为什么numpy.linalg.solve()比numpy.linalg.inv()提供更精确的矩阵求逆? [英] Why does numpy.linalg.solve() offer more precise matrix inversions than numpy.linalg.inv()?

问题描述

我不太理解为什么numpy.linalg.solve()给出更精确的答案，而numpy.linalg.inv()却有些分解，给出了(我相信是)估计值.

I do not quite understand why numpy.linalg.solve() gives the more precise answer, whereas numpy.linalg.inv() breaks down somewhat, giving (what I believe are) estimates.

对于一个具体的例子，我正在求解方程式C^{-1} * d，其中C表示矩阵，而d是向量数组.为了便于讨论，C的尺寸为形状(1000,1000)，d的形状为(1,1000).

For a concrete example, I am solving the equation C^{-1} * d where C denotes a matrix, and d is a vector-array. For the sake of discussion, the dimensions of C are shape (1000,1000) and d is shape (1,1000).

numpy.linalg.solve(A, b)求解x的等式A*x=b，即x = A^{-1} * b.因此，我可以通过以下方式解决此等式

numpy.linalg.solve(A, b) solves the equation A*x=b for x, i.e. x = A^{-1} * b. Therefore, I could either solve this equation by

(1)

inverse = numpy.linalg.inv(C)
result = inverse * d

或(2)

numpy.linalg.solve(C, d)

方法(2)给出的结果要精确得多.为什么是这样?

Method (2) gives far more precise results. Why is this?

究竟发生了什么，使得一个比另一个更好"?

What exactly is happening such that one "works better" than the other?

推荐答案

np.linalg.solve(A, b)不会不计算 A 的逆.而是调用 gesv LAPACK例程之一，该例程首先将 A 使用LU分解，然后使用正向和反向替换求解 x (请参见此处).

np.linalg.solve(A, b) does not compute the inverse of A. Instead it calls one of the gesv LAPACK routines, which first factorizes A using LU decomposition, then solves for x using forward and backward substitution (see here).

np.linalg.inv使用相同的方法通过求解 A·A中的 A ^-1 来计算 A 的逆. ^-1 = I 其中， I 是身份*.分解步骤与上述步骤完全相同，但是需要更多的浮点运算来求解 A ^-1 (一个 n×n 矩阵)，而不是 x (长为 n 的向量).此外，如果您随后想通过身份 A ^-1 ·b = x 获得 x ，那么额外的矩阵乘法将导致更多的浮动点操作，因此会降低性能并增加数值误差.

np.linalg.inv uses the same method to compute the inverse of A by solving for A^-1 in A·A^-1 = I where I is the identity*. The factorization step is exactly the same as above, but it takes more floating point operations to solve for A^-1 (an n×n matrix) than for x (an n-long vector). Additionally, if you then wanted to obtain x via the identity A^-1·b = x then the extra matrix multiplication would incur yet more floating point operations, and therefore slower performance and more numerical error.

不需要计算 A ^-1 的中间步骤-直接获取 x 更快，更准确.

There's no need for the intermediate step of computing A^-1 - it is faster and more accurate to obtain x directly.

* inv的相关来源为此处.不幸的是，由于它是C模板，所以要理解它有点棘手.需要注意的重要一点是，身份矩阵正作为参数B传递给LAPACK求解器.

* The relevant bit of source for inv is here. Unfortunately it's a bit tricky to understand since it's templated C. The important thing to note is that an identity matrix is being passed to the LAPACK solver as parameter B.

这篇关于为什么numpy.linalg.solve()比numpy.linalg.inv()提供更精确的矩阵求逆?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！