## anonymous one year ago is there any method of constructing a nilpotent matrix of order 3,whose every entry is nonzero??

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1. JoshDanziger23

kartikpandey, According to Wikipedia the canonical form for the nilpotent matrix of order 3 is $S=\left[\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{matrix}\right]$and this certainly satisfies $$S^3=0$$ (with $$S$$ and $$S^2$$ both non-zero). We can construct similar matrices $$A$$ using any invertible $$P$$ by setting $$A=P ^{-1}SP$$; then $$A^3=P ^{-1}SPP ^{-1}SPP ^{-1}SP=P ^{-1}S^3P=0$$, so $$A$$ will be nilpotent. Trying the first invertible matrix that came into my head $P=\left[\begin{matrix}1& 2& 3 \\ 4& 5& 6 \\ 7& 8& 10\end{matrix}\right]$ gives $A=\left[\begin{matrix}-12&-14&\frac{-52}{3}\\23&26&\frac{98}{3}\\-10&-11&-14\end{matrix}\right]$which seems to be nilpotent, order 3, with no zeroes. Best wishes Josh