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I've just began to learn mathematics at university, and I have met this problem which make me stuck at finding a way to solve, that is:

Find all matrix $A$ with size $m$ that product of ($n$) matrices $\underbrace {A \times A \times \cdot \cdot \cdot \times A} _{{n-\text{times}}} = I_m$.
($I_m$ is the identity matrix with size $m$)

Can you help me solve this?

Davood
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Hiếu Phạm Ngọc
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    Write $A$ as $P(D+N)P^{-1}$ (its Jordan form), where $D$ is diagonal, $N$ has only a few $1$'s (maybe none) in the super diagonal and zeros everywhere else, and $P$ is some invertible matrix. Then $I=A^n=P(D+N)^nP^{-1}$ implies that $I=(D+N)^n$. Now use that multiplying $N$ by itself just moves the $1$'s one diagonal up. – Hellen Sep 30 '17 at 11:00
  • (n) matrices? What do you mean? – J Tg Sep 30 '17 at 11:01
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    General method for solution of $A^n=I$? It depends on $m$ and $n$ .. – Widawensen Sep 30 '17 at 11:09
  • @Widawensen It depends on $m$ and $n$ only in a trivial way. – Hellen Sep 30 '17 at 11:11
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    @Hellen What do you mean Hellen by the word 'trivial' in this context? – Widawensen Sep 30 '17 at 11:15
  • @Widawensen It means that $n$ only appears in some denominators and $m$ practically doesn't need to be written. – Hellen Sep 30 '17 at 11:31
  • \begin{eqnarray} \sqrt{\left[ \begin{array}{cc} A & B \ C & D\ \end{array} \right]}=\left[ \begin{array}{cc} \sqrt{A-\frac{BC}{A+D \pm 2 \sqrt{AD-BC}} } & \frac{B}{\sqrt{A+D \pm 2 \sqrt{AD-BC}}} \ \frac{C}{\sqrt{A+D \pm 2 \sqrt{AD-BC}}} & \sqrt{D-\frac{BC}{A+D \pm 2 \sqrt{AD-BC}} } \ \end{array} \right] \end{eqnarray} – Donald Splutterwit Sep 30 '17 at 11:31
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    Let me complete the computation. Since $I=(D+N)^n=D^n+\sum_{k=1}^{n}\binom{n}{k}D^{n-k}N^k$ we must have $N=0$. Notice that $\binom{n}{k}D^{n-k}N^k$ can only have non-zero entries in the $k$-th diagonal above the main diagonal. Therefore $I=D^n$. The solutions of these are the diagonal matrices $D$ with $n$-th roots of unity in the diagonal. Therefore $A=PDP^{-1}$, with $D$ diagonal with $n$-th roots of unity in the diagonal and $P$ any invertible matrix. – Hellen Sep 30 '17 at 11:50
  • @Widawensen The argument computes and proves these are all the solutions. – Hellen Sep 30 '17 at 12:00
  • @Widawensen Yes, as there are infinitely many choices for the invertible matrix $P$ in the parametrization $A=PDP^{-1}$. So? – Hellen Sep 30 '17 at 12:03
  • I have a question to @Hellen, why must we have N = 0? – Hiếu Phạm Ngọc Sep 30 '17 at 15:10
  • @HiếuPhạmNgọc Each term in the summation only has non-zero entries in different diagonals, term $k$ in the diagonal $k$ above the main diagonal. Therefore they don't cancel each other, unless they are zero. They need to be zero above the main diagonal to sum up to $I$. – Hellen Sep 30 '17 at 16:50

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