Question involving nilpotent or self-inverse matrices.

Question

Let $A\in M_n(\mathbb{R})$, for $n\geq 2$. Which of the following statements are true?

1. $A^{2n}=0 \implies A^n=0$
2. $A^2=I \implies A=\pm I$
3. $A^{2n}=I \implies A^{n}=\pm I$

Answer 1

1. Let $A=U^{-1}(D+N)U$ be the Jordan form for $A$ over the complex numbers, where $D$ is diagonal and $N$ is zero everywhere excep for ones right above the main diagonal. Then $A^{2n}=U^{-1}(D+N)^{n}U=U^{-1}(D^{2n}+...)U$, where all matrices in $...$ are strictly upper triangular. Therefore $D=0$. But then $A^n=U^{-1}N^nU=0$ since $N^{n}=0$.

2. Take a diagonal matrix of size $2$ with a $1$ and a $-1$ in the diagonal.

3. Take the matrix $\left(\begin{matrix}0&0&0&...&0&1\\1&0&0&...&0&0\\0&1&0&...&0&0\\.&.&.&...&.&.\\0&0&0&...&1&0\end{matrix}\right)$ of size $2n$ for $n>1$. For $n=1$ use the previous example. The reason why this matrix works is because its characteristic polynomial is $x^{2n}-1$. Therefore its eigenvalues are the $2n$ roots of unity and in particular different. Therefore, over the complex numbers you can diagonalize it. The equations $A^{2n}=I$ and $A^n=\pm I$ remain the same by passing to the diagonalization. Now, while the diagonal becomes all ones when raised to the power $2n$, not all of them turn real when raised to the power $n$. That is why $A^n$ is not even going to be diagonal.

   If you want to avoid the complex numbers argument you can just notice that this matrix gives a permutation of the vector when applied to the standard basis. It rotates them $e_1\to e_2\to e_3\to...\to e_n\to e_1$. And this applying this permutation over and over doesn't return to the identity until you do the full $2n$ applications.

Answer 2

For 2) Try$$A=\begin{pmatrix}3&8\\-1&-3\end{pmatrix}$$

Answer 3

For $2$ see this my answer, the procedure is similar (description is too long to repeat it again - I hope it is unnecessary) - there are infinitely many matrices $ 2 \times 2$ which give $A^2=I$

The first statement seems to be true as multiplication by $A$ must be at least $-1$ rank lowering operation so if it succeeded to lower it by $2n-1 $ to zero then also it will succeed to lower rank by $n-1$ multiplications (nilpotent matrix has additionally all eigenvalues equal to $0$ so its rank is the most $n-1$ ). If the $n \times n$ matrix is nilpotent then always $A^n =0$.

For 3) rotation in 3d space about $z$ axis by $60^\circ$ is a good example where $R^6=I$ but $R^3 \neq \pm I$