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If you read this and you’re curious, there’s a proof that is fairly easy to follow if you understand eigenvectors. Write the operation that takes the two last elements of the sequence and produces the following two, notice it’s linear, then analyze the eigenvalues of the associated matrix and relate the original operation to the power method.

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You actually don't need eigenvectors for this proof. Sketch: Write x = lim a_n/a_{n-1} where a_n is the nth term of Fibonacci. Replace a_n with a_{n-1} + a_{n-2}, which will give you a quadratic equation in x. Solving this quadratic equation gives you the golden ratio.

I believe that gives you the result starting from Fibonacci, but not that any starting pair of numbers eventually lands at the golden ratio.

No, it works for any pair of numbers, as long as they are not both zero. I am only using the recursive relation of the Fibonacci sequence, not the starting terms.

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