. There exists a nondiagonal matrix $A\in \mathcal {M}_n (\mathbb{R}) $ s.t. $A^{k+1}=I_n $ and $I_n-A $ invertible?

Question

Let $k\in \mathbb {N} $. There exists a nondiagonal matrix $A\in \mathcal {M}_n (\mathbb{R}) $ s.t. $A^{k+1}=I_n $ and $I_n-A $ invertible?

Answer 1

• If $k$ is odd, then you can always take $A = -I_n$ as in the other answer.
• If $k$ is even and $n$ is even too, then you can take the block matrix, where $B$ appears $n/2$ times: $$A = \begin{pmatrix} B & 0 & 0 &\dots & 0 \\ 0 & B & 0 & \dots & 0 \\ \vdots & \ddots & \ddots & \ddots& \vdots \\ 0 && 0 & B &0\\ 0 & \dots & 0 & 0 & B \end{pmatrix},$$ and the rotation matrix $B$ is given by: $$B = \begin{pmatrix} \cos\bigl(\frac{2\pi}{k+1)}\bigr) & \sin\bigl(\frac{2\pi}{k+1}\bigr) \\ -\sin\bigl(\frac{2\pi}{k+1}\bigr) & \cos\bigl(\frac{2\pi}{k+1}\bigr) \end{pmatrix}.$$ The eigenvalues of $A$ are all nonreal, hence $A-I_n$ is invertible.
• If $k$ is even and $n$ is odd, then $A$ must have a real eigenvalue $\lambda$, because its polynomial characteristic has odd degree (and every real-valued odd-degree polynomial has a real root). But then $\lambda^{k+1} = 1$, hence $\lambda = 1$. Therefore $A - I_n$ has $0$ as an eigenvalue and isn't invertible.

Answer 2

If $k $ is odd, you can take $A=-I $. So the answer in this case is yes.

If $k $ is even and $n$ is odd, then $A^{k+1} $ has a real eigenvalue $\lambda $ such that $\lambda ^{k+1}=1$. So $\lambda=1$, and so $0$ is an eigenvalue of $I-A $. So the answer in this case is no.

For $k$ even and and $n$ even, see the example in Najib's answer.