Power of a matrix, given its jordan form

Can someone please explain how to find the power of a matrix $A$, given $A=MJM^{-1}$ where the matrix $J$ is in the Jordan canonical form? Or else please explain how to find the powers of a matrix $J$ that is in the Jordan canonical form.

• The $k$th power $J^k$ of a block diagonal matrix $J := J_1 \oplus \cdots \oplus J_n$ is just the block diagonal matrix $J_1^k \oplus \cdots \oplus J_n^k$, so it's enough to understand how to take powers of a single Jordan block. If you work this out by hand, say for a Jordan block of size $3$ of eigenvalue $\lambda$, you'll see the easy pattern for yourself. – Travis Willse Aug 27 '14 at 8:03

The $k$-th power of $A$ in $A=MJM^{-1}$ Jordan normal form is $A^k=MJ^kM^{-1}$. Since $J$ is block-diagonal you get the $k$-th power of $J$ as
$$J^k=\begin{bmatrix}J_{m_1}^k(\lambda_1) & 0 & 0 & \cdots & 0 \\0 & J_{m_2}^k(\lambda_2) & 0 & \cdots & 0 \\\vdots & \cdots & \ddots & \cdots & \vdots \\0 & \cdots & 0 & J_{m_{s-1}}^k(\lambda_{s-1}) & 0 \\0 & \cdots & \cdots & 0 & J_{m_s}^k(\lambda_s)\end{bmatrix}$$
And last but not least the $k$-th power of $m_i \times m_i$ Jordan blocks is
$$J_{m_i}^k(\lambda_i)=\begin{bmatrix}\lambda_i^k & {k \choose 1}\lambda_i^{k-1} & {k \choose 2}\lambda_i^{k-2} & \cdots & {k \choose m_i-1}\lambda_i^{k-m_i+1} \\0 & \lambda_i^k & {k \choose 1}\lambda_i^{k-1} & \cdots & {k \choose m_i-2}\lambda_i^{k-m_i+2} \\\vdots & \vdots & \ddots & \ddots & \vdots \\0 & 0 & \cdots & \lambda_i^k & {k \choose 1}\lambda_i^{k-1} \\0 & 0 & \cdots & 0 & \lambda_i^k\end{bmatrix},$$ where $k \geq m_i-1$.