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: $A^2 = u^tvu^tv= u^t (vu^t)v\\
A^n = u^tvu^tv= u^t (vu^t)^{n-1}v\\
(vu^t) = 3\\
A^n = 3^{n-1} A$
A: The value for $A=u^tv$ you got is not correct.
The correct value is
$$A=u^tv=\begin{bmatrix}  1  \\ 2 \\ 3 \end{bmatrix}\begin{bmatrix} 1 & \dfrac { 1 }{ 2 }  & \dfrac { 1 }{ 3 }  \end{bmatrix}=\begin{bmatrix} 1 & \dfrac { 1 }{ 2 }  & \dfrac { 1 }{ 3 }  \\ 2 & 1 & \dfrac { 2 }{ 3 }  \\ 3 & \dfrac { 3 }{ 2 }  & 1 \end{bmatrix}$$ 
Now you could use diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PDP^{-1}$ with $B$ as a diagonal matrix and $P^{-1}$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PD^2P^{-1}$. Similarly for $A^n=PD^nP^{-1}$
$1.$Find the characteristic polynomial $p(t)$ of $A$.
$2.$Find eigenvalues $λ$ of the matrix $A$ and their algebraic multiplicities from the characteristic polynomial $p(t)$.
$3.$For each eigenvalue $λ$ of $A$, find a basis of the eigenspace $E_λ$.
If there is an eigenvalue $λ$ such that the geometric multiplicity of $λ$, dim$(E_λ)$, is less than the algebraic multiplicity of $λ$, then the matrix $A$ is not diagonalizable. If not, $A$ is diagonalizable, and proceed to the next step.
$4.$If we combine all basis vectors for all eigenspaces, we obtained n linearly independent eigenvectors $v1,v2,…,vn$.
$5.$Define the nonsingular matrix $P=[v_1v_2…v_n]$.
$6.$Define the diagonal matrix $D$, whose $(i,i)$-entry is the eigenvalue λ such that the $i^{th}$ column vector $v_i$ is in the eigenspace $E_λ$
$7.$Then the matrix $A$ is diagonalized as $P^{-1}AP=D$
