Find $A^n , n \ge 1$ Let $u = (1, 2, 3)$ and $v = (1, \frac{1}{2},\frac{1}{3})$ · Let $A= u^tv.$ Find $A^n , n \ge  1$
My attempts  :$ A = u^tv$  = $\begin{bmatrix}  1  \\ 2 \\ 3 \end{bmatrix}[ 1     \frac{1}{2}     \frac{1}{3} ]$.
$A = \begin{bmatrix}  1 &1& \frac{2}{3} \\ 2& 1& \frac{2}{3}\\3& \frac{3}{2} & 1 \end{bmatrix}$.
i don't  know  How  to  find  $A^n   $?

$A^2 = u^tvu^tv= u^t (vu^t)v\\
A^n = u^tvu^tv= u^t (vu^t)^{n-1}v\\
(vu^t) = 3\\
A^n = 3^{n-1} A$
