Computing the powers of matrices

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To compute A^25 for the diagonalizable matrix A with the characteristic polynomial p(λ)=(λ-1)^3, it is established that A has a single eigenvalue of 1. By diagonalizing A, it can be expressed as A = I_3, where I_3 is the identity matrix. Consequently, raising A to any power results in A^25 being equal to I_3. The final answer is A^25 = [1 0 0; 0 1 0; 0 0 1].
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Suppose p(λ)=(λ-1)^3 for some diagonalizable matrix A. Calculate A^25.

I'm confused as to how to approach this question without A being given. I thought perhaps I could use the characteristic equation in some way although I am still unsure. I think I could start with using λ=1. Would my matrix then be [ 1 0 0; 0 1 0; 0 0 1] then I would do [1^25 0 0; 0 1^25 0; 0 0 1^25], and finally I would arrive at my answer which would be A^25= [ 1 0 0; 0 1 0; 0 0 1]?

Thank you.
 
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Yes, it's correct.

You know that ##P^{-1} A P = I_3## for some matrix ##P## because ##A## can be diagonalised and has only the eigenvalue ##3##. Consequently ##A = I_3## and the result follows.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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