For $A\in \mathbb{M}_n(\mathbb{R})$, which of the following are true? 
For $A\in \mathbb{M}_n(\mathbb{R})$, $ n\ge 2$ which of the following statements are true?
  $$\text{a. If}\quad A^{2n}=0, \text{then}\quad A^n=0$$
  $$\text{b. If}\quad A^{2}=I, \text{then}\quad A=\pm I$$
  $$\text{c. If}\quad A^{2n}=I, \text{then}\quad A^n=\pm I$$

I could check that b. is false by taking $A$ to be 2*2 matrix with off-diagonal elements as 1 and diagonal elements as 0.
Other options are true I think if we argue with minimal polynomial, but I'm not very sure.
 A: The first statement is true, and has a proof via eigenvalues and Cayley-Hamilton. The last is false; try to generalize your example for the second problem.
A: for option c) consider n=3 and look at  the below matrix 
$$
        \begin{pmatrix}
        1 & 0 &0  \\
        0 & 1 &1\\
        0 &-1 &0    
\end{pmatrix}
$$
This matrix $A^6$ =I but $A^3$$ \not= I or -I$.you can check .so option c is not correct.The idea is $A^6 -I=0 $ factor as $(A^3-I)(A^3+I)=0$ further it can be factor as $(A-I)(A^2+A+I)(A+I)(A^2-A+I)=0$ Now consider first and forth factor (you can also consider middle two also) and form a matrix whose characteristic polynomial is this.this wiil work .
for option a) it is correct as minimal polynomial must divide anhilating polynomial $A^{2n}=0$ and minimal and characteristic polynomial has same root
A: If $A\in \mathbb{M}_n(\mathbb{R})$ and $A^{2n}=0$, its minimum polinomial $\mu(\lambda)$ divides to $\lambda^{2n}$. Necessarily, its characteristic polynomial is $\chi (\lambda)=\lambda^n$, so $A^n=0.$
