In linear algebra I have an equation $x_n = Ax_0$. I know the values of $x$ for any given value of $n$, and I know $x_0$. Both are $2\times 1$ matrices. How do I solve for $A$? The answer should be a $2\times2$ matrix with entries $[1 1; 1 0]$. This is for Lucas Numbers.

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$\mathbf x_n = A \mathbf x_{n-1}\\ \mathbf x_n = A^n \mathbf x_0$

That is:

$\mathbf x_n = \begin{bmatrix}x_{n+1}\\x_{n}\end{bmatrix}\\ \mathbf x_{n+1} = \begin{bmatrix}x_{n+1}+x_n\\x_{n+1}\end{bmatrix}\\ \mathbf x_{n+1} = \begin{bmatrix}1&1\\1&0\end{bmatrix}\mathbf x_n\\ $

How do you find $A^n$? Diagonalize $A, A = P^{-1} D P \implies A^n =P^{-1} D^n P$

Doug M

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