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
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
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.