# Eigen Diagonalization

• Nov 4th 2007, 10:45 AM
Thomas
Eigen Diagonalization
Let $M=\left( \begin{array}{ccc} 5 & 2 \\ -1 & 2 \end{array} \right)$.

Find formulas for the entries of $M^n$, where $n$ is a positive integer.

$M^n=\left( \begin{array}{ccc} ? & ? \\ ? & ? \end{array} \right)$.

I'm not sure how to go about this. Could someone give push me in the right direction?
• Nov 4th 2007, 07:42 PM
topsquark
Quote:

Originally Posted by Thomas
Let $M=\left( \begin{array}{ccc} 5 & 2 \\ -1 & 2 \end{array} \right)$.

Find formulas for the entries of $M^n$, where $n$ is a positive integer.

$M^n=\left( \begin{array}{ccc} ? & ? \\ ? & ? \end{array} \right)$.

I'm not sure how to go about this. Could someone give push me in the right direction?

Probably the simplest thing to do would be to calculate what multiplying M by an arbitrary 2 x 2 matrix would give:
$M = \left ( \begin{matrix} 5 & 2 \\ -1 & 2 \end{matrix} \right )$

$M \times \left ( \begin{matrix} a & b \\ c & d \end{matrix} \right ) = \left ( \begin{matrix} 5a + 2c & 5b + 2d \\ -a + 2c & -b + 2d \end{matrix} \right )$

Now you can set up a recursion: for $M^2$ a = 5, b = 2, c = -1, and d = 2. This generates the next set of a, b, c, and d.

So $a_n = 5a_{n - 1} + 2c_{n - 1}$, $b_n = 5b_{n - 1} + 2d_{n - 1}$, etc.

Now solve the system of recursions. (That step is beyond me, I'm afraid!)

-Dan
• Nov 7th 2007, 09:14 PM
ThePerfectHacker
That is not how to do it topsquark. You need to diagnolize the matrix.
Meaning,
$AMA^{-1} = D$.
Where $A$ is some invertible matrix and $D$ is a diagnol matrix. Once you do that the above equation is easy to handle.
Bring everything to power of $n$,
$(AMA^{-1})^n = D^n$
Thus,
$AM^nA^{-1} = D^n$
And,
$\boxed{M^n = A^{-1}D^nA}$
Where $A$ can be found from its eigenvectors and $D$ from its eigenvalues.
(Note, computing a power of a diagnol is easy. Just raise each diagnol entry to that exponent).