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Hello guys can someone help me with that please ? And thanks for that

$A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}, \tag 1$

and we have

$B = A − I. \tag 2$

Calculate $B^n$ for $n ∈ N$ and deduct from that the expression of $A^n$.

Robert Lewis
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Aita KANE
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  • the matrix is equal to that:

    A = (1 1 1) (0 1 1) (0 0 1)

     – Aita KANE Apr 17 '18 at 21:44
  • the last bracket means the end of the first line . Actually A is a matrix (3x3) – Aita KANE Apr 17 '18 at 21:46
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    This will help you with the formatting. https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Doug M Apr 17 '18 at 21:47
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    Aita, suggest you find $B^2$ and $B^3$ yourself. Easier than you might think – Will Jagy Apr 17 '18 at 21:50
  • Thanks Doug M and Will yes i already found them and i have for$B^2$ = $$ \begin{matrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ \end{matrix} $$ and $B^3 = 0$ – Aita KANE Apr 17 '18 at 21:53
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    Aita, since $I$ commutes with anything, it means you can calculate $A^n = (I+B)^n$ by ordinary binomial expansion – Will Jagy Apr 17 '18 at 21:55
  • Aita, let me emphasize that, if we have two square matrices that do NOT commute, we cannot use binomials. If $EF \neq FE,$ then $(E+F)^2 = E^2 + EF + FE + F^2,$ and we cannot combine the mixed terms. However, in your case, also the case of Jordan Normal Form, also the Jordan-Chevalley decomposition, we get to use binomial expansion, and this ends in a fixed number of terms because one of the matrices is "nilpotent." – Will Jagy Apr 17 '18 at 22:08
  • alright guys thank you for help – Aita KANE Apr 17 '18 at 22:28

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