Find an invertible non-diagonal $3 \times 3$ matrix $D$ such that $D^3 = D$.

Question

Find an invertible non-diagonal $3 \times 3$ matrix $D$ such that $D^3 = D$.

I have forgotten how to solve this kind of question, can somebody give me some hint or idea how to start?

Answer 1

Since $D^3-D=0$, the minimal polynomial $\mu_D(x)$ divides $x^3-x=x(x^2-1)$, but $D$ is invertible hence $\mu_D(x)$ divides $x^2-1=(x-1)(x+1)$. Since the matrix $D$ is not diagonal, $D\ne\pm I$ hence $\mu_D(x)\ne x\pm1$, hence $\mu_D(x)=(x-1)(x+1)$. No factor of $\mu_D(x)$ has degree greater than $1$ hence $D$ is diagonalizable, with eigenvalues $1$ and $-1$, that is:

The solutions $D$ are the non-diagonal matrices similar to the diagonal matrix with diagonal $(1,1,-1)$ or to the diagonal matrix with diagonal $(1,-1,-1)$.

Example: $D=\begin{pmatrix}1&0&0\\2&1&2\\-2&0&-1\end{pmatrix}$.

Answer 2

An idempotent matrix ($A^2=A$) matrix also has $A^n=A$, so therefore this question reduces to the following question:

Constructing idempotent matrices

Answer 3

The equation $D^3 = D$ is invariant under similarity, i.e. if $D$ is a solution then so is $SDS^{-1}$ for any nonsingular $S$. Find a diagonal matrix that satisfies the equation (and is not a multiple of the identity) and take some random $S$.