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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.

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    a) and c) have the same "If" is that correct? – 2017-01-24
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    In c), the change should be "If $A^{2n}=I,$ then $A^{n}=\pm I$". This question appeared in NBHM 2017. I am pretty sure OP wants to ask that question. – 2017-01-24
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    Your counterexample is a special case or the fact that a symmetry with respect to a hyperplane violates b. That fact also provides a counterexample to c. – 2018-02-16

3 Answers 3

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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.

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    sorry for the typo, please see c) now – 2017-01-24
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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

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    sorry for the typo, please see c) now – 2017-01-24
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    The matrix you are considering for c) satisfies $A^2=-I$ – 2017-01-24
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    oh sorry .I have edited .take a look@MathMan – 2017-01-24
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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 $A^n=0.$

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    sorry for the typo, please see c) now – 2017-01-24
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    Sorry, I'm interested in how can one deduce that necessarily the characteristic polynomial is $\lambda^n$ given that the minimal polynomial divides $\lambda^{2n}$. Thanks in advance. – 2017-01-24
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    If $\mu(\lambda)$ divides to $\lambda^{2n}$ then, $\mu(\lambda)=\lambda^q.$ Besides, $\mu(\lambda)$ and $\chi(\lambda)$ have the same irreducible factors and $A$ has order $n$ so, $\chi(\lambda)=\lambda^n.$ – 2017-01-24