Powers of Bidiagonal Toeplitz Matrix












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Consider the following bidiagonal $n times n$ Toeplitz matrix $A$, where $0 < p < 1$:
$$A=begin{bmatrix}
1-p & 0 & 0 & cdots & 0\
p & 1-p & 0 && vdots \
0 & ddots & ddots & ddots & 0 \
vdots && p & 1-p & 0\
0 & cdots & 0 & p & 1-p
end{bmatrix}$$



What is $A^m$ for any $m ge 2$? It's easy to show what the matrix is when $n = 2$ for all $m$, but not for general $n$. I've seen several papers on powers of tridiagonal Toeplitz matrices but they assume that the off-by-1 diagonals are all nonzero, but the "upper" diagonal here is all 0s.






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    Consider the following bidiagonal $n times n$ Toeplitz matrix $A$, where $0 < p < 1$:
    $$A=begin{bmatrix}
    1-p & 0 & 0 & cdots & 0\
    p & 1-p & 0 && vdots \
    0 & ddots & ddots & ddots & 0 \
    vdots && p & 1-p & 0\
    0 & cdots & 0 & p & 1-p
    end{bmatrix}$$



    What is $A^m$ for any $m ge 2$? It's easy to show what the matrix is when $n = 2$ for all $m$, but not for general $n$. I've seen several papers on powers of tridiagonal Toeplitz matrices but they assume that the off-by-1 diagonals are all nonzero, but the "upper" diagonal here is all 0s.






    matrices







    share|cite|improve this question

























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      0







      Consider the following bidiagonal $n times n$ Toeplitz matrix $A$, where $0 < p < 1$:
      $$A=begin{bmatrix}
      1-p & 0 & 0 & cdots & 0\
      p & 1-p & 0 && vdots \
      0 & ddots & ddots & ddots & 0 \
      vdots && p & 1-p & 0\
      0 & cdots & 0 & p & 1-p
      end{bmatrix}$$



      What is $A^m$ for any $m ge 2$? It's easy to show what the matrix is when $n = 2$ for all $m$, but not for general $n$. I've seen several papers on powers of tridiagonal Toeplitz matrices but they assume that the off-by-1 diagonals are all nonzero, but the "upper" diagonal here is all 0s.






      matrices







      share|cite|improve this question













      Consider the following bidiagonal $n times n$ Toeplitz matrix $A$, where $0 < p < 1$:
      $$A=begin{bmatrix}
      1-p & 0 & 0 & cdots & 0\
      p & 1-p & 0 && vdots \
      0 & ddots & ddots & ddots & 0 \
      vdots && p & 1-p & 0\
      0 & cdots & 0 & p & 1-p
      end{bmatrix}$$



      What is $A^m$ for any $m ge 2$? It's easy to show what the matrix is when $n = 2$ for all $m$, but not for general $n$. I've seen several papers on powers of tridiagonal Toeplitz matrices but they assume that the off-by-1 diagonals are all nonzero, but the "upper" diagonal here is all 0s.






      matrices




      matrices






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      Ryan

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      3811219






















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          You can rewrite $A$ as
          $$A = (1-p)I + p D $$
          and note that $D^k$ corresponds to non-null elements on the $k^{th}$ sub-diagonal.



          Then, noting that multiplication is commutative for $I$ and the $D^k$ matrices:
          $$A^m = sum_k {m choose k} (1-p)^k p^{m-k} D^{m-k} = sum_k {m choose k} (1-p)^{m-k} p^{k} D^{k}$$



          Note: $D^k = 0$ for $k ge n$






          share|cite|improve this answer























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            You can rewrite $A$ as
            $$A = (1-p)I + p D $$
            and note that $D^k$ corresponds to non-null elements on the $k^{th}$ sub-diagonal.



            Then, noting that multiplication is commutative for $I$ and the $D^k$ matrices:
            $$A^m = sum_k {m choose k} (1-p)^k p^{m-k} D^{m-k} = sum_k {m choose k} (1-p)^{m-k} p^{k} D^{k}$$



            Note: $D^k = 0$ for $k ge n$






            share|cite|improve this answer




























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              You can rewrite $A$ as
              $$A = (1-p)I + p D $$
              and note that $D^k$ corresponds to non-null elements on the $k^{th}$ sub-diagonal.



              Then, noting that multiplication is commutative for $I$ and the $D^k$ matrices:
              $$A^m = sum_k {m choose k} (1-p)^k p^{m-k} D^{m-k} = sum_k {m choose k} (1-p)^{m-k} p^{k} D^{k}$$



              Note: $D^k = 0$ for $k ge n$






              share|cite|improve this answer


























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                2






                You can rewrite $A$ as
                $$A = (1-p)I + p D $$
                and note that $D^k$ corresponds to non-null elements on the $k^{th}$ sub-diagonal.



                Then, noting that multiplication is commutative for $I$ and the $D^k$ matrices:
                $$A^m = sum_k {m choose k} (1-p)^k p^{m-k} D^{m-k} = sum_k {m choose k} (1-p)^{m-k} p^{k} D^{k}$$



                Note: $D^k = 0$ for $k ge n$






                share|cite|improve this answer














                You can rewrite $A$ as
                $$A = (1-p)I + p D $$
                and note that $D^k$ corresponds to non-null elements on the $k^{th}$ sub-diagonal.



                Then, noting that multiplication is commutative for $I$ and the $D^k$ matrices:
                $$A^m = sum_k {m choose k} (1-p)^k p^{m-k} D^{m-k} = sum_k {m choose k} (1-p)^{m-k} p^{k} D^{k}$$



                Note: $D^k = 0$ for $k ge n$







                share|cite|improve this answer














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                Damien

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