Ivan Shishkin, Rye (1878)

Problems/Linear algebraExercise

My first nilpotent matrices

by Sequoia·translated by visitor·
22
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This score reflects both the level of the required concepts and the difficulty of the solution.Ce score tient compte à la fois du niveau des notions nécessaires et de la difficulté de la résolution.

  1. 1–10First steps / middle schoolPremiers pas / collège
  2. 11–25Beginner / high schoolDébutant / lycée
  3. 26–50Intermediate / undergraduateIntermédiaire / licence
  4. 51–70Advanced / graduateAvancé / master
  5. 71–90Expert / specializedExpert / spécialisé
  6. 91–100Research levelNiveau recherche
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Show that the matrices (0100)\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}, (010001000)\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 &0\end{pmatrix}, and (−11−11)\begin{pmatrix} -1 & 1 \\ -1 & 1 \end{pmatrix} are nilpotentFR. Find their nilpotency indicesFR.

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Solution by darktoasterFR

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Premieˋre matrice\textbf{Première matrice}
Par calcul matriciel :(0100)2=(0000)=02\begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} ^2 = \begin{pmatrix} 0 & 0 \\ 0&0 \end{pmatrix} = \textbf{0}_{2}
Par ailleurs (0100)1=(0100)≠02\begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix}^{1} = \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} \ne \textbf{0}_{2}
Donc p=2p= 2 est le plus petit entier naturel tel que (0100)p=02\begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix}^{p} = \textbf{0}_{2} . L’indice de nilpotence de (0100)\begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} est 22.
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Deuxieˋme matrice\textbf{Deuxième matrice}
(010001000)2=(001000000)≠03\begin{pmatrix} 0 & 1 & 0 \\ 0&0 & 1 \\ 0 & 0 &0 \end{pmatrix} ^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0&0 & 0 \\ 0 & 0 &0 \end{pmatrix} \ne \textbf{0}_{3}

(010001000)3=(010001000)2(010001000)=(001000000)(010001000)=03\begin{pmatrix} 0 & 1 & 0 \\ 0&0 & 1 \\ 0 & 0 &0 \end{pmatrix} ^3 = \begin{pmatrix} 0 & 1 & 0 \\ 0&0 & 1 \\ 0 & 0 &0 \end{pmatrix}^{2} \begin{pmatrix} 0 & 1 & 0 \\ 0&0 & 1 \\ 0 & 0 &0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0&0 & 0 \\ 0 & 0 &0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ 0&0 & 1 \\ 0 & 0 &0 \end{pmatrix} = \textbf{0}_{3}

L’indice de nilpotence de (010001000)\begin{pmatrix} 0 & 1 & 0 \\ 0&0 & 1 \\ 0 & 0 &0 \end{pmatrix} est 33.

Troisieˋme matrice\textbf{Troisième matrice}

(−11−11)2=(0000)=02\begin{pmatrix} -1 & 1 \\ -1&1 \end{pmatrix} ^2 = \begin{pmatrix} 0 & 0 \\ 0&0 \end{pmatrix} = \textbf{0}_{2}
Par ailleurs (−11−11)1=(−11−11)≠02\begin{pmatrix} -1 & 1 \\ -1&1 \end{pmatrix} ^1 = \begin{pmatrix} -1 & 1 \\ -1&1 \end{pmatrix} \ne \textbf{0}_{2}
Donc l’indice de nilpotence de (−11−11)\begin{pmatrix} -1 & 1 \\ -1&1 \end{pmatrix} est 22.

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