本来想做绕轴旋转，绕轴旋转是在垂直于轴向量的空间里旋转，但是n维空间里与某个向量垂直的空间为n-1维，而旋转只在二维空间里有定义。所以这里就改成了，向任意方向旋转。

计算单位向量 X = ( x 1 , x 2 , ⋯ , x n ) X=(x_1,x_2,\cdots,x_n) X=(x1​,x2​,⋯,xn​)旋转到单位向量 V = ( v 1 , v 2 , ⋯ , v n ) V=(v_1,v_2,\cdots,v_n) V=(v1​,v2​,⋯,vn​)的旋转矩阵 R R R。

旋转坐标系

任意的旋转都可以看作绕着一个轴，在某个平面上的旋转。

不失一般性，假定向量 V = ( v 1 , v 2 , ⋯ , v n ) V=(v_1,v_2,\cdots,v_n) V=(v1​,v2​,⋯,vn​)在 v 1 × v 2 v_1\times v_2 v1​×v2​张成的平面上旋转，用矩阵乘法可表示为： V ′ = R 1 V V'=R_1V V′=R1​V其中旋转矩阵 R 1 R_1 R1​定义为

R 1 = ( cos ⁡ α − sin ⁡ α sin ⁡ α cos ⁡ α 1 ⋱ 1 ) R_1=\left( \begin{array}{ccc} \cos\alpha& -\sin\alpha& & & \\ \sin\alpha& \cos\alpha& & & \\ & &1 & & \\ & & & \ddots & \\ & & & & 1 \\ \end{array} \right) R1​=⎝⎜⎜⎜⎜⎛​cosαsinα​−sinαcosα​1​⋱​1​⎠⎟⎟⎟⎟⎞​

其中 cos ⁡ α = v 2 v 1 2 + v 2 2 \cos\alpha=\frac{v_2}{\sqrt{v_1^2+v_2^2}} cosα=v12​+v22​ ​v2​​， sin ⁡ α = v 1 v 1 2 + v 2 2 \sin\alpha=\frac{v_1}{\sqrt{v_1^2+v_2^2}} sinα=v12​+v22​ ​v1​​。

由此我们将向量 V V V旋转到与 ( v 1 , 0 , ⋯ , 0 ) (v_1,0,\cdots,0) (v1​,0,⋯,0)正交。得 V ′ = ( 0 , v 1 2 + v 2 2 , v 3 , v 4 , ⋯ , v n ) V'=(0,\sqrt{v_1^2+v_2^2}, v_3,v_4,\cdots,v_n) V′=(0,v12​+v22​ ​,v3​,v4​,⋯,vn​)。

以此类推，将 V V V旋转到 ( 0 , ⋯ , 0 , 1 ) (0,\cdots,0,1) (0,⋯,0,1)需要 n − 1 n-1 n−1次旋转。将这些旋转变换组合为一个旋转矩阵 R 0 = R n − 1 R n − 2 ⋯ R 1 R_0=R_{n-1}R_{n-2}\cdots R_1 R0​=Rn−1​Rn−2​⋯R1​（注意，乘法不满足交换律，顺序不能颠倒）。顺便我们可以计算出这个旋转的逆矩阵 R 0 − 1 = R 1 ⊤ ⋯ R n − 2 ⊤ R n − 1 ⊤ R_0^{-1}=R_1^\top\cdots R_{n-2}^\top R_{n-1}^\top R0−1​=R1⊤​⋯Rn−2⊤​Rn−1⊤​。

旋转向量

现在我们要做的是将 R 0 X R_0X R0​X旋转到坐标轴 ( 0 , ⋯ , 0 , 1 ) (0,\cdots,0,1) (0,⋯,0,1)，并求旋转矩阵。做法与上面相同。最后可以求得，在旋转后的坐标系下，旋转矩阵为 R X R_X RX​: R 0 V = R X R 0 X R_0V=R_XR_0X R0​V=RX​R0​X

虽然 R R R和 R X R_X RX​表示的旋转角度相同，但是这两个旋转所在的坐标系是不同的，还需要将 R X R_X RX​的坐标轴旋转回去：

R = R 0 − 1 R X R 0 R=R_0^{-1}R_XR_0 R=R0−1​RX​R0​

代码实现

import numpy as npdef Rot_map(V):assert len(V.shape) == 1assert np.linalg.norm(V) - 1 < 1e-8n_dim = V.shape[0]Rot = np.eye(n_dim)Rot_inv = np.eye(n_dim)for rotate in range(n_dim-1):rot_mat = np.eye(n_dim)rot_norm = np.sqrt(V[rotate]**2 + V[rotate+1]**2)cos_theta = V[rotate+1]/rot_normsin_theta = V[rotate]/rot_normrot_mat[rotate,rotate] = cos_thetarot_mat[rotate,rotate+1] = - sin_thetarot_mat[rotate+1,rotate] = sin_thetarot_mat[rotate+1,rotate+1] = cos_thetaV = np.dot(rot_mat, V)Rot = np.dot(rot_mat, Rot)Rot_inv = np.dot(Rot_inv,rot_mat.transpose())return Rot, Rot_invn_dim = 512X = np.random.rand(n_dim)X = X / np.linalg.norm(X)V = np.random.rand(n_dim)V = V / np.linalg.norm(V)R_0, R_0_inv = Rot_map(V)R_X, _ = Rot_map(np.dot(R_0, X))R = np.dot(np.dot(R_0_inv, R_X), R_0)assert np.linalg.norm(np.dot(R,X) - V) < 1e-8

扩展

这个方法可以用来在正交于某个向量的空间中随机采样。例如，在 n n n维空间中，采样与 V = ( v 1 , v 2 , ⋯ , v n ) V=(v_1,v_2,\cdots,v_n) V=(v1​,v2​,⋯,vn​)正交的向量，则代码如下：

import numpy as npdef Rot_map(V):assert len(V.shape) == 1assert np.linalg.norm(V) - 1 < 1e-8z_dim = V.shape[0]Rot = np.eye(z_dim)Rot_inv = np.eye(z_dim)for rotate in range(z_dim-1):rot_mat = np.eye(z_dim)rot_norm = np.sqrt(V[rotate]**2 + V[rotate+1]**2)cos_theta = V[rotate+1]/rot_normsin_theta = V[rotate]/rot_normrot_mat[rotate,rotate] = cos_thetarot_mat[rotate,rotate+1] = - sin_thetarot_mat[rotate+1,rotate] = sin_thetarot_mat[rotate+1,rotate+1] = cos_thetaV = np.dot(rot_mat, V)Rot = np.dot(rot_mat, Rot)Rot_inv = np.dot(Rot_inv,rot_mat.transpose())return Rot, Rot_invz_dim = 512n_samples = 10000V = np.random.rand(z_dim)V = V / np.linalg.norm(V)R_0, R_0_inv = Rot_map(V)samples = np.random.rand(z_dim-1, n_samples)samples = np.concatenate((samples, np.zeros((1, n_samples))))samples = np.dot(R_0_inv, samples)assert np.mean(np.abs(np.dot(V, samples))) < 1e-8

进一步，我们采样与一组向量 V s Vs Vs正交的向量：

import numpy as npdef gs(X, row_vecs=True, norm = True):'''/iizukak/1287876/edad3c337844fac34f7e56ec09f9cb27d4907cc7'''if not row_vecs:X = X.TY = X[0:1,:].copy()for i in range(1, X.shape[0]):proj = np.diag((X[i,:].dot(Y.T)/np.linalg.norm(Y,axis=1)**2).flat).dot(Y)Y = np.vstack((Y, X[i,:] - proj.sum(0)))if norm:Y = np.diag(1/np.linalg.norm(Y,axis=1)).dot(Y)if row_vecs:return Yelse:return Y.Tdef Rot_map(V):assert len(V.shape) == 1assert np.linalg.norm(V) - 1 < 1e-8z_dim = V.shape[0]Rot = np.eye(z_dim)Rot_inv = np.eye(z_dim)for rotate in range(z_dim-1):rot_mat = np.eye(z_dim)rot_norm = np.sqrt(V[rotate]**2 + V[rotate+1]**2)cos_theta = V[rotate+1]/rot_normsin_theta = V[rotate]/rot_normrot_mat[rotate,rotate] = cos_thetarot_mat[rotate,rotate+1] = - sin_thetarot_mat[rotate+1,rotate] = sin_thetarot_mat[rotate+1,rotate+1] = cos_thetaV = np.dot(rot_mat, V)Rot = np.dot(rot_mat, Rot)Rot_inv = np.dot(Rot_inv,rot_mat.transpose())return Rot, Rot_invz_dim = 10 # number of dimensionsn_vectors= 2 # number of Vsassert n_vectors < z_dimn_samples = 10 # number of samples to orthogonal VsVs = np.random.rand(n_vectors, z_dim)# Gram-Schmidt Orthogonizationorth_Vs = gs(Vs)orth_Vs = orth_Vs.TRots = np.eye(z_dim)Rot_invs = np.eye(z_dim)for idx in range(n_vectors):V = orth_Vs[:, idx]R_0 = np.eye(z_dim)R_0_inv = np.eye(z_dim)R_0[0:z_dim-idx,0:z_dim-idx], R_0_inv[0:z_dim-idx,0:z_dim-idx] = Rot_map(V[range(z_dim-idx)])orth_Vs = R_0.dot(orth_Vs)Rots = np.dot(R_0, Rots)Rot_invs = np.dot(Rot_invs, R_0_inv)samples = np.random.rand(z_dim-n_vectors, n_samples)samples = np.concatenate((samples, np.zeros((n_vectors, n_samples))))samples = np.dot(Rot_invs, samples)assert np.mean(np.abs(np.dot(Vs, samples))) < 1e-8

用矩阵旋转太慢了，上述采样操作可以用矩阵分解解决(在任意向量的正交空间中采样)：

import numpy as npz_dim = 10 # number of dimensionsn_vectors= 2 # number of Vsassert n_vectors < z_dimn_samples = 20 # number of samples to orthogonal VsVs = np.random.rand(n_vectors, z_dim)Vs = np.concatenate((Vs,np.zeros((z_dim-n_vectors, z_dim))))U, s, V = np.linalg.svd(Vs)directions = np.eye(z_dim)directions[range(n_vectors),range(n_vectors)] = 0directions = np.dot(np.dot(U, directions), V)samples = np.random.rand(z_dim, n_samples)samples = np.dot(directions.T, samples)assert np.mean(np.abs(np.dot(Vs, samples))) < 1e-8