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Find the minimal polynomial of $A=\begin{pmatrix} 0 &0 & 0 &1\\ 1 & 0 & 0&0\\ 0 &1 &0 &0\\ 0 &0 & 1&0\end{pmatrix}$

Since this is a square matrix, the minimal polynmial will be such that $p(A)=0$. How do I find such a polynomial?

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  • $\begingroup$ That's not a square matrix, it's $5 \times 4$. And it's not lower triangular either. $\endgroup$ – Robert Israel Nov 18 '15 at 20:28
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    $\begingroup$ Hint: This is a permutation matrix $\endgroup$ – Omnomnomnom Nov 18 '15 at 20:35
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    $\begingroup$ $\phantom{}$ $A^4 = I$. $\endgroup$ – Jack D'Aurizio Nov 18 '15 at 21:08
  • $\begingroup$ @JackD'Aurizio thank you for the observation, can you explain how I can use that fact? $\endgroup$ – grayQuant Nov 18 '15 at 21:27
  • $\begingroup$ @grayQuant: In addition to Jack's point, compute $A^n$ for $0\le n\le 3$ and observe that no two of them has a nonzero element in the same position -- so they are certainly linearly independent. $\endgroup$ – Henning Makholm Nov 18 '15 at 21:37
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Note the eigenvalues of $A$ are $\pm 1,\pm i$, each of which has the multiplicity 1. So the minimal polynomial is just the characteristic polynomial $p(x)=x^4-1$.

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