Let $B$ be an $n \times n$ matrix. Does $B^3 = O_n$ imply $B = O_n$? If so, why?
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3Not true when $n>1$. Look up "nilpotent matrix" from your textbook or from the internet. – user1551 Oct 21 '19 at 14:04
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Thanks, Didn't learn that yet. – Dvir Peretz Oct 21 '19 at 14:06
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2Put all entries zero, except for the one in the top right corner. – conditionalMethod Oct 21 '19 at 14:06
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I suppose that $B_n$ is a $n\times n$-matrix. Did you mean $N=n$? – Dietrich Burde Oct 21 '19 at 14:19
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No, take the matrix $$ \begin{pmatrix}0 &\ldots &0 &1 &0 \\ \vdots & & &0 &1 \\ \vdots & & & &0 \\ \vdots & & & &\vdots \\ 0 &\ldots &\ldots &\ldots &0 \end{pmatrix} $$
Tuvasbien
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For $n=2$ or $n\ge 2$ we can take $\begin{pmatrix} 0 & 1 \cr 0 & 0 \end{pmatrix}$ for your example (which is for $n>2$). – Dietrich Burde Oct 21 '19 at 14:17
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@DietrichBurde how can we take his example for $n \ge 2$? I can't read your comment. – mathworker21 Nov 09 '19 at 01:34