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Is there a matrix $A\in \mathbb{C}^{4\times 4}$ such that: $$A^4=\left(\begin{array}{cccc} 0 & 2 & -1 & 1\\ 0 & 0 & 3 &1\\ 0 & 0& 0 & 4\\ 0 & 0 & 0 & 0 \\ \end{array} \right) ~?$$

Any hint is appreciated. Thanks a lot!

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    $\begingroup$ What do you know about nilpotent matrices? $\endgroup$ – Daniel Fischer Dec 23 '16 at 21:34
  • $\begingroup$ I could say that $B^4=0$, for $B$ the given matrix. $\endgroup$ – Nikolaos Skout Dec 23 '16 at 21:36
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    $\begingroup$ Good. So if there is an $A$ with $A^4 = B$, then what follows about $A$? $\endgroup$ – Daniel Fischer Dec 23 '16 at 21:39
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No. Suppose $A$ is such a matrix. Then, $(A^4)^4=A^{16}=0$ so the minimal polynomial $m(x)$ of $A$ divides $x^{16}$. Since $A$ is $4\times 4$, its minimal polynomial is of degree $\leq 4$ so $m(x)$ divides $x^4$. Then, $A^4=0$, contradiction.

Added: The underlying idea behind this proof is that if an $n\times n$ matrix is nilpotent (i.e. $A^m=0$ for some $m>0$), then in fact $A^n=0$.

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If an $n\times n$ matrix satisfies $A^m=0$ for some $m$ then it satisfies $A^n=0$.

notice that if $A^4$ was equal to that matrix then $(A^4)^4=0$, but then $A^4=0$.

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