如何用简洁命令构造由基矩阵幂次构成的分块矩阵？

Great question! When you need to construct a block matrix B = [A; A²; A³; ...; Aⁿ] from a base matrix A, there are cleaner, more elegant alternatives to explicit for loops. Here are my go-to approaches:

1. One-Liner with arrayfun + cell2mat #

While you mentioned arrayfun, you can condense it into a single, readable line without a full loop. This is perfect for small to moderate values of n:

n = 3; % Replace with your desired positive integer
A = [1, 2; -1, 3];
B = cell2mat(arrayfun(@(k) A^k, 1:n, 'UniformOutput', false));

How it works:

• arrayfun(@(k) A^k, 1:n, 'UniformOutput', false) generates a cell array where each element is A^k for k from 1 to n.
• cell2mat then vertically stacks all these matrices into your desired block matrix B.

2. Efficient Recursive Approach (Cayley-Hamilton Theorem) #

For larger values of n, computing each matrix power individually can be inefficient. Instead, use the Cayley-Hamilton Theorem, which states that any matrix satisfies its own characteristic equation. This lets us express higher powers of A as linear combinations of lower powers, reducing redundant calculations:

n = 5; % Example large n
A = [1, 2; -1, 3];

% Get coefficients of A's characteristic polynomial
char_poly = poly(A);

% Initialize cell array to hold matrix powers
powers = cell(1, n);
powers{1} = A;
if n >= 2, powers{2} = A^2; end

% Recursively compute higher powers using Cayley-Hamilton
for k = 3:n
    % Combine lower powers using the characteristic equation
    powers{k} = -char_poly(1:end-1) * cell2mat(powers(k-1:-1:k-length(char_poly)+1));
end

% Stack into the final block matrix
B = cell2mat(powers);

This method shines when n is large because it avoids recalculating matrix products from scratch for each power.

Quick Note on Element-Wise vs Matrix Powers #

Just to clarify: since you mentioned scalar A.^(1:n), remember that A^k is matrix multiplication (what you need here), while A.^k is element-wise exponentiation—make sure you use the correct operator for your use case!

内容的提问来源于stack exchange，提问作者ThomasIsCoding