Thread: Finding a formula for a matrix raised to the nth power

1. Finding a formula for a matrix raised to the nth power

One problem I've been having some trouble with is finding an expression for $A^n$ where $A=\left(\begin{array}{cc}1&4\\2&3\end{array}\right )$

I've found a similar problem on Wikibooks/Wikipedia, but those aren't the best examples. I just need someone to explain a streamlined process of how to solve one of these (using the concepts of diagonalizability). Any help?

2. Step 1)
Find the eigenvalues. Find the roots of the characteristic polynomial: $det(A-xI)$.
In this case I got 5 and -1.
Step 2)
Find a basis for the nullspace of $A-\lambda I$. In this case you have two distinct eigenvalues, so you will get an eigenvector for each one.

I got $(1,-1)^T$ for $\lambda_1=5$
and $(2,-1)^T$ for $\lambda_1=-1$

Step 3) these eigenvectors will give you the matrices you need to conjugate A by to get it into its diagonal form (or its Jordan Canonical Form which is as close to diagonal as possible).

Let $P=\begin{pmatrix}1 &2 \\ -1 &-2 \end{pmatrix}$

Then $P^{-1}AP=D:=\begin{pmatrix}5 &0 \\ 0 &-1 \end{pmatrix}$

Step 4) $A=PDP^{-1}$ start raising it to powers
$A^n=(PDP^{-1})^n=PDP^{-1}PDP^{-1}...PDP^{-1}=PD^nP^{-1}$

Fortunately powers of diagonal matrices are very easy to calculate, you just raise the elements on the diagonal to the power.

$D^n=\begin{pmatrix}5^n &0 \\ 0 & (-1)^n \end{pmatrix}$

Side note, sometimes if you get repeated eigenvalues the Jordan Form will have 1s on the first superdiagonal. This is okay because it is the sum of a diagonal and nilpotent matrix (the one with 0s everywhere except possibly the 1s on some of the super diagonal). You can expand this using binomial theorem and eventually those terms with the nilpotent matrix will disappear and it will cut down a lot on the computations.

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how to find general formula for matrix

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