행렬식, 역행렬, 전치행렬

행렬의 전치

import numpy as np

A = np.array([[1, 2, 3],
              [4, 5, 6]])
print('A shape:', A.shape)       # (2, 3)
print('A.T shape:', A.T.shape)   # (3, 2)
print(A.T)

전치의 실제 활용

import numpy as np

A = np.array([[1.0, 2.0],
              [3.0, 4.0],
              [5.0, 6.0]])  # shape (3, 2)

ATA = A.T @ A   # shape (2, 2), symmetric
print('A.T @ A:')
print(ATA)
print('Is symmetric:', np.allclose(ATA, ATA.T))

행렬식: 의미

import numpy as np

# Non-singular matrix
A = np.array([[3.0, 1.0],
              [2.0, 4.0]])
print('det(A):', np.linalg.det(A))   # 3*4 - 1*2 = 10

# Singular matrix (row 2 = 2 * row 1)
B = np.array([[1.0, 2.0],
              [2.0, 4.0]])
print('det(B):', np.linalg.det(B))   # ~0

np.linalg.inv()로 역행렬 계산하기

import numpy as np

A = np.array([[3.0, 1.0],
              [2.0, 4.0]])

A_inv = np.linalg.inv(A)
print('Inverse:')
print(A_inv)

# Verify: A @ A_inv should be identity
I = A @ A_inv
print('A @ A_inv (should be I):')
print(np.round(I, 10))

직접 역행렬을 계산할 때의 수치적 불안정성

import numpy as np

A = np.array([[1.0, 2.0],
              [1.0001, 2.0]])
print('Condition number:', np.linalg.cond(A))
# Very high condition number -> near-singular, unstable inversion

b = np.array([3.0, 3.0001])
# Prefer solve over inv @ b
print('Solution via solve:', np.linalg.solve(A, b))

직교 행렬과 전치

import numpy as np

# 90-degree rotation matrix is orthogonal
theta = np.pi / 2
R = np.array([[np.cos(theta), -np.sin(theta)],
              [np.sin(theta),  np.cos(theta)]])

print('R.T @ R (should be identity):')
print(np.round(R.T @ R, 10))
print('Is orthogonal:', np.allclose(R.T @ R, np.eye(2)))

정방 행렬이 아닌 행렬의 의사역행렬

import numpy as np

# Over-determined system: 3 equations, 2 unknowns
A = np.array([[1.0, 1.0],
              [1.0, 2.0],
              [1.0, 3.0]])
b = np.array([2.0, 3.0, 5.0])

# Least-squares solution using pseudo-inverse
x = np.linalg.pinv(A) @ b
print('Best-fit solution:', x)
# Or use lstsq directly
x2, _, _, _ = np.linalg.lstsq(A, b, rcond=None)
print('lstsq solution:', x2)

행렬식의 부호와 부피 배율

import numpy as np

# Identity: det = 1 (no scaling, no flip)
I = np.eye(3)
print('det(I):', np.linalg.det(I))    # 1.0

# Scaling by 2 in all directions: det = 8 = 2^3
S = 2 * np.eye(3)
print('det(2I):', np.linalg.det(S))   # 8.0

# Reflection: det = -1
F = np.array([[-1.0, 0.0],
              [ 0.0, 1.0]])
print('det(reflection):', np.linalg.det(F))

안정성을 위한 np.linalg.slogdet() 사용

import numpy as np

A = np.random.rand(100, 100)

# Direct det might underflow for large matrices
sign, logdet = np.linalg.slogdet(A)
print('Sign:', sign)
print('Log|det|:', logdet)
print('Actual det (may be ~0 due to float):', np.linalg.det(A))
print('Reconstructed |det|:', np.exp(logdet))

실전: 손으로 2x2 풀기와 NumPy 비교

import numpy as np

a, b, c, d = 3.0, 1.0, 2.0, 4.0
A = np.array([[a, b], [c, d]])
det = a*d - b*c

# Manual inverse
A_inv_manual = (1/det) * np.array([[d, -b], [-c, a]])
print('Manual inv:')
print(A_inv_manual)

# NumPy inverse
print('np.linalg.inv:')
print(np.linalg.inv(A))
print('Match:', np.allclose(A_inv_manual, np.linalg.inv(A)))

파이프라인에서 전치와 역행렬 결합하기

import numpy as np

# Design matrix for simple linear regression (with bias)
X = np.column_stack([np.ones(5), np.array([1., 2., 3., 4., 5.])])
y = np.array([2.1, 3.9, 6.2, 8.1, 9.8])

# OLS via normal equations: beta = (X.T @ X)^-1 @ X.T @ y
beta = np.linalg.inv(X.T @ X) @ X.T @ y
print('Coefficients (intercept, slope):', np.round(beta, 3))

# Verify with lstsq
beta2, _, _, _ = np.linalg.lstsq(X, y, rcond=None)
print('lstsq result:', np.round(beta2, 3))

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