anonymous
  • anonymous
is there any method of constructing a nilpotent matrix of order 3,whose every entry is nonzero??
MIT 18.06 Linear Algebra, Spring 2010
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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JoshDanziger23
  • 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

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