# Thread: jordan in power n

1. ## jordan in power n

A is similar to $\displaystyle G=diag(J_{2}(-1),J_{2}(1))$
find the jordan form of $\displaystyle A^{n}$
i have the formula that $\displaystyle A^{n}$ similar to $\displaystyle G^{n}=diag(J_{2}^{n}(-1),J_{2}^{n}(1))$
so $\displaystyle J_{2}^{n}(\lambda)=\left(\begin{array}{cc}\lambda^ {n} & n\lambda^{n-1}\\ 0 & \lambda^{n}\end{array}\right)$
but in the book its $\displaystyle \left(\begin{array}{cc}\lambda^{n} & 1\\0 & \lambda^{n}\end{array}\right)$
why?

2. ## Re: jordan in power n

You are correct. If your text says that $\displaystyle \begin{pmatrix}\lambda & 1 \\ 0 & \lambda\end{pmatrix}^n= \begin{pmatrix}\lambda^n & 1 \\ 0 & \lambda^n\end{pmatrix}$, it is wrong.

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