Binary numbers

[David C. Ullrich]

In article <960onh$ov3$1 at nnrp1.deja.com>,
  sampe99 at my-deja.com wrote:
>
>
> > How are you planning on deciding which one of the many n-th roots
> > of your matrix is the "right" one? (Is there some reason to
> > think that your matrix _has_ an n-th root which is also a
> > transition matrix???)
>
> All matrices with negative or complex elements can be excluded and
> those left can be ranked after certain criterias. The question if
there
> exist a transition matrix or not among the roots is still left
> unanswered. My intuitive feeling is that there exists one
> unique "correct" matrix

Ah.

> after my experiments on 3x3 and 4x4 matrices
> (calculations by hand).
>
> If I have a transistion matrix and want to transform it to a longer
> time horizon I take the n:th power of it (under certain assumptions).

A lot of things have n-th powers (for integers n) but no square
roots. I _think_ that a "transition matrix" is an nxn matrix
with all entries non-negative and with each row summing to 1.
Assuming so: A = [[0,1], [1,0]] is a transition matrix with
no real square root, much less a square root which is a transition
matrix.

(A has eigenvalues 1 and -1, so any square root of A will
have one real eigenvalue and one complex eigenvalue. A
real matrix can have complex eigenvalues but they come
in conjugate pairs, so the square roots of A cannot be real
matrices.)

> Sent via Deja.com
> http://www.deja.com/
>

--
Oh, dejanews lets you add a sig - that's useful...


Sent via Deja.com
http://www.deja.com/