# Non-diagonal n-th roots of the identity matrix

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.)

• 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. Jul 1 at 20:26
• The problem I see is that a matrix can have n different nth roots of a matrix. How do you decide which you want? Jul 1 at 20:38
• @GeorgeIvey "The problem I see is that a matrix can have n different nth roots of a matrix." More, usually, and occasionally fewer. Jul 1 at 20:41
• The companion matrix of the $n$th cyclotomic polynomial has order $\phi(n)$, which is less than $n$. Does that work for you?
– lhf
Jul 1 at 20:48
• 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 $A^n = I$ Jul 1 at 21:19

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

• Yours is the most general example, even in dimension $n$. Can you prove it?
– Ruy
Jul 2 at 0:26
• Thanks. Please check my updated solution for the proof. 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 $$0. You can build new matrices using any invertible matrix $$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.)

• 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. 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.