 상삼각행렬: + - 📌 원문 — 식 (5) (p.76): [🔗 PDF 위치](((6a636514-b25c-4535-a0de-ec772ee116c6))) + - ![소거 예제 식 (5) — E32 적용, U 완성 (p.76)](../assets/2장_p76_소거예제_식5_E32.png) - 진행 순서 (행 교환이 필요 없는 경우): - ① 첫 번째 방정식으로 1열의 피벗 아래 성분들을 0으로. - ② **새로운** 두 번째 방정식으로 2열의 피벗 아래 성분들을 0으로. diff --git a/pages/2.2 소거행렬과 역행렬.md b/pages/2.2 소거행렬과 역행렬.md index 82e3686..6ddae2c 100755 --- a/pages/2.2 소거행렬과 역행렬.md +++ b/pages/2.2 소거행렬과 역행렬.md @@ -19,6 +19,8 @@ deck:: Logseq/전공공부/선형대수학 - ⚠️ $E_{32}$ 는 $A$ 의 원래 2행이 아니라 **소거가 진행된 뒤의 새로운 2행**에 $\ell_{32}$ 를 곱해 뺀다 (p.86 EX1). - $E$ 를 직접 곱해보면 좌하단에 복잡한 항이 생긴다 (p.91 식 (10)): - $$E = E_{32}E_{31}E_{21} = \begin{bmatrix} 1 & & \\ -\ell_{21} & 1 & \\ (\ell_{32}\ell_{21} - \ell_{31}) & -\ell_{32} & 1 \end{bmatrix}$$ + - 📌 원문 — 식 (10) (p.91): [🔗 PDF 위치](((6a63652b-602e-4795-920c-257281d9a689))) + - ![E = E32 E31 E21 의 복잡한 곱 — 식 (10) (p.91)](../assets/2장_p91_E_식10.png) - 💡 $(3,1)$ 위치의 $\ell_{32}\ell_{21}$ 항이 생기는 이유: 1행이 2행을 바꾸고, 그 바뀐 2행이 다시 3행을 바꾸는 **연쇄 효과** 때문. 그래서 $E$ 자체는 쓸모 있는 공식이 안 되고, 역행렬 $L$ 쪽이 깨끗하다 (아래). - **$E_{ij}$ 의 역행렬 — 빼면, 역은 더한다** - $E_{ij}$ 가 "$i$ 행에서 $j$ 행 $\times \ell_{ij}$ 를 뺀다"면, $E_{ij}^{-1}$ 는 "$i$ 행에 $j$ 행 $\times \ell_{ij}$ 를 **더한다**". diff --git a/pages/2.3 행렬 계산과 A = LU.md b/pages/2.3 행렬 계산과 A = LU.md index 96adff5..f73c89c 100755 --- a/pages/2.3 행렬 계산과 A = LU.md +++ b/pages/2.3 행렬 계산과 A = LU.md @@ -28,8 +28,12 @@ deck:: Logseq/전공공부/선형대수학 - 📌 원문 (p.99): ((6a635be3-d31e-4142-a915-420dc9b58289)) - 소거 중 피벗 행에서 빼는 것은 $A$ 의 원래 행이 아니라 이미 완성된 **$U$ 의 행**이다 (피벗 행은 그 뒤로 다시 바뀌지 않으므로). - $$(U \text{의 3행}) = (A \text{의 3행}) - \ell_{31}(U \text{의 1행}) - \ell_{32}(U \text{의 2행}) \tag{3}$$ + - 📌 원문 — 식 (3) (p.100): [🔗 PDF 위치](((6a63653f-b4da-4d0f-a4e6-a661af2898b6))) + - ![A=LU 증명 식 (3) (p.100)](../assets/2장_p100_ALU증명_식3.png) - 이 식을 이항하면: - $$(A \text{의 3행}) = \ell_{31}(U \text{의 1행}) + \ell_{32}(U \text{의 2행}) + 1 \cdot (U \text{의 3행}) \tag{4}$$ + - 📌 원문 — 식 (4) (p.100): [🔗 PDF 위치](((6a636543-5209-4619-b32e-f0af8ee44f26))) + - ![A=LU 증명 식 (4) (p.100)](../assets/2장_p100_ALU증명_식4.png) - 식 (4)가 정확히 행렬곱 $A = LU$ 의 3행이다 — $L$ 의 3행 $[\ell_{31} \ \ \ell_{32} \ \ 1]$ 이 그대로 읽힌다. 모든 크기 $n$ 에서 성립. - **증명 2 — (열)×(행) 랭크 1 행렬로 보기** (p.100~101) - 소거를 "$A$ 에서 (L의 열)·(U의 행) 인 랭크 1 행렬을 빼는 것"으로 본다. diff --git a/pages/2.5 미분과 유한차분행렬.md b/pages/2.5 미분과 유한차분행렬.md index b9df419..26117fd 100755 --- a/pages/2.5 미분과 유한차분행렬.md +++ b/pages/2.5 미분과 유한차분행렬.md @@ -44,6 +44,8 @@ deck:: Logseq/전공공부/선형대수학 - **$K_n$ 의 정확한 형태** 🧩 (고정–고정 경계, 2차차분행렬) - **주대각선은 전부 $2$, 주대각선 바로 위·아래의 대각선은 전부 $-1$, 그 외는 전부 $0$.** - $$K_{ii} = 2, \qquad K_{i,i+1} = K_{i+1,i} = -1, \qquad \text{나머지 } 0$$ + - 📌 원문 — 식 (8) $K_4$ (p.122): [🔗 PDF 위치](((6a63655a-721b-4fd9-9d4e-104ea606096c))) + - ![고정된 끝점이 있는 2차차분행렬 K4 — 식 (8) (p.122)](../assets/2장_p122_K4_식8.png) - 💡 부호가 $-1, 2, -1$ 인 이유: 근사하는 대상이 $+\dfrac{d^2u}{dx^2}$ 가 아니라 $-\dfrac{d^2u}{dx^2}$ 이기 때문 (물리 문제에서 $-u''$ 꼴이 표준이라 부호를 반대로 뒤집은 것). - **$K$ 의 5가지 성질** (p.122~124) — 눈에 보이는 3가지 + 숨은 2가지 - ❶ **대칭행렬**이다: $K_{ij} = K_{ji}$, $K^{\mathrm{T}} = K$. @@ -61,9 +63,16 @@ deck:: Logseq/전공공부/선형대수학 - **$K_n$ (고정–고정)**: 양 끝 모두 벽에 고정. 위에서 본 그대로 — 대각 $2, 2, \cdots, 2$. - **$T_n$ (자유–고정)**: 위 끝 자유, 아래 끝 고정. **$K$ 에서 $(1,1)$ 성분 하나만 $2 \to 1$ 로** 바뀐다. - $$T_4 = \begin{bmatrix} \mathbf{1}&-1&0&0 \\ -1&2&-1&0 \\ 0&-1&2&-1 \\ 0&0&-1&2 \end{bmatrix}$$ + - 📌 원문 — 식 (10) $T_4$ (p.125): [🔗 PDF 위치](((6a63656f-97f2-40c9-a8c5-26180f7f9782))) + - ![자유-고정 경계 조건의 T4 — 식 (10) (p.125)](../assets/2장_p125_T4_식10.png) - 1행의 $1, -1$ 은 **자유 경계 조건** $\frac{du}{dx} = 0$ 을 나타낸다. - $T$ 는 주대각선이 일정하지 않으므로 더 이상 퇴플리츠 행렬이 아니지만, 여전히 **양의 정부호**다 (모든 피벗이 1). - - $T$ 의 아름다운 분해: $T = LL^{\mathrm{T}}$ ($L$ = 후진차분행렬). $L^{-1}, U^{-1}, T^{-1}$ 모두 1로 채워진 삼각형/사각형 꼴 (p.125 식 (12)). + - $T$ 의 아름다운 분해: $T = LU = LL^{\mathrm{T}}$ ($L$ = 후진차분행렬, 모든 피벗이 1이라 $K$ 보다 간단히 분해된다): + - 📌 원문 — 식 (11) $T = LU$ (p.125): [🔗 PDF 위치](((6a636573-14ee-44d8-b3ad-7c38a6d221bc))) + - ![T의 LU 분해 — 식 (11) (p.125)](../assets/2장_p125_T_LU분해_식11.png) + - $L^{-1}, U^{-1}, T^{-1}$ 는 모두 1로 채워진 삼각형/사각형 꼴의 아름다운 형태다: + - 📌 원문 — 식 (12) 역행렬들 (p.125): [🔗 PDF 위치](((6a636579-147e-4d4a-8f81-87ca3fe5306a))) + - ![U, L, T의 역행렬 — 식 (12) (p.125)](../assets/2장_p125_역행렬들_식12.png) - **$B_n$ (자유–자유)**: 양 끝 모두 자유. **$K$ 에서 $(1,1)$ 과 $(n,n)$ 둘 다 $2 \to 1$ 로** 바뀐다. - $$B_3 = \begin{bmatrix} \mathbf{1}&-1&0 \\ -1&2&-1 \\ 0&-1&\mathbf{1} \end{bmatrix}$$ - **$B$ 는 특이행렬(비가역)이다!** diff --git a/pages/hls__2장_선형방정식_Ax_=_b_풀기_1784805780272_0.md b/pages/hls__2장_선형방정식_Ax_=_b_풀기_1784805780272_0.md index c984308..8f33ab7 100755 --- a/pages/hls__2장_선형방정식_Ax_=_b_풀기_1784805780272_0.md +++ b/pages/hls__2장_선형방정식_Ax_=_b_풀기_1784805780272_0.md @@ -304,4 +304,81 @@ 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