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Is there a better/closed form for the Cauchyproduct $$A^k + A^{k-1}(A+I)/2 + A^{k-2}((A+I)/2)^2 + ... +( (A+I)/2)^k$$ ?Here $I$ is the identity and $A$ the upper subdiagonal filled with the unit ($a_{r,c}= \delta_{r+1,c}$). Clearly $A$ and $I$ commute so we can do some reformulations, but unfortunately $A$ is not invertible.

$$A=\small {\begin{bmatrix} 0&1&0&.&.&... \\ .&0&1&0&.&... \\ .&.&0&1&0&... \\ ...\end{bmatrix} }$$

Gottfried Helms
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  • Do you mean $A_{ij}=1$ if $j\gt i$ and $A_{ij}=0$ if $j\leqslant i$? – Did Sep 18 '13 at 07:22
  • @Did: I've updated the question. It is also answered (so simple...). Thanks anyway! – Gottfried Helms Sep 18 '13 at 08:22
  • Next time you must deal with matrix A, you might want to use the phrase Jordan matrix. – Did Sep 18 '13 at 08:24
  • @Did: I see what you mean. Well, here the matrix $A$ occured in a context, where it might be understood as the matrix $A$ whose powerseries $I + A + A^2 + A^3 + ... = B$ where $B$ is the the upper triangular unit-matrix. So I didn't consider and was far off the eigen-/Jordan decomposition here. (Here is the source of my problem:http://math.stackexchange.com/questions/486807/how-much-weight-is-on-each-person-in-a-human-pyramid/495492#495492) The matrix $A$ here is there the matrix $L$ and here: $(A+I)$ is there: $M$ – Gottfried Helms Sep 18 '13 at 08:57

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When $A$ is strictly triangular and $B=\frac{A+I}2$, the matrix difference $A-B=\frac{A-I}2$ is invertible. Since $A$ and $B$ commute, $A^k+A^{k-1}B+\ldots+AB^{k-1}+B^k=(A-B)^{-1}(A^{k+1}-B^{k+1})$.

user1551
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