# jordan in power n

• Oct 17th 2011, 06:18 AM
transgalactic
jordan in power n
A is similar to $G=diag(J_{2}(-1),J_{2}(1))$
find the jordan form of $A^{n}$
i have the formula that $A^{n}$ similar to $G^{n}=diag(J_{2}^{n}(-1),J_{2}^{n}(1))$
so $J_{2}^{n}(\lambda)=\left(\begin{array}{cc}\lambda^ {n} & n\lambda^{n-1}\\
0 & \lambda^{n}\end{array}\right)$

but in the book its $\left(\begin{array}{cc}\lambda^{n} & 1\\0 & \lambda^{n}\end{array}\right)$
why?
• Oct 17th 2011, 08:34 AM
HallsofIvy
Re: jordan in power n
You are correct. If your text says that $\begin{pmatrix}\lambda & 1 \\ 0 & \lambda\end{pmatrix}^n= \begin{pmatrix}\lambda^n & 1 \\ 0 & \lambda^n\end{pmatrix}$, it is wrong.

It's easy to check with n= 2:
$\begin{pmatrix}\lambda & 1 \\ 0 & \lambda\end{pmatrix}^2= \begin{pmatrix}\lambda & 1 \\ 0 & \lambda\end{pmatrix}\begin{pmatrix}\lambda & 1 \\ 0 & \lambda\end{pmatrix}= \begin{pmatrix}\lambda^2 & 2\lambda \\ 0 & \lambda^2\end{pmatrix}$
as you say.