From 6a0d886f5241a3234389ce0952af8be0e7710782 Mon Sep 17 00:00:00 2001
From: wizardforcel <562826179@qq.com>
Date: Thu, 11 Oct 2018 21:59:19 +0800
Subject: [PATCH] 10

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+# 十、实现 QR 分解
+
+我们在计算特征值时使用 QR 分解并计算最小二乘回归。 它是数值线性代数中的重要组成部分。
+
+“数值线性代数中的一种算法比其他算法更重要：QR 分解。” --Trefethen，第 48 页
+
+回想一下，对于任何矩阵`A`，`A = QR`，其中`Q`是正交的，`R`是上三角。
+
+提醒：我们在上一课中看到的 QR 算法使用 QR 分解，但不要混淆二者。
+
+### NumPy 中
+
+```py
+import numpy as np
+
+np.set_printoptions(suppress=True, precision=4)
+
+n = 5
+A = np.random.rand(n,n)
+npQ, npR = np.linalg.qr(A)
+```
+
+检查`Q`是正交的：
+
+```py
+np.allclose(np.eye(n), npQ @ npQ.T), np.allclose(np.eye(n), npQ.T @ npQ)
+
+# (True, True)
+```
+
+检查`R`是三角。
+
+```py
+npR
+
+'''
+array([[-0.8524, -0.7872, -1.1163, -1.2248, -0.7587],
+       [ 0.    , -0.9363, -0.2958, -0.7666, -0.632 ],
+       [ 0.    ,  0.    ,  0.4645, -0.1744, -0.3542],
+       [ 0.    ,  0.    ,  0.    ,  0.4328, -0.2567],
+       [ 0.    ,  0.    ,  0.    ,  0.    ,  0.1111]])
+'''
+```
+
+当向量`b`投影到直线`a`上时，其投影`p`是`b`沿着直线`a`的一部分。
+
+让我们看看 [沉浸式线性代数在线版](http://immersivemath.com/ila/index.html)的[第 3.2.2 节：投影](http://immersivemath.com/ila/ch03_dotproduct/ch03.html)的交互图。
+
+![](img/projection_line.png)
+
+> 来源：[沉浸式数学](http://immersivemath.com/ila/ch03_dotproduct/ch03.html)
+
+以下是将向量投影到平面上的样子：
+
+![](img/projection.png)
+
+> 来源：[最小二乘回归的线性代数视角](https://medium.com/@andrew.chamberlain/the-linear-algebra-view-of-least-squares-regression-f67044b7f39b)
+
+当向量`b`投影到直线`a`上时，其投影`p`是`b`沿着直线`a`的一部分。 所以`p`是`a`的一些倍数。 设 ![\mathbf{p} = \hat{x}\mathbf{a}](img/tex-a32cff3b4bdd4f46771f4cc817a60308.gif) 其中 ![\hat{x}](img/tex-2a95aaaf954c2187999c6357b04a58dd.gif) 是标量。
+
+### 正交性
+
+投影的关键是正交性：从`b`到`p`的直线（可以写成 ![\mathbf{b} - \hat{x}\mathbf{a}](img/tex-bb4e955e77268f56bb0aa2b892c69ea5.gif)）垂直于`a`。
+
+这意味着：
+
+![\mathbf{a} \cdot (\mathbf{b} -  \hat{x}\mathbf{a}) = 0](img/tex-69d4b6b19f75f7e5b87932c6e2d651ae.gif)
+
+所以：
+
+![\hat{x} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a} \cdot \mathbf{a}}](img/tex-e188ba54d741c47cccdec77b8ef7c5e5.gif)
+
+## Gram-Schmidt
+
+### 经典的 Gram-Schmidt（不稳定）
+
+对于每列`j`，计算单一投影：
+
+![v_j = P_ja_j](img/tex-efd8ffa0fdc4e994808665f543b0fc78.gif)
+
+其中 ![P_j](img/tex-7f57ce5c29b329529f4e3f9a3765b114.gif) 与 ![q_1, ..., q_{j-1}](img/tex-e2d55fffd53a113312172c3c24cc951c.gif) 的跨度正交的空间。
+
+```py
+def cgs(A):
+    m, n = A.shape
+    Q = np.zeros([m,n], dtype=np.float64)
+    R = np.zeros([n,n], dtype=np.float64)
+    for j in range(n):
+        v = A[:,j]
+        for i in range(j):
+            R[i,j] = np.dot(Q[:,i], A[:,j])
+            v = v - (R[i,j] * Q[:,i])
+        R[j,j] = np.linalg.norm(v)
+        Q[:, j] = v / R[j,j]
+    return Q, R
+
+Q, R = cgs(A)
+
+np.allclose(A, Q @ R)
+
+# True
+```
+
+检查`Q`是酉矩阵。
+
+```py
+np.allclose(np.eye(len(Q)), Q.dot(Q.T))
+
+# True
+
+np.allclose(npQ, -Q)
+
+# True
+
+R
+
+'''
+array([[ 0.02771,  0.02006, -0.0164 , ...,  0.00351,  0.00198,  0.00639],
+       [ 0.     ,  0.10006, -0.00501, ...,  0.07689, -0.0379 , -0.03095],
+       [ 0.     ,  0.     ,  0.01229, ...,  0.01635,  0.02988,  0.01442],
+       ..., 
+       [ 0.     ,  0.     ,  0.     , ...,  0.     , -0.     , -0.     ],
+       [ 0.     ,  0.     ,  0.     , ...,  0.     ,  0.     , -0.     ],
+       [ 0.     ,  0.     ,  0.     , ...,  0.     ,  0.     ,  0.     ]])
+'''
+```
+
+Gram-Schmidt 应该让你想起来一点 Arnoldi 迭代（用于将矩阵转换为海森堡形式），因为它也是一个结构化的正交化。
+
+### 改进版 Gram-Schmidt
+
+经典（不稳定的）Gram-Schmidt：对于每列`j`，计算单一投影：
+
+![v_j = P_ja_j](img/tex-efd8ffa0fdc4e994808665f543b0fc78.gif)
+
+其中 ![P_j](img/tex-7f57ce5c29b329529f4e3f9a3765b114.gif) 与 ![q_1, ..., q_{j-1}](img/tex-e2d55fffd53a113312172c3c24cc951c.gif) 的跨度正交的空间。
+
+改进版 Gram-Schmidt：对于每列`j`，计算`n - 1`个投影：
+
+![P_j = P_{\perp q_{j-1}\cdots\perp q_{2}\perp q_{1}}](img/tex-6f2e0e91ebd226538ef465b081f4468f.gif)
+
+```py
+import numpy as np
+n = 3
+A = np.random.rand(n,n).astype(np.float64)
+
+def cgs(A):
+    m, n = A.shape
+    Q = np.zeros([m,n], dtype=np.float64)
+    R = np.zeros([n,n], dtype=np.float64)
+    for j in range(n):
+        v = A[:,j]
+        for i in range(j):
+            R[i,j] = np.dot(Q[:,i], A[:,j])
+            v = v - (R[i,j] * Q[:,i])
+        R[j,j] = np.linalg.norm(v)
+        Q[:, j] = v / R[j,j]
+    return Q, R
+
+def mgs(A):
+    V = A.copy()
+    m, n = A.shape
+    Q = np.zeros([m,n], dtype=np.float64)
+    R = np.zeros([n,n], dtype=np.float64)
+    for i in range(n):
+        R[i,i] = np.linalg.norm(V[:,i])
+        Q[:,i] = V[:,i] / R[i,i]
+        for j in range(i, n):
+            R[i,j] = np.dot(Q[:,i],V[:,j])
+            V[:,j] = V[:,j] - R[i,j]*Q[:,i]
+    return Q, R
+
+Q, R = mgs(A)
+
+np.allclose(np.eye(len(Q)), Q.dot(Q.T.conj()))
+
+# True
+
+np.allclose(A, np.matmul(Q,R))
+
+# True
+```
+
+## Householder
+
+### 引言
+
+![\begin{array}{ l | l | c } \hline Gram-Schmidt & Triangular\, Orthogonalization & A R_1 R_2 \cdots R_n = Q  \\ Householder  & Orthogonal\, Triangularization & Q_n \cdots Q_2 Q_1 A = R  \\ \hline \end{array}](img/tex-3fee5b26f9c6ce2571715fb9b282f431.gif)
+
+Householder 反射产生更接近正交的矩阵`Q`，具有舍入误差
+
+Gram-Schmidt 可以部分停止，留下`A`的前`n`列的简化 QR。
+
+### 初始化
+
+```py
+import numpy as np
+n = 4
+A = np.random.rand(n,n).astype(np.float64)
+
+Q = np.zeros([n,n], dtype=np.float64)
+R = np.zeros([n,n], dtype=np.float64)
+
+A
+
+'''
+array([[ 0.5435,  0.6379,  0.4011,  0.5773],
+       [ 0.0054,  0.8049,  0.6804,  0.0821],
+       [ 0.2832,  0.2416,  0.8656,  0.8099],
+       [ 0.1139,  0.9621,  0.7623,  0.5648]])
+'''
+
+from scipy.linalg import block_diag
+
+np.set_printoptions(5)
+```
+
+### 算法
+
+我添加了更多的计算和更多的信息，因为它说明了算法的工作原理。 此版本也返回 Householder 反射。
+
+```py
+def householder_lots(A):
+    m, n = A.shape
+    R = np.copy(A)
+    V = []
+    Fs = []
+    for k in range(n):
+        v = np.copy(R[k:,k])
+        v = np.reshape(v, (n-k, 1))
+        v[0] += np.sign(v[0]) * np.linalg.norm(v)
+        v /= np.linalg.norm(v)
+        R[k:,k:] = R[k:,k:] - 2*np.matmul(v, np.matmul(v.T, R[k:,k:]))
+        V.append(v)
+        F = np.eye(n-k) - 2 * np.matmul(v, v.T)/np.matmul(v.T, v)
+        Fs.append(F)
+    return R, V, Fs
+```
+
+检查`R`是上三角。
+
+```py
+R
+
+'''
+array([[-0.62337, -0.84873, -0.88817, -0.97516],
+       [ 0.     , -1.14818, -0.86417, -0.30109],
+       [ 0.     ,  0.     , -0.64691, -0.45234],
+       [-0.     ,  0.     ,  0.     , -0.26191]])
+'''
+```
+
+作为检查，我们将使用分块矩阵`F`计算 ![Q^T](img/tex-8b7f1b39b8e0258e80aea39ca1c24265.gif) 和`R`。矩阵`F`是 householder 反射。
+
+请注意，这不是一种处理`Q`的有效计算方式。在大多数情况下，你实际上并不需要`Q`。例如，如果你使用 QR 来求解最小二乘，则只需要`Q * b`。
+
++   对于隐式计算乘积`Q * b`或`Qx`的技巧，请参阅 Trefethen 第 74 页。
++   请参阅[这些讲义](http://www.cs.cornell.edu/~bindel/class/cs6210-f09/lec18.pdf)，了解 Householder 的不同实现，它同时计算`Q`，作为`R`的一部分。
+
+```py
+QT = np.matmul(block_diag(np.eye(3), F[3]), 
+               np.matmul(block_diag(np.eye(2), F[2]), 
+                         np.matmul(block_diag(np.eye(1), F[1]), F[0])))
+
+F[1]
+
+'''
+array([[-0.69502,  0.10379, -0.71146],
+       [ 0.10379,  0.99364,  0.04356],
+       [-0.71146,  0.04356,  0.70138]])
+'''
+
+block_diag(np.eye(1), F[1])
+
+'''
+array([[ 1.     ,  0.     ,  0.     ,  0.     ],
+       [ 0.     , -0.69502,  0.10379, -0.71146],
+       [ 0.     ,  0.10379,  0.99364,  0.04356],
+       [ 0.     , -0.71146,  0.04356,  0.70138]])
+'''
+
+block_diag(np.eye(2), F[2])
+
+'''
+array([[ 1.     ,  0.     ,  0.     ,  0.     ],
+       [ 0.     ,  1.     ,  0.     ,  0.     ],
+       [ 0.     ,  0.     , -0.99989,  0.01452],
+       [ 0.     ,  0.     ,  0.01452,  0.99989]])
+'''
+
+block_diag(np.eye(3), F[3])
+
+'''
+array([[ 1.,  0.,  0.,  0.],
+       [ 0.,  1.,  0.,  0.],
+       [ 0.,  0.,  1.,  0.],
+       [ 0.,  0.,  0., -1.]])
+'''
+
+np.matmul(block_diag(np.eye(1), F[1]), F[0])
+
+'''
+array([[-0.87185, -0.00861, -0.45431, -0.18279],
+       [ 0.08888, -0.69462,  0.12536, -0.70278],
+       [-0.46028,  0.10167,  0.88193, -0.00138],
+       [-0.14187, -0.71211,  0.00913,  0.68753]])
+'''
+
+QT
+
+'''
+array([[-0.87185, -0.00861, -0.45431, -0.18279],
+       [ 0.08888, -0.69462,  0.12536, -0.70278],
+       [ 0.45817, -0.112  , -0.88171,  0.01136],
+       [ 0.14854,  0.71056, -0.02193, -0.68743]])
+'''
+
+R2 = np.matmul(block_diag(np.eye(3), F[3]), 
+               np.matmul(block_diag(np.eye(2), F[2]),
+                         np.matmul(block_diag(np.eye(1), F[1]), 
+                                   np.matmul(F[0], A))))
+
+np.allclose(A, np.matmul(np.transpose(QT), R2))
+
+# True
+
+np.allclose(R, R2)
+
+# True
+```
+
+这是 Householder 的简洁版本（尽管我创建了一个新的`R`，而不是覆盖`A`，并原地计算它）。
+
+```py
+def householder(A):
+    m, n = A.shape
+    R = np.copy(A)
+    Q = np.eye(m)
+    V = []
+    for k in range(n):
+        v = np.copy(R[k:,k])
+        v = np.reshape(v, (n-k, 1))
+        v[0] += np.sign(v[0]) * np.linalg.norm(v)
+        v /= np.linalg.norm(v)
+        R[k:,k:] = R[k:,k:] - 2 * v @ v.T @ R[k:,k:]
+        V.append(v)
+    return R, V
+
+RH, VH = householder(A)
+```
+
+检查`R`是对角的。
+
+```py
+RH
+
+'''
+array([[-0.62337, -0.84873, -0.88817, -0.97516],
+       [-0.     , -1.14818, -0.86417, -0.30109],
+       [-0.     , -0.     , -0.64691, -0.45234],
+       [-0.     ,  0.     ,  0.     , -0.26191]])
+'''
+
+VH
+
+'''
+[array([[ 0.96743],
+        [ 0.00445],
+        [ 0.2348 ],
+        [ 0.09447]]), array([[ 0.9206 ],
+        [-0.05637],
+        [ 0.38641]]), array([[ 0.99997],
+        [-0.00726]]), array([[ 1.]])]
+'''
+
+np.allclose(R, RH)
+
+# True
+
+def implicit_Qx(V,x):
+    n = len(x)
+    for k in range(n-1,-1,-1):
+        x[k:n] -= 2*np.matmul(v[-k], np.matmul(v[-k], x[k:n]))
+
+A
+
+'''
+array([[ 0.54348,  0.63791,  0.40114,  0.57728],
+       [ 0.00537,  0.80485,  0.68037,  0.0821 ],
+       [ 0.2832 ,  0.24164,  0.86556,  0.80986],
+       [ 0.11395,  0.96205,  0.76232,  0.56475]])
+'''
+```
+
+经典和改良的 Gram-Schmidt 都需要`2mn^2`个浮点运算。
+
+### 陷阱
+
+有些事情需要注意：
+
++   当你复制值时 VS 当你有两个指向同一内存位置的变量时
++   长度为`n`的向量与`1 x n`矩阵之间的差异（`np.matmul`以不同方式处理它们）
+
+## 类比
+
+
+| | `A=QR` | `A=QHQ*` |
+| --- | --- | --- |
+| 正交结构化 | Householder | Householder |
+| 结构化正交 | Gram-Schmidt | Arnoldi |
+
+Gram-Schmidt 和 Arnoldi：连续的三角运算，可以部分停止，前`n`列是正确的。
+
+Householder：连续的正交运算。 在存在舍入误差的情况下产生更接近正交的`A`。
+
+请注意，要计算海森堡化简`A = QHQ *`，将 Householder 反射应用于`A`的两侧，而不是仅应用于一侧。
+
+## 示例
+
+以下示例来自 Trefethen 和 Bau 的第 9 讲，尽管从 MATLAB 翻译成 Python。
+
+### 示例：经典与改进的 Gram-Schmidt
+
+这个例子是 Trefethen 第 9 节的实验 2。 我们想要构造一个方阵`A`，它具有随机奇异向量和广泛变化的奇异值，间隔为 ![2^{-1}](img/tex-3522f8b8d6b2b912a41177e213f96dd2.gif) 和 ![2^{-(n + 1)}](img/tex-9eee2c1c197de7cf3b522d1cf624846f.gif) 之间的 2 的倍数。
+
+```py
+import matplotlib.pyplot as plt
+from matplotlib import rcParams
+%matplotlib inline
+
+n = 100
+U, X = np.linalg.qr(np.random.randn(n,n))   # 将 U 设为随机正交矩阵
+V, X = np.linalg.qr(np.random.randn(n,n))   # 将 V 设为随机正交矩阵
+S = np.diag(np.power(2,np.arange(-1,-(n+1),-1), dtype=float))  # 将 S 设为对角矩阵 w/ exp 
+                                                              # 值在 2^-1 和 2^-(n+1) 之间
+
+A = np.matmul(U,np.matmul(S,V))
+
+QC, RC = cgs(A)
+QM, RM = mgs(A)
+
+plt.figure(figsize=(10,10))
+plt.semilogy(np.diag(S), 'r.', basey=2, label="True Singular Values")
+plt.semilogy(np.diag(RM), 'go', basey=2, label="Modified Gram-Shmidt")
+plt.semilogy(np.diag(RC), 'bx', basey=2, label="Classic Gram-Shmidt")
+plt.legend()
+rcParams.update({'font.size': 18})
+```
+
+![](img/10-1.png)
+
+```py
+type(A[0,0]), type(RC[0,0]), type(S[0,0])
+
+# (numpy.float64, numpy.float64, numpy.float64)
+
+eps = np.finfo(np.float64).eps; eps
+
+# 2.2204460492503131e-16
+
+np.log2(eps), np.log2(np.sqrt(eps))
+
+# (-52.0, -26.0)
+```
+
+### 示例：正交性的数值损失
+
+这个例子是 Trefethen 第 9 节的实验 3。
+
+```py
+A = np.array([[0.70000, 0.70711], [0.70001, 0.70711]])
+
+A
+
+'''
+array([[ 0.7    ,  0.70711],
+       [ 0.70001,  0.70711]])
+'''
+```
+
+Gram-Schmidt：
+
+```py
+Q1, R1 = mgs(A)
+```
+
+Householder：
+
+```py
+R2, V, F = householder_lots(A)
+Q2T = np.matmul(block_diag(np.eye(1), F[1]), F[0])
+```
+
+NumPy 的 Householder：
+
+```py
+Q3, R3 = np.linalg.qr(A)
+```
+
+检查 QR 分解是否能用：
+
+```py
+np.matmul(Q1, R1)
+
+'''
+array([[ 0.7   ,  0.7071],
+       [ 0.7   ,  0.7071]])
+'''
+
+np.matmul(Q2T.T, R2)
+
+'''
+array([[ 0.7   ,  0.7071],
+       [ 0.7   ,  0.7071]])
+'''
+
+np.matmul(Q3, R3)
+
+'''
+array([[ 0.7   ,  0.7071],
+       [ 0.7   ,  0.7071]])
+'''
+```
+
+检查`Q`多么接近完美正交。
+
+```py
+np.linalg.norm(np.matmul(Q1.T, Q1) - np.eye(2))  # 改进的 Gram-Schmidt
+
+# 3.2547268868202263e-11
+
+np.linalg.norm(np.matmul(Q2T.T, Q2T) - np.eye(2))  # 我们的 Householder 实现
+
+# 1.1110522984689321e-16
+
+np.linalg.norm(np.matmul(Q3.T, Q3) - np.eye(2))  # Numpy（它使用 Householder）
+
+# 2.5020189909116529e-16
+```
+
+GS（Q1）不如 Householder（Q2T，Q3）稳定。
diff --git a/img/10-1.png b/img/10-1.png
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-- 
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