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Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
Calculate \(\displaystyle A^{17}\) where
.

Progress,
I have multiplicate without adding them together to see a pattern and I can se at \(\displaystyle A^{17}\)
on that matrice where it's 6's it will be \(\displaystyle 6^{17}\) and rest I cant se any pattern those riight side of the triangle, cause the left will be zero

Regards,
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hello MHB,
Calculate \(\displaystyle A^{17}\) where
.

Progress,
I have multiplicate without adding them together to see a pattern and I can se at \(\displaystyle A^{17}\)
on that matrice where it's 6's it will be \(\displaystyle 6^{17}\) and rest I cant se any pattern those riight side of the triangle, cause the left will be zero

Regards,
Cayley Hamilton theorem might be useful. Cayley
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
A little exploration shows:

If [tex]A = \left( \begin{array}{ccc}
a & 1 & 0 \\
0 & a & 1 \\
0 & 0 & a \end{array} \right)[/tex]

Then [tex]A^{17} = \left( \begin{array}{ccc}
a^{17} & 17a^{16} & 8\cdot 17\cdot a^{15} \\
0 & a^{17} & 17a^{16} \\
0 & 0 & a^{17} \end{array} \right)[/tex]

Upper triangular with only '1' off the diagonal? That HAS to be rather tractable.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
In general, if we have a Jordan block $J(\lambda)$ of order $n$ we can express $$J(\lambda)=\begin{bmatrix} \lambda & 1 & 0 &\ldots & 0 & 0 & 0\\ 0 & \lambda & 1 &\ldots & 0&0&0 \\0 & 0 & \lambda &\ldots & 0&0&0 \\\vdots&&&&&&\vdots \\ 0 &0 & 0 &\ldots & \lambda & 1&0\\0 &0 &0 &\ldots &0&\lambda & 1\\0 & 0 &0&\ldots & 0&0&\lambda\end{bmatrix}=\lambda I+N$$ where $N$ is nilpotent of order $n$. As $(\lambda I_n)N=N(\lambda I_n)$, we can apply the binomial theorem $$J(\lambda)^m=(\lambda I+N)^m=(\lambda I)^{m}+\binom{m}{1}(\lambda I)^{m-1}N+\binom{m}{2}(\lambda I)^{m-2}N^2+\ldots\\=\lambda^mI+m\lambda^{m-1} N+\frac{m(m-1)}{2}\lambda^{m-2}N^2+\ldots$$ If $n=3$, $N$ is nilpotent of order $2$: $$N=\begin{bmatrix}{0}&{1}&{0}\\{0}&{0}&{1}\\{0}&{0}&{0}\end{bmatrix},N^2=\begin{bmatrix}{0}&{0}&{1}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{bmatrix},N^3=0$$ $$J(\lambda)^m=\lambda^mI+m\lambda^{m-1} N+\frac{m(m-1)}{2}\lambda^{m-2}N^2$$ $$\begin{bmatrix}{\lambda}&{1}&{0}\\{0}&{\lambda}&{1}\\{0}&{0}&{\lambda}\end{bmatrix}^m=\lambda^m \begin{bmatrix}{1}&{0}&{0}\\{0}&{1}&{0}\\{0}&{0}&{1}\end{bmatrix}+m\lambda^{m-1}\begin{bmatrix}{0}&{1}&{0}\\{0}&{0}&{1}\\{0}&{0}&{0}\end{bmatrix}+\frac{m(m-1)}{2}\lambda^{m-2}\begin{bmatrix}{0}&{0}&{1}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{bmatrix}$$ $$=\begin{bmatrix}{\lambda^m}&{m\lambda^{m-1}}&{\frac{m(m-1)}{2}}\lambda^{m-2}\\{0}&{\lambda^m}&{m\lambda^{m-1}}\\{0}&{0}&{\lambda^m}\end{bmatrix}$$