I can't find a formula for : $$ A =\begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \\ \end{pmatrix}^n $$ I tried to separate $A = I + J$ with $J$ nilpotent but I didn't success.
Can you give me a hint?
Thanks.
Asked
Active
Viewed 84 times
2 Answers
2
You ask for a hint...
$A^2 = \begin{pmatrix} 1 & 3 & 1 \\ 0 & 4 & 5 \\ 0 & 0 & 9 \end{pmatrix}$
$A^3 = \begin{pmatrix} 1 & 7 & 6 \\ 0 & 8 & 19 \\ 0 & 0 & 27 \end{pmatrix}$
$A^4 = \begin{pmatrix} 1 & 15 & 25 \\ 0 & 16 & 65 \\ 0 & 0 & 81 \end{pmatrix}$
$A^5 = \begin{pmatrix} 1 & 31 & 90 \\ 0 & 32 & 211 \\ 0 & 0 & 243 \end{pmatrix}$
Diagonal elements are easy, other elements are fairly obvious...
Joffan
- 39,627
2
After computing a few terms, we see that $$ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \\ \end{pmatrix}^n = \begin{pmatrix} 1 & 2^n-1 & a_n \\ 0 & 2^n & 3^n-2^n \\ 0 & 0 & 3^n \\ \end{pmatrix} $$ where, $a_n=S(n+1,3)$, Stirling numbers of second kind (A000392). In this case, as noted by @Joffan, $$ a_n = \dfrac{(3^n-2^n)-(2^n-1)}{2}= \dfrac{3^n-2^{n+1}+1}{2} $$
All this comes from expanding $$ \begin{pmatrix} 1 & a & b \\ 0 & c & d \\ 0 & 0 & e \\ \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \\ \end{pmatrix} = \begin{pmatrix} 1 & 2a+1 & a+3b \\ 0 & 2c & c+3d \\ 0 & 0 & 3e \\ \end{pmatrix} $$ and noting that the relations $a=c-1, d=e-c, 2b=d-a$ are preserved.
lhf
- 216,483