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Show that the matrices ( 0 1 0 0 ) \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} ( 0 0 1 0 ) , ( 0 1 0 0 0 1 0 0 0 ) \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 &0\end{pmatrix} 0 0 0 1 0 0 0 1 0 , and ( − 1 1 − 1 1 ) \begin{pmatrix} -1 & 1 \\ -1 & 1 \end{pmatrix} ( − 1 − 1 1 1 ) are nilpotentFR . Find their nilpotent index .
Solutions 1 Reveal solutions Are you sure? Give it a try first. Premi e ˋ re matrice \textbf{Première matrice} Premi e ˋ re matrice Par calcul matriciel :( 0 1 0 0 ) 2 = ( 0 0 0 0 ) = 0 2 \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} ^2 = \begin{pmatrix} 0 & 0 \\ 0&0 \end{pmatrix} = \textbf{0}_{2} ( 0 0 1 0 ) 2 = ( 0 0 0 0 ) = 0 2 Par ailleurs ( 0 1 0 0 ) 1 = ( 0 1 0 0 ) ≠ 0 2 \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix}^{1} = \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} \ne \textbf{0}_{2} ( 0 0 1 0 ) 1 = ( 0 0 1 0 ) = 0 2 Donc p = 2 p= 2 p = 2 est le plus petit entier naturel tel que ( 0 1 0 0 ) p = 0 2 \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix}^{p} = \textbf{0}_{2} ( 0 0 1 0 ) p = 0 2 . L’indice de nilpotence de ( 0 1 0 0 ) \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} ( 0 0 1 0 ) est 2 2 2 .\newline Deuxi e ˋ me matrice \textbf{Deuxième matrice} Deuxi e ˋ me matrice ( 0 1 0 0 0 1 0 0 0 ) 2 = ( 0 0 1 0 0 0 0 0 0 ) ≠ 0 3 \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} 0 0 0 1 0 0 0 1 0 2 = 0 0 0 0 0 0 1 0 0 = 0 3
( 0 1 0 0 0 1 0 0 0 ) 3 = ( 0 1 0 0 0 1 0 0 0 ) 2 ( 0 1 0 0 0 1 0 0 0 ) = ( 0 0 1 0 0 0 0 0 0 ) ( 0 1 0 0 0 1 0 0 0 ) = 0 3 \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} 0 0 0 1 0 0 0 1 0 3 = 0 0 0 1 0 0 0 1 0 2 0 0 0 1 0 0 0 1 0 = 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 = 0 3
L’indice de nilpotence de ( 0 1 0 0 0 1 0 0 0 ) \begin{pmatrix} 0 & 1 & 0 \\ 0&0 & 1 \\ 0 & 0 &0 \end{pmatrix} 0 0 0 1 0 0 0 1 0 est 3 3 3 .
Troisi e ˋ me matrice \textbf{Troisième matrice} Troisi e ˋ me matrice
( − 1 1 − 1 1 ) 2 = ( 0 0 0 0 ) = 0 2 \begin{pmatrix} -1 & 1 \\ -1&1 \end{pmatrix} ^2 = \begin{pmatrix} 0 & 0 \\ 0&0 \end{pmatrix} = \textbf{0}_{2} ( − 1 − 1 1 1 ) 2 = ( 0 0 0 0 ) = 0 2 Par ailleurs ( − 1 1 − 1 1 ) 1 = ( − 1 1 − 1 1 ) ≠ 0 2 \begin{pmatrix} -1 & 1 \\ -1&1 \end{pmatrix} ^1 = \begin{pmatrix} -1 & 1 \\ -1&1 \end{pmatrix} \ne \textbf{0}_{2} ( − 1 − 1 1 1 ) 1 = ( − 1 − 1 1 1 ) = 0 2 Donc l’indice de nilpotence de ( − 1 1 − 1 1 ) \begin{pmatrix} -1 & 1 \\ -1&1 \end{pmatrix} ( − 1 − 1 1 1 ) est 2 2 2 .