diagonalization exp(A) counter exemple We consider a finite vector space E over a field K. We consider u in End(E). We can show that if the characteristic polynomial of A splits then we have the following equivalence : A diagonalizable iff exp(A) diagonalizable. 
I am searching a counter exemple in the case of a non splitting characteristic polynomial, ie a matrix such that exp(A) is diagonalizable but A is not.  
Thankyou for your answers !
 A: trying to guess what you might be asking, take
$$ A =
\left(
\begin{array}{rr}
0 & \pi \\
- \pi & 0
\end{array}
\right).
 $$
Then
$$ e^A =
\left(
\begin{array}{rr}
-1 & 0 \\
0 & -1
\end{array}
\right).
 $$
If 
$$ C =
\left(
\begin{array}{rr}
0 & 1 \\
- 1 & 0
\end{array}
\right)
 $$
then $C^2 = -I$ and,  for real $t,$ 
$$  e^{tC} = I + tC - I t^2/2 - C t^3/6 + I t^4/24 +C t^5/120 - I t^6/720-C t^7/5040...$$
$$ e^{tC} = (I - I t^2/2  + I t^4/24  - I t^6/720...) + (tC  - C t^3/6  +C t^5/120 -C t^7/5040...)  $$
$$ e^{tC} = I(1 -  t^2/2  +  t^4/24  -  t^6/720...) + C(t  -  t^3/6  + t^5/120 - t^7/5040...)  $$
$$ e^{tC} = I \cos t + C \sin t $$
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The other possible interpretation has to do with Jordan blocks. My best guess is that exponentiating does not change the arrangment of Jordan blocks, just the diagonal elements. For example, if
$$ F =
\left(
\begin{array}{rr}
\lambda & 1 \\
0 & \lambda
\end{array}
\right)
 $$
then
$$ e^F =
\left(
\begin{array}{rr}
e^\lambda & e^\lambda \\
 0 & e^\lambda
\end{array}
\right)
 $$
But then
$$ 
\left(
\begin{array}{rr}
e^{-\lambda} & 0 \\
0 & 1
\end{array}
\right)
\left(
\begin{array}{rr}
e^\lambda & e^\lambda \\
0 & e^\lambda
\end{array}
\right)
\left(
\begin{array}{rr}
e^\lambda & 0 \\
0 & 1
\end{array}
\right) =
\left(
\begin{array}{rr}
e^\lambda & 1 \\
0 & e^\lambda
\end{array}
\right)
 $$
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Also did 3 by 3
$$ G =
\left(
\begin{array}{rrr}
\lambda & 1 & 0 \\
0 & \lambda & 1 \\
0 & 0 & \lambda
\end{array}
\right)
 $$
then
$$ e^G =
\left(
\begin{array}{rrr}
e^\lambda & e^\lambda & \frac{1}{2} e^\lambda \\
 0 & e^\lambda & e^\lambda \\
0 & 0 & e^\lambda
\end{array}
\right)
 $$
But then
$$ 
\left(
\begin{array}{rrr}
e^{-2 \lambda} &  -\frac{1}{2} e^{-\lambda} & 0 \\
0 &   e^{- \lambda} & 0 \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{rrr}
e^\lambda & e^\lambda & \frac{1}{2} e^\lambda \\
 0 & e^\lambda & e^\lambda \\
0 & 0 & e^\lambda
\end{array}
\right)
\left(
\begin{array}{rrr}
e^{2 \lambda} &  \frac{1}{2} e^\lambda & 0 \\
0 &   e^\lambda & 0 \\
0 & 0 & 1
\end{array}
\right) =
\left(
\begin{array}{rrr}
e^\lambda & 1 & 0\\
0 & e^\lambda & 1 \\
0 & 0 & e^\lambda
\end{array}
\right)
 $$
