Similar Matrices Minimal Polynomial of A^3

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SUMMARY

The minimal polynomial of the matrix A^3, given its characteristic polynomial F(x)=(x-1)(x-2)(x-3)(x-4), is derived from the diagonal matrix B with eigenvalues 1, 2, 3, and 4. The diagonalization confirms that A is diagonalizable, leading to A^3 having eigenvalues 1, 8, 27, and 64. Consequently, the minimal polynomial of A^3 is g(x)=(1-x)(8-x)(27-x)(64-x), which matches its characteristic polynomial due to the distinct eigenvalues.

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Homework Statement



Let F(x)=(x-1)(x-2)(x-3)(x-4) be the charecteristic polynomial of A. Find the minimal polynomial of A^3

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The Attempt at a Solution


A is similar to a diagnol matrix with 1,2,3,4 on the diagnol, let's call it B.
We know that A is diagnizable because The degree of the charecteristic polynomial of a matrix is always equal to its size, and if a matrix of size n has n eigenvalues then it si diagnizable
So [tex]A=P^{-1}BP[/tex]
[tex]A^2=P^{-1}BPP^{-1}BP=P^{-1}B^2P[/tex]
[tex]A^3=P^{-1}B^2PP^{-1}BP=P^{-1}B^3P[/tex]

And so B^3 is a diagnizable amtrix with 1, 8, 27, 64 on its diagnol, which means that its charecteristic polynomial is g(x)=(1-x)(8-x)(27-x)(64-x) and because each value in the charecteristic polynomial must apear in the minimal polynomial g(x) si also the minimal polynomial of A^3.
Is that correct?




 
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that sounds reasonable to me, to see it another way, assume u is an eiegnevector of A with e-val lambda, and consider the action of A^3
[tex]A^3u = AAAu = AA\lambda u= A\lambda^2 u= \lambda^3 u[/tex]

so lambda^3 is an eigenvalue of A^3, wth eignevector u
 

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