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I am given the following matrix $$A=\begin{bmatrix} 0 & 0 & 4 & 1\\ 0& 0 & 1 & 4\\ 4 & 1 & 0 &0\\ 1 & 4 & 0 & 0 \end{bmatrix}$$ And I have to find the minimal polynomial of the matrix. The characteristic polynomial is $$K(\lambda)=-(\lambda -5)(\lambda +5)(\lambda-3)(\lambda +3)$$ The minimal polynomial $m(\lambda)$ divides the characteristic polynomial. I know that the characteristic polynomial is the minimal, but how do i eliminate the possibilities of the linear, quadratic and qubic factors in the polynomial. When do i know the minimal is actually the characteristic polynomial?

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    $\begingroup$ If the characteristic polynomial has distinct roots, then it is the minimal polynomial. $\endgroup$ – Lord Shark the Unknown Jul 10 '17 at 17:16
  • $\begingroup$ I think you can just set all the exponents of the factored characteristic polynomial to 1 and then you get the minimal polynomial. Going the other way means figuring out which exponents the factors should have. $\endgroup$ – mathreadler Jul 10 '17 at 17:20
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    $\begingroup$ Neat observation: the matrix can be written as $$ \pmatrix{0&1\\1&0} \otimes \pmatrix{4&1\\1&4} $$ (where $\otimes$ denotes the Kronecker product) which makes the eigenvalue computation pretty quick. $\endgroup$ – Omnomnomnom Jul 10 '17 at 17:27
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    $\begingroup$ The point is that all roots of the characteristic polynomial are eigenvalues and therefore must be roots of the minimal polynomial, because if $A v = \lambda v$ and $v \ne 0$, $0 = P(A) v = P(\lambda) v$ implies $P(\lambda) = 0$. $\endgroup$ – Robert Israel Jul 10 '17 at 17:57
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The key fact is that the minimal polynomial and the characteristic polynomial have the same irreducible factors.

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    $\begingroup$ Indeed the characteristic polynomial divides some power of the minimal polynomial and the minimal polynomial divides the characteristic polynomial, hence the claim. $\endgroup$ – Alex Ortiz Jul 10 '17 at 17:23
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In this case you can do easily without the characteristic polynomial. It is easy to see that even powers of $A$ will have their nonzero entries in the $2\times2$ blocks at the top left and bottom right, and the odd powers of $A$ have them (like $A$ itself) in the bottom left and top right $2\times2$ blocks. You are looking for a power$~A^k$ that equals a linear combination of lower powers, and by what I just remarked the exponents of the contributing lower powers will have the same parity as$~k$. Since $A$ is nonsingular a minimal polynomial with only odd degree terms is not possible (it would have $0$ as a root), so the smallest $k$ for which a relation exists will be even. Also you know that the minimal polynomial has degree at most$~4$, the size of the matrix. So all there is to it is compute $A^0=I$, $A^2$ and $A^4$ and find a linear combination.

Concretely $A^2$ has as top left block $B=(\begin{smallmatrix}17&8\\8&17\end{smallmatrix})$ (repeated at the bottom right), and $A^4$ has as as top left block $B^2=(\begin{smallmatrix}353&272\\272&353\end{smallmatrix})$. Now $B^2$ minus $272/8=34$ times $B$ is a multiple of the identity, or you can apply the Cayley-Hamilton theorem for $B$; either way you find $B^2-34B+225I_2=0$ whence $A^4-34A^2+225I_4=0$, and $X^4-34X^2+225$ is the minimal polynomial you were after.

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