ij-th Component for **Matrix Power** $ A^n = AAA..A$
Here I simply do the Power Operation of a Matrix A.
Is there any general form to express ij-th component of this n-th powered Matrix?
$ (A^n)_{ij} = A_{i?}+...+A_{?j}?$
 A: Suppose $A$ has size $m\times m$.
Without further structure, the direct formula is:
$$ (A^n)_{i,j} = \sum_{l_1=1}^m\sum_{l_2=1}^m\cdots\sum_{l_{n-1}=1}^m A_{i,l_1}A_{l_2,l_3}\cdot\ldots\cdot A_{l_{n-2},l_{n-1}}A_{l_{n-1},j}= \sum_{l_1,\ldots,l_{n-1}=1}^m A_{i,l_1}A_{l_2,l_3}\cdot\ldots\cdot A_{l_{n-2},l_{n-1}}A_{l_{n-1},j}.$$
This can easily be proved by induction over $n\geq 1$.
For $n=1$, it is clear. Suppose it is true for $n$, then
\begin{align*}
(A^{n+1})_{i,j} &= \sum_{l_n=1}^m (A^n)_{i,l_n}A_{l_n,j}\\
&= \sum_{l_n=1}^m \bigg(\sum_{l_1,\ldots,l_{n-1}=1}^m A_{i,l_1}A_{l_2,l_3}\cdot\ldots\cdot A_{l_{n-2},l_{n-1}}A_{l_{n-1},l_n}\bigg)A_{l_n,j}\\
&= \sum_{l_1,\ldots,l_{n}=1}^m A_{i,l_1}A_{l_2,l_3}\cdot\ldots\cdot A_{l_{n-1},l_{n}}A_{l_{n},j}.
\end{align*}
A: Consider the diagonalization of $A=PVP^{-1}$ where $V$ contains the eigenvalues of $A$ and $P$ the eigenvectors. Then, $A^n=PV^nP^{-1}$. From this, you can read the $ij$th component of $A^n$.
A: If we can diagonalize the matrix $A$ as $A = PDP^{-1}$, where $D$ is the diagonal matrix of eigenvalues and $P$ is the matrix of eigenvectors, then we can express the matrix $A^n$ as:
$A^n = PD^nP^{-1}$
hence :
the ij th elementment of $A^n$ let say $b_ij$ is:
$b_{ij} = \sum_{k=1}^p c_k \lambda_k^n (P^{-1})_{ki} P_{kj}$
where $c_k$ are constants that depend on the initial conditions, $(P^{-1})_{ki}$ is the $(k,i)$th element of the inverse matrix $P^{-1}$, and $P_{kj}$ is the $(k,j)$th element of $P$.
