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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 $A$?


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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"? – joriki Feb 21 '12 at 10:05
By $k=1,N$, do you mean $k=1,\dotsc,N$? What's $N$? – joriki Feb 21 '12 at 10:15
@Joriki: A is defined as a 2 x 2 matrix, not given, sorry – Riccardo.Alestra Feb 21 '12 at 10:16
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 $N$. – joriki Feb 21 '12 at 10:43
@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 $A^2=I+A$! – Jyrki Lahtonen Feb 21 '12 at 13:44
up vote 2 down vote accepted

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.

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