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Let the matrix $A=(a_{ij})=Mat(2, R)$, satisfies $A^{2011}=0$

Prove that: $A^2=0$

Thanks in advance!

Did
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Iloveyou
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4 Answers4

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Consider the minimal polynomial $p(x)$ of $A$. By hypothesis, $p(x) \mid x^{2011}$, whence $p(x) = x^k$ for some $k\in \mathbb{N}$. Also $deg(p(x)) \leq 2$, and so $p(x) = x$ or $p(x) = x^2$.

In either case, $A^2 = 0$.

Prahlad Vaidyanathan
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The purpose of this exercise, of course, is to illustrate that if $A$ is an $n\times n$ nilpotent matrix, then $A^n=0$. However, since $n=2$ here, we have a more elementary proof.

As $A^{2011}=0$, $A$ is singular. So, its two rows are linearly dependent and hence $A=uv^\top$ for some vectors $u$ and $v$. It follows that for $m\ge2$, $A^m=(uv^\top)^m=uv^\top uv^\top\cdots uv^\top=(v^\top u)^{m-1}A$. So, ...

user1551
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Hint: Try to reason on possible characteristic polynomial of A.

Ivan Di Liberti
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$A^n=O_2=> det(A^n)=0=>det(A)=0=>A^2=Tr(A).A=>A^n=[Tr(A)]^{n-1}.A=>[Tr(A)]^{n-1}.A=O_2.$

I. $Tr(A)=0=>A^2=Tr(A).A=O_2;$

II. $A=O_2=>A^2=O_2.$

medicu
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    The matrix $\begin{pmatrix} 0&1\1&0\end{pmatrix}$ has vanishing trace but does not square to zero. – Rasmus Nov 24 '13 at 08:47
  • Case I uses again the partial result that det(A)=0. – Did Nov 24 '13 at 10:03
  • Rasmus, please carefully read the statement of the problem in question. Your example does not satisfy the condition of enunciation:$A^n=O_2$ – medicu Nov 24 '13 at 10:08
  • $Det(A)=-1\neq 0$, thus, i.e. the matrix $\begin{pmatrix}0&1\1&0\end{pmatrix}$ does not satisfy the condition – Iloveyou Nov 24 '13 at 13:24