最陡的下降吐出了不合理的大价值

Question

我对最陡下降求解Ax = b的实现显示出一些奇怪的行为:对于任何足够大的矩阵(〜10 x 10,到目前为止只测试了方形矩阵),返回x包含所有巨大的值(大约为1x10^10).

def steepestDescent(A, b, numIter=100, x=None):
    """Solves Ax = b using steepest descent method"""
    warnings.filterwarnings(action="error",category=RuntimeWarning)

    # Reshape b in case it has shape (nL,)
    b = b.reshape(len(b), 1)

    exes = []
    res = []

    # Make a guess for x if none is provided
    if x==None:
        x = np.zeros((len(A[0]), 1))
        exes.append(x)

    for i in range(numIter):
        # Re-calculate r(i) using r(i) = b - Ax(i) every five iterations
        # to prevent roundoff error. Also calculates initial direction
        # of steepest descent.
        if (numIter % 5)==0:
            r = b - np.dot(A, x)
        # Otherwise use r(i+1) = r(i) - step * Ar(i)
        else:
            r = r - step * np.dot(A, r)

        res.append(r)

        # Calculate step size. Catching the runtime warning allows the function
        # to stop and return before all iterations are completed. This is
        # necessary because once the solution x has been found, r = 0, so the
        # calculation below divides by 0, turning step into "nan", which then
        # goes on to overwrite the correct answer in x with "nan"s
        try:
            step = np.dot(r.T, r) / np.dot( np.dot(r.T, A), r )
        except RuntimeWarning:
            warnings.resetwarnings()
            return x
        # Update x
        x = x + step * r
        exes.append(x)

    warnings.resetwarnings()
    return x, exes, res

(exes并res返回进行调试)

我认为问题必须是计算r或step(或更深层次的问题),但我无法弄清楚它是什么.

Answer 1

该代码似乎是正确的。例如，以下测试对我有用（linalg.solve 和 最陡峭的下降在大多数情况下都给出了接近的答案）：

import numpy as np

n = 100
A = np.random.random(size=(n,n)) + 10 * np.eye(n)
print(np.linalg.eig(A)[0])
b = np.random.random(size=(n,1))
x, xs, r = steepestDescent(A,b, numIter=50)
print(x - np.linalg.solve(A,b))

问题出在数学上。如果A是正定矩阵，则该算法保证收敛到正确的解。通过将 10 * 单位矩阵添加到随机矩阵，我们增加了所有特征值为正的概率

如果您使用大型随机矩阵进行测试（例如A = random.random(size=(n,n))，您几乎可以肯定具有负特征值，并且算法将不会收敛。