Calculating the Minimal Polynomial for a Given Matrix A

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To calculate the minimal polynomial for the given matrix A with a characteristic polynomial of (1-λ)⁵, one must identify the monic polynomial of lowest degree that satisfies M(A)=0. It is confirmed that the minimal polynomial's irreducible factors must divide the characteristic polynomial. The minimal polynomial is determined to be (1-λ)², as (I-A)²=0 while (I-A)≠0. The approach of testing powers of (I-A) to find the minimal polynomial is valid and aligns with its definition. This method effectively confirms the minimal polynomial for the matrix A.
gottfried
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I've been given a matrix A and calculated the characteristic polynomial. Which is (1-λ)5. Given this how does one calculate the minimal polynomial?

Also just to check, is it correct that the minimal polynomial is the monic polynomial with lowest degree that satisfies M(A)=0 and that all the irreducible factors of the minimal polynomial divide the characteristic polynomial?

Given this I think the minimal polynomial is (1-λ)2 since (I-A)≠0 and (I-A)2=0 but this method to figure it out seems a little ad hoc.

A=
[1 1 0 0 0]
[0 1 0 0 0]
[0 0 1 1 0]
[0 0 0 1 0]
[0 0 0 0 1]
 
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No, that's perfecty valid. The characteristic polynomial is (1- \lambda)^5 and the minimal polynomial is the polynomial, p, of lowest degree that is a factor of the characteristic polynomial and such that p(A)= 0. The obvious thing to do is to start with the factor of lowest degree, 1- x, that is a factor of that and see if I- A= 0. Since it does not, try (I- A)^2. I presume that did gives 0 but if it had not, you would then try (I- A)^3 and so on. There is nothing "Ad hoc" about using the definition of something.
 
Thanks for clearing it up:)
 
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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