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Good afternoon. My question is, if there is any known formula for the expression $A^n \mathbf{x}^n + A^{n-1} \mathbf{x}^{n-1} + \ldots+ A \mathbf{x} + I,$ where $A$ is a matrix and $\mathbf{x}$ is a column vector. By $\mathbf{x}^n$, I denote the element-wise $n$-th power of the column vector $\mathbf{x}$.

What I am looking for is something like a closed-form formula. It seems to be likely that something like that exists, since the expression resembles the geometric sum.

Thanks a lot in advance.

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    I agree with J.M. and Gerry Myerson. In the context of the question it seems to be illogical to calculate $\mathbf{x}^n$ by element-wise and $A^n$ in a usual way. Instead of it$I$would consider $A^n$ in the same way, i.e., $A^n$ is taken element-wise. We don't know the origin of the question. Does (a) it have a motivation from a special research area or (b) you try to generalize some theory? If (a) then say some words about it because it is very interesting (at least for me but I'm not expert in linalg), if (b) then some words about what you want to generalize.2012-08-16

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Edit: The following answer assumes that all components of the vector ${\bf x}$ are $=x$, $x\in{\mathbb R}$.

Let $[1\ 1\ \ldots\ 1]'=:{\bf e}$. For $x\in{\mathbb R}$ your "powers" ${\bf x}^k$ can be written as ${\bf x}^k=\bigl[{\rm diag}(x,x,\ldots ,x)\bigr]^k {\bf e}\ ,$ and as $A$ commutes with $[{\rm diag}(x,x,\ldots, x)\bigr]$ we have $A^k{\bf x}^k=(xA)^k {\bf e}\qquad(k\geq 0)\ .$ By the formula for the sum of a geometric series it follows that $\sum_{k=0}^n A^k{\bf x}^k=\bigl(I-(xA)^{n+1}\bigr)(I- xA)^{-1}\ {\bf e}\ ,$ which makes sense for all $x$ of sufficiently small absolute value.

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    @Patrick Da Silva: Whatever $A$, for sufficiently small $|x|$ the transformation $(I-xA)^{-1}$ is well defined, e.g., by the geometric series.2012-08-16