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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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    Presumably $I$ is supposed to be a column vector2012-08-16
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    Why don't you try a simple example, say a 2-by-2 matrix with small entries and $x=(1,2)$, and see whether any pattern emerges?2012-08-16
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    "...since the expression resembles the geometric sum." - maybe, but taking the componentwise power of a vector isn't as natural as taking powers of a matrix, so I'm not so optimistic.2012-08-16
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    @Henry By $I$ I denote the identity matrix ($I = A^0 \mathbf{x}^0$). It is not very important there, anyway.2012-08-16
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    @GerryMyerson I have tried that, but I could not see anything that would resemble a pattern. So I became interested, if there is a known (perhaps less trivial) formula for this...2012-08-16
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    @042: What Henry meant is that it does not make much sense to add a vector to a matrix (unless you interpret the vector as a matrix which has all but one rows 0... but then your element-wise power would be even more confusing than it is as it stands). Also, $A^{0}x^{0}\neq I$, at least not with the usual definition of $I$...2012-08-16
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    I'm with J. M. - $x^n$ is unnatural, and only looks like but doesn't really act like $A^n$, so I don't expect much. But I haven't actually tried it, so I don't know. And I think $I$ has to be the all-ones vector for there to be any chance of having a formula.2012-08-16
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    @tomasz Yes, of course, my error.2012-08-16
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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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    The components of $x$ aren't necessarily all the same2012-08-16
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    The difficulty of components of $x$ not being all the same can be handled at the expense of introducing an extra summation, i.e. expressing $x$ as a linear combination of the standard basis vectors $e_i$.2012-08-16
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    @hardmath This will work.2012-08-16
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    How do you even define $(I - xA)^{-1}$? This seems to me like a sketchy answer. Or at least it's not clear enough to me ; the notation is troubling. I think what would make this answer understandable is if we could see how the operations are defined componentwise because right now it's just hard to guess. On your defense, maybe it is also OP's job to clarify his notation.2012-08-16
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    @Cocopuffs: You are of course right. I shall correspondingly edit my answer.2012-08-16
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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