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We define a $"2 \times 2"$matrix $A$. The following recurrence equation is given:$$A^{k+1}=\frac{A^k}{k}+I,(k=1,N)$$where $I$ is the identity matrix.

How can I find the matrix $A$?

Thanks

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    You say that $A$ is given, then ask how to find it. Did you mean "Let $A$ be a $2\times2$ matrix"?2012-02-21
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    By $k=1,N$, do you mean $k=1,\dotsc,N$? What's $N$?2012-02-21
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    @Joriki: A is defined as a 2 x 2 matrix, not given, sorry2012-02-21
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    @Joriki: yes k = 1,..,N where N is a natural number greather than 12012-02-21
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    Are you sure you mean $A^k$ and not $A_k$? This will be impossible for sufficiently large $N$. The matrix $A$ has at least one eigenvalue $a$, and this must satisfy all equations $a^{k+1}=a^k/k+1\;(k=1,\dotsc,N)$, which is impossible for sufficiently large $N$.2012-02-21
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    You say the recurrence is "given". Who gave it? Whether it was homework or not, please make sure you have reported the equation correctly, since (as joriki notes) it doesn't make much sense the way it looks now.2012-02-21
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    @Gerry Merson: suppose $N=4$ In this case, you have:$$A^4=\frac{1}{3}[\frac{1}{2}A^2+I]+I$$ and $A^2=\frac{1}{1}[A+I]$ and so on...2012-02-21
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    I don't know who Gerry Merson is, but (as joriki noted) the procedure you indicate will give you infinitely many equations for $A$, one for each $N$, and there will be no matrix $A$ that satisfies all of them. Your statement is broken. It's time for you to look where you got it, and fix it. Where did you get it?2012-02-21
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    @Gerry Myerson: in fact there are infinite solutions depending on N. The problem is: is it possible to find $A$ as a function of N, assuming $A$ a $2\times2$ matrix?2012-02-21
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    I get $\small A^{k+1}= A \cdot \frac1{k!} + I \cdot {1!+2!+3!+...+k! \over k!} $2012-02-21
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    @Riccardo.Alestra: No! There are several problems remaining in your question. For starters, it is **not** a recurrence relation at all! No matrix $A$ will satisfy all of them. The equation that we get from $k=1$ tells us that $A^2=A+I$. As a consequence of this equation we get that $$A^3=A^2A=(I+A)A=A+A^2=A+(A+I)=2A+I.$$ But you also insist that we have the equation ($k=2$) $$A^3=(A^2/2)+I=\frac{I+A}2+I=\frac32 I+ \frac12 A.$$ Putting these two equations together we get $$\frac32 I+\frac12 A=A^3=I+2A$$ implying that $A=I/3$. But that matrix does not satisfy the equation $A^2=I+A$!2012-02-21
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    @Riccardo.Alestra: In other words: When you write $A^2$ we assume that this is a power of the matrix, so $A^2=AA$, $A^3=AAA$ et cetera. If that is not what you want, you have to use subscripts instead of superscripts. The latter make more sense as exponents here.2012-02-21
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    So, the answer is: there is no matrix $A$ satisfying the relation:$$A^{k+1}=\frac{A^k}{k}+I$$. Good.2012-02-21
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    @Riccardo.Alestra: The answer given by my calculation was that there is no matrix $A$ that satisfies the relation $$A^{k+1}=\frac{A^k}k+I$$ for both $k=1$ and $k=2$. It doesn't say anything about a single equation. If your source is convince about the existence of solutions, then we still don't know, what is really being asked!2012-02-21
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    @JyrkiLahtonen: the equations comes from an autoregressive model, but I see that my interpretetion of the physical phenomenon is wrong. Thanks.2012-02-22

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Based on the comments, I'm going to hazard a guess that the problem is actually,

Given a positive integer $k$, how can we find a $2\times2$ matrix $A$ such that $A^{k+1}=(1/k)A^k+I$.

The algebraic equation $kx^{k+1}-x^k-1=0$ is guaranteed to have solutions (indeed, at least one positive real solution); let $\alpha$ be a solution to this algebraic equation. Then $$A=\pmatrix{\alpha&0\cr0&\alpha\cr}$$ is a solution to $A^{k+1}=(1/k)A^k+I$.

Notice that $\alpha$ depends on $k$; we are getting a different matrix $A$ for each positive integer $k$. My apologies if this was not the question OP had in mind.