Q. Are there $n$-th root analogs of this non-diagonal cube-root of the $3 \times 3$ identity matrix?

\begin{align*} \left( \begin{array}{ccc} 0 & 0 & -i \\ i & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right)^3 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) \end{align*}

I am looking for $A^n=I$, where $I$ is any dimension $\le n$.

(A naive question: I am not an expert in this area.)

  • 3
    $\begingroup$ I'm not sure what features you want of your "analogue" here. Are you just looking for a non-diagonal $A$ such that $A^n = I$? Does it have to be $3 \times 3$? Or maybe $n \times n$? Does it have to match some feature of how the non-zero terms are laid out (e.g. one non-zero term in each row/column)? It would be good to get some more information about what you'd like to see in an answer. $\endgroup$ Jul 1 at 20:26
  • $\begingroup$ The problem I see is that a matrix can have n different nth roots of a matrix. How do you decide which you want? $\endgroup$ Jul 1 at 20:38
  • $\begingroup$ @GeorgeIvey "The problem I see is that a matrix can have n different nth roots of a matrix." More, usually, and occasionally fewer. $\endgroup$ Jul 1 at 20:41
  • $\begingroup$ The companion matrix of the $n$th cyclotomic polynomial has order $\phi(n)$, which is less than $n$. Does that work for you? $\endgroup$
    – lhf
    Jul 1 at 20:48
  • 3
    $\begingroup$ Hi, Joseph. The permutation matrix of a cyclic permutation always works. Apparently you can alter the final pair of 1's into $i$ and $-i$ and still get $A^n = I$ $\endgroup$
    – Will Jagy
    Jul 1 at 21:19

4 Answers 4


We want to solve $A^n = I $ where $A$ and $I$ are $m \times m$

Let $ A = P \Lambda P^{-1} $

where $\Lambda = \text{diag}(\omega_1, \omega_2, \dots , \omega_m) $

where the $\omega_k$'s are $n$-roots of unity. That is, $\omega_k = \exp\big(\dfrac{i 2 \pi j}{n} \big)$ ,$ k = 1,2,\dots,m $ and $ j \in \{ 0, 1, 2, ...., n-1 \} $

and $P$ is any invertible $ m \times m $ matrix.

Then $A^n = \big(P \Lambda P^{-1}\big)\big(P \Lambda P^{-1}\big)\dots \big(P \Lambda P^{-1}\big) = P^{-1} \Lambda^n P = P^{-1} I P = I $

  • $\begingroup$ Yours is the most general example, even in dimension $n$. Can you prove it? $\endgroup$
    – Ruy
    Jul 2 at 0:26
  • $\begingroup$ Thanks. Please check my updated solution for the proof. $\endgroup$ Jul 2 at 1:13

You can take a rotation matrix that rotates $\phi=2\pi/n$ around some axis, for example in 3 dimensions:

$$R_\phi=\begin{pmatrix} \cos \phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi &0 \\ 0 & 0 & 1\\ \end{pmatrix}$$

This won't work for $n=2$ though, because $R_\phi$ is diagonal in that case like $R_\phi=\operatorname{diag}(-1,-1,1)$.

Then $R_\phi^n=I$ and $R_\phi^k\neq I$ for any $0<k<n$. You can build new matrices using any invertible matrix $A$ and conjugate like

$$R_{\phi,A}:= AR_\phi A^{-1} \tag 1$$

then obviously:

$$R_{\phi,A}^n = (AR_\phi A^{-1})^n = A R_\phi^nA^{-1} = I$$

(I am not sure whether this could fix the case $n=2$ and produce some valid (non-diagonal) results.)

  • $\begingroup$ You can use Wolfram Alpha to check that {{1,1,0},{0,0,1},{1,0,0}}^-1 * {{-1,0,0},{0,-1,0},{0,0,1}} * {{1,1,0},{0,0,1},{1,0,0}} is not diagonal. Of course, for n=2 you can also use a reflection instead of a rotation, I.e. exchange two basis vectors. $\endgroup$
    – Carsten S
    Jul 2 at 14:39

If $A^n = I$, i.e., $A^n - I = 0$, then the minimal polynomial $m_A$ of $A$ divides $p(x) := x^n - 1$. (For $n > 1$) one nondiagonal solution is the companion matrix of $p$ itself, namely, $$C_p := \pmatrix{\cdot & 1 \\ I_{n - 1} & \cdot}$$ (indeed, $p = m_{C_p}$).

Notice that $C_p$ is the permutation matrix for the permutation (in fact $n$-cycle) $$\pmatrix{1 & 2 & \cdots & n - 1 & n}$$ of $n$ elements. More generally, the permutation matrix of any permutation of $n$ elements of order dividing $n$ (and not the identity permutation) yields another solution.

There are uncountably many solutions for any $n > 1$: Given any solution $A$ and any invertible matrix $S$, $SAS^{-1}$ is another solution provided it is not diagonal (since diagonality is a closed condition that by hypothesis is not always satisfied).


The most general solution of the equation $A^n=1$ among $m\times m$ complex matrices is $$ A=SDS^{-1}, $$ where $S$ is an invertible matrix and $D$ is a diagonal matrix whose diagonal entries are $n^{th}$ roots of unity.

The reason is that the minimal polynomial of such a matrix $A$ divides the polynomial $x^n-1$, and hence has distinct roots, which in turn implies that $A$ is diagonalizable.


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