矩阵功率和 [英] Matrix power sum

问题描述

什么是计算矩阵比如A ^ I + A的总和的最好办法^（I + 1）+ A ^ I + 2 ........ A ^ n表示非常大的n？

What is the best way to calculate sum of matrices such as A^i + A^(i+1) + A^i+2........A^n for very large n?

我没有想到的两种可能的方式：

1）使用对数矩阵幂（LME）的A ^我，然后通过乘法计算以后的矩阵。

1) Use logarithmic matrix exponentiation(LME) for A^i, then calculate the subsequent matrices by multiplying by A.

问题：并没有真正利用LME算法，因为我用它只能实现低功耗！

Problem: Doesn't really take advantage of the LME algorithm as i am using it only for the lowest power!!

2）使用LME查找^ n和memoize的中间计算。

2)Use LME for finding A^n and memoize the intermediate calculations.

问题：需要太多的空间大的n

Problem: Too much space required for large n.

有没有第三条路？

推荐答案

注意：

A + A^2 = A(I + A)
A + A^2 + A^3 = A(I + A) + A^3
A + A^2 + A^3 + A^4 = (A + A^2)(I + A^2)
                    = A(I + A)(I + A^2)

让

B(n) = A + ... + A^n

我们有：

B(1) = A
B(n) = B(n / 2) * (I + A^(n / 2)) if n is even
B(n) = B(n / 2) * (I + A^(n / 2)) + A^n if n is odd

所以，你会做的步骤的对数数和没有必要计算逆。

So you will be doing a logarithmic number of steps and there is no need to compute inverses.

虽然直接实施将导致（log n）的^ 2 的因素，你可以在保持日志N 通过计算 A的权力为你计算 B 。

While a direct implementation will lead to a (log n)^2 factor, you can keep it at log n by computing the powers of A as you compute B.

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