Prove $A$ is diagonalizable given the following conditions. Say $A$ is a complex $n\times n$ matrix. $A$ is invertible, and moreover, supposer that the set $\{||A^n||:n=1,2,...\}$ is bounded. Prove that $A$ is diagonalizable. 
My attempt was to do the following. Say $J$ is the Jordan Canonical form of $A$. Since norm is invariant under similar matrices, we have that $J^n$ is bounded as well. Let $B_1,..,B_k$ be the Jordan blocks appearing in $J$. Assume for sake of contradiction that there is a $j$ such that $B_j$ has size bigger than $1$. Then 
$B_j=\lambda I +S$, where $S$ is the matrix with $1$'s in the entries above the diagonal and $0$ elsewhere. 
I also know that $\lambda\neq 0$, but I dont know how to finish the problem. Any thoughts?
 A: This is false as stated.
Note that 
$$  
 \left(  \begin{array}{rr}
  \frac{1}{b} &  0   \\
   0 & 1     
\end{array} 
  \right)  \cdot 
 \left(  \begin{array}{rr}
  a &  b  \\
   0 & a      
\end{array} 
  \right)  \cdot 
 \left(  \begin{array}{rr}
  b &  0   \\
   0 &1      
\end{array} 
  \right) 
 =
 \left(  \begin{array}{rr}
  a &  1   \\
   0 & a      
\end{array} 
  \right).
  $$
So that shows you how to find the Jordan form of such a 2 by 2 matrix, as soon as $b \neq 0.$
Next, take your 
$$ A \; = \;  
 \left(  \begin{array}{rr}
  \lambda &  1   \\
   0 & \lambda     
\end{array} 
  \right).
$$
We calculate 
$$ A^n \; = \;  
 \left(  \begin{array}{rr}
  \lambda^n &  n \, \lambda^{n-1}   \\
   0 & \lambda^n     
\end{array} 
  \right).
$$
In turn, the Jordan form of $A^n$ is
$$ J_n \; = \;  
 \left(  \begin{array}{rr}
  \lambda^n &  1   \\
   0 & \lambda^n     
\end{array} 
  \right).
$$ 
So, if $| \lambda | \leq 1,$ the norm of $A^n$ stays bounded. Depending on the norm you mean, it may require $| \lambda | < 1$ for the norm to stay bounded.
A: Take $J = \begin{bmatrix} \frac{1}{2} && 1 \\ 0 && \frac{1}{2} \end{bmatrix}$. Then a slightly tedious computation gives $J^n = \begin{bmatrix} \frac{1}{2^n} && n\frac{1}{2^{n-1}}  \\ 0 && \frac{1}{2^n} \end{bmatrix}$.
Since $J^n \to 0$, it is clear that the sequence is bounded, but $J$ is not diagonalizable.
Answer to modified question (purely for my own interest):
Since $A^k v = \lambda^k v$ for an eigenvector $v$, it is clear that if $A^k$ is bounded for all integers $k$, then $|\lambda| = 1$, ie, all eigenvalues lie on the unit circle. Suppose $A$ is not diagonalizable, then $A$ has a Jordon normal form with a Jordan chain at least 2 long, ie, there exists unit vectors $v_1,v_2$ such that $Av_1 = \lambda v_1$ and $A v_2 = v_1 + \lambda v_2$. Another tedious computation shows that $A^k v_2 = k \lambda^{k-1} v_1 + \lambda^k v_2$, which contradicts $A^k$ being bounded since $\|A^k\| \geq \| A^k v_2 \| \geq k - 1$. Hence all Jordan chains are of length 1, which implies that $A$ is diagonalizable.
