# Is there a generalized formula to solve the $n$th exponent of an $m$ dimensional square matrix?

I was trying to do something like a general formula, and testing, I came with this:

$$\left( \begin{array}{cc} a & b \\ a' & b' \\ \end{array} \right)^2=\left( \begin{array}{ccc} a^2+a'b & b(a+b') \\ a'(a+b') & b'^2+a'b \\ \end{array} \right)$$

I've also tried with a $3$x$3$ matrix, but i don't get anything.

I was trying to do something that can solve matrices like this with only one formula:

$$\left( \begin{array}{ccccc} a_{11}&a_{12}&a_{13}&…&a_{1j}\\ a_{21}&a_{22}&a_{23}&…&a_{2j}\\ a_{31}&a_{32}&a_{33}&…&a_{3j}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ a_{i1}&a_{i2}&a_{i3}&\dots&a_{ij} \end{array} \right)^{n}$$

And that's what I want to know.

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## 1 Answer

Yes. You compute the Jordan form of the matrix which is related to $A$ by a similarity transformation $J =P^{-1}AP$ thus $A = PJP^{-1}$ and it's easy to show $A^k = PJ^kP^{-1}$. It turns out the $J^k$ is simple to calculate. For example, if $A$ is diagonalizable then $J = diag(\lambda_1,\dots , \lambda_n)$ hence $J^k = diag(\lambda_1^k,\dots , \lambda_n^k)$. So, look up the Jordan form of a matrix to answer this question in general.

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