如何求解超定线性系统X * A = B，A, B，并且我需要找到X？
对于方形矩阵来说，解决方案似乎很简单(https://stackoverflow.com/a/18848074/1179925)：
X * A = B
M * A * Inv(A) = B * Inv(A)
M = B * Inv(A)
但是如何处理非平方矩阵呢？
示例代码：
def to_homogeneous(_pts):
n = _pts.shape[0]
_pts = _pts.transpose()
pts = np.ones((3,n), np.float32)
pts[:2,:] = _pts
return pts
def get_random_affine_matrix():
src_tri = np.random.rand(3,2) * np.random.randint(1,10)
src_tri = src_tri.astype(np.float32)
src_tri = to_homogeneous(src_tri)
dst_tri = np.random.rand(3,2) * np.random.randint(1,10)
dst_tri = dst_tri.astype(np.float32)
dst_tri = to_homogeneous(dst_tri)
m = dst_tri @ np.linalg.inv(src_tri)
print('-'*60)
print('src_tri.shape', src_tri.shape)
print('src_tri:')
print(src_tri)
print('-'*60)
print('dst_tri.shape', dst_tri.shape)
print('dst_tri:')
print(dst_tri)
print('-'*60)
print('m.shape', m.shape)
print('m:')
print(np.round(m, 5))
return m
m = get_random_affine_matrix()
src_pts = np.random.rand(4,2) * np.random.randint(1,10)
src_pts = src_pts.astype(np.float32)
src_pts = to_homogeneous(src_pts)
dst_pts = m @ src_pts
print('-'*60)
print('src_pts.shape', src_pts.shape)
print('src_pts:')
print(src_pts)
print('-'*60)
print('dst_pts.shape', dst_pts.shape)
print('dst_pts:')
print(dst_pts)
m = dst_pts @ np.linalg.inv(src_pts) # Gives LinAlgError: Last 2 dimensions of the array must be square
输出：
------------------------------------------------------------
src_tri.shape (3, 3)
src_tri:
[[2.7440674 3.0138168 2.118274 ]
[3.5759468 2.724416 3.2294705]
[1. 1. 1. ]]
------------------------------------------------------------
dst_tri.shape (3, 3)
dst_tri:
[[0.5950692 0.5453126 1.6243374 ]
[0.11342596 0.95533025 0.9599543 ]
[1. 1. 1. ]]
------------------------------------------------------------
m.shape (3, 3)
m:
[[-1.42684 -0.39356 5.91778]
[-0.68516 -1.20574 6.30521]
[ 0. 0. 1. ]]
------------------------------------------------------------
src_pts.shape (3, 4)
src_pts:
[[8.33037 0.7841637 7.4935784 7.830109 ]
[0.63932455 0.18196557 7.003411 8.807565 ]
[1. 1. 1. 1. ]]
------------------------------------------------------------
dst_pts.shape (3, 4)
dst_pts:
[[-6.2199397 4.7272906 -7.5306525 -8.7208805 ]
[-0.17327118 5.5485325 -7.2733746 -9.679293 ]
[ 1.0000011 1.0000001 1.0000015 1.0000017 ]]
LinAlgError: Last 2 dimensions of the array must be square
我认为最好的解决方案是使用numpy的lstsq函数：
像这个一样重写你的系统
#X*A = B <=> A.T*X.T = B.T
让我们使用
Xt = np.linalg.lstsq(A.T, B.T)
X = Xt.T
这是一个更好的解决方案，无论是在速度和准确性方面使用反演和伪反演，除非你明确需要这些，我强烈建议你不要使用它们
我能够使用伪逆来解决它：
# X * A = B
# X * A * A.tr() = B * A.tr()
# X = B * A.tr() * Inv(A * A.tr())
m2 = dst_pts @ src_pts.transpose() @ np.linalg.inv(src_pts @ src_pts.transpose())