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https://issues.apache.org/jira/browse/MATH-435?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=12929987#action_12929987
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Mikkel Meyer Andersen edited comment on MATH-435 at 11/9/10 1:55 AM:
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Of course numerical instability still occurs for large powers. One possibility
is to extract scalars of the type 10eX for integer X, i.e. both positive and
negative, and then return X together with the scaled result. This requires a
bit more bookkeeping, but might be a good idea. Of course, if the matrix is to
be used directly, this doesn't change much, but for further processing this
actually might save the day.
was (Author: mikl):
Of course numerical instability still occurs for large powers. One
possibility is to extract scalars of the type 10eX for integer X, i.e. both
positive and negative, and then X together with the scaled result. This
requires a bit more bookkeeping, but might be a good idea. Of course, if the
matrix is to be used directly, this doesn't change much, but for further
processing this actually might save the day.
> Efficient matrix power
> ----------------------
>
> Key: MATH-435
> URL: https://issues.apache.org/jira/browse/MATH-435
> Project: Commons Math
> Issue Type: Improvement
> Reporter: Mikkel Meyer Andersen
> Assignee: Mikkel Meyer Andersen
> Attachments: MATH435-patch1
>
> Original Estimate: 0.07h
> Remaining Estimate: 0.07h
>
> For symmetric matrices A it is easy to find A^n also for large n by making an
> eigenvalue/-vector decomposition.
> In general, if the structure of the matrix is not know and the n'th power is
> needed, A*A*...*A is way too inefficient. By using a binary representation
> and powers of 2, powers can be found far faster similar to finding 5^14 as
> 5^14 = 5^8 * 5^4 = ((5^2)^2)^2 * (5^2)^2 = x3 * x2 where x1 = 5^2, x2 = x1^2,
> and x3 = x2^2, thus saving a lot of computations.
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