Vectorization of matrices power

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Hi all,
Let A a matrix n*n and let N an integer. I wish to create a matrix of power of the form AN = [I A A^2,...,A^N], where I=eye(n), without resorting to a "for" cycle. I have tried to take a look at several commands, including cumprod, kron, etc, and trying to combine them, but I failed. I made it only for the scalar case, i.e. n=1.
After that, I wish to create a matrix
AA = [I 0 0 0;A I 0 0;A^2 A I 0;A^3 A^2 A I]
without using any "for" cycle. I have noticed that the first column of AA is equal to AN' (if it may help). Any hints? Thanks.
  1 个评论
Matt J
Matt J 2013-11-12
It is doubtful that a for-loop is to be feared here. Surely you can't be doing this for N very large?

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采纳的回答

Matt J
Matt J 2013-11-12
>> [AN,T]=matpowers(diag([2,3]),3),
AN =
1 0 2 0 4 0 8 0
0 1 0 3 0 9 0 27
T =
1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
2 0 1 0 0 0 0 0
0 3 0 1 0 0 0 0
4 0 2 0 1 0 0 0
0 9 0 3 0 1 0 0
8 0 4 0 2 0 1 0
0 27 0 9 0 3 0 1
function [AN,T]=matpowers(A,c)
persistent pcell N
if isempty(pcell), pcell={eye(size(A))}; end
if nargin>1, N=c; end
if ~isempty(N) && N>0
pcell=[pcell,{pcell{end}*A}];
N=N-1;
AN=matpowers(A);
else
AN=pcell; pcell=[]; N=[];
end
if nargout>1
z=[AN,{zeros(size(A))}];
T=toeplitz(1:c+1,[1,ones(1,c)*(c+2)]);
T=cell2mat(z(T)) ;
AN=cell2mat(AN);
end

更多回答(2 个)

Sean de Wolski
Sean de Wolski 2013-11-12
This should give you the tools you need:
x = (1:3).^(1:3)
xm = tril(toeplitz(x))

Ubaldo Tiberi
Ubaldo Tiberi 2013-11-13
Thanks for the reply.
Sean, unfortunately it seems that your solution works only with scalar values and not for matrices.
Matt, thanks for the impressive answer! However, I was wondering if I can get the same result without using a function, but just by combining elementary Matlab operations such as blkdiag, kron, cumprod, etc. But I do like this function! However, how would you write a "for" cycle that solves my problem?
Thanks again!
  1 个评论
Matt J
Matt J 2013-11-13
编辑:Matt J 2013-11-13
but just by combining elementary Matlab operations such as blkdiag, kron, cumprod
If A is symmetric you might be able to. But I think a for-loop will be the most efficient, regardless. Are you sure you need to build these matrices explicitly? They contain a lot of redundant data. What are you planning to use them for?

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