Matrix Powers - General Expression for Mk to the Nth Power

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The discussion focuses on deriving a general expression for the matrix Mk, defined as (k+1 k-1) (k-1 k+1), raised to the nth power. The proposed expression is M^{N}_{K} = 2^{n-1}([(k+1) +(k - 1)\sum^{n}_{x=1} k^{x}] [(k-1) +(k - 1)\sum^{n}_{x=1} k^{x}])([(k - 1) +(k - 1)\sum^{n}_{x=1} k^{x}] [(k + 1) +(k - 1)\sum^{n}_{x=1} k^{x}]). The discussion emphasizes that the matrix Mk can be diagonalized due to its symmetric nature, leading to the expression M^n = S^(-1).D^n.S, where D is the diagonal matrix. This systematic approach is crucial for accurate computation of Mk to the nth power.

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This problem first appeared on another thread under Statistics and probability. I found it when I got the same problem, which is as follows

Consider matrices in the form (k+1 k-1)
(k-1 k+1)

We will call this matrix Mk, find a general expression for Mk to the nth power in terms of k and n.


I tried several different matrices of this form, the general expression i came up with is attached in the thumbnail, but I still tried making it somewhat clear with latex.

M^{N}_{K} = 2^{n-1}\left( [(k+1) +(k - 1)\sum^{n}_{x=1} k^{x}] [(k-1) +(k - 1)\sum^{n}_{x=1} k^{x}] \right)
\left([(k - 1) +(k - 1)\sum^{n}_{x=1} k^{x}] [(k + 1) +(k - 1)\sum^{n}_{x=1} k^{x}] \right)
 

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I don't think that's quite right, if that was your question. But there is a systematic way to do this. M can be diagonalized since it's symmetric. Find the matrix which diagonalizes it so M=S^(-1).D.S where D is diagonal. Then M^n=S^(-1).D^n.S.
 
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