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Suppose I looked at the equation in $Z^n=1$. Now to construct a matrix which satisfies this equation may seem simple if I just put the roots of the above equation in the diagonal like this: $$ \begin{pmatrix} 1 & 0& 0 & \cdots & 0 \\ 0 & \omega_1& 0 & \cdots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \omega_{n-1} \end{pmatrix} $$

where $\{1,\omega_1,\omega_2,...,\omega_{n-1}\}$, are the $n$ roots of unity. Now the problem arises if the question is slightly changed to a real matrix (except $I_n$). Will there be a real matrix except ($I_n$) which will satisfy this equation??

More generally: Given an equation what guarantees that a matrix will exist which will satisfy the given equation?? Will there always be such a matrix??

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  • $\begingroup$ Could you perhaps clarify your last couple questions? $\endgroup$ Commented May 30, 2014 at 6:40

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Consider any real polynomial $P(X)=X^n-a_{n-1}X^{n-1}-\cdots-a_1X-a_0$, than the matrix $$ A=\left[\matrix{0&0&\cdots&0&a_0\cr 1&0&\cdots&0&a_1\cr 0&1&\cdots&0&a_2\cr \vdots& & & &\vdots\cr 0&0&\cdots&1&a_{n-1}\cr }\right] $$ Satisfies $P(A)=0$. This is called the Companion Matrix.

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There are many examples of such matrices, but one class that springs to mind is cyclic permutation matrices: any $n\times n$ cyclic permutation matrix raised to the $n$-th power gives you back the identity (because it's the equivalent of applying the permutation $n$ times which gives you back the identity permutation).

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Consider the real $n\times n$ matrix $A$ defined by $$a_{i,j}=\cases{1&if $i=j+1$\cr 1&if $i=1$, $j=n$\cr 0&otherwise.\cr}$$ Then $A^n=I$.

Proof. This is the companion matrix of the polynomial $z^n-1$. It therefore has $n$ distinct eigenvalues, namely, the $n$th roots of $1$, so it is similar to a diagonal matrix $D$ with these roots on the diagonal. So $A^n=PD^nP^{-1}=PIP^{-1}=I$.

Comment. This is also a permutation matrix and is therefore a special case of Peter Woolfitt's answer.

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