An Example of a non-diagonalizable $4 \times 4$ skew-symmetric matrix with non-zero eigenvalues I looked around this site to see if there is any question that addresses my concern, but so far, I couldn't find any. I apologize that if this ends up being a duplicate, but I have been looking for a while. I am having much difficulties in concocting an example of a $4 \times 4$ skew-symmetric matrix with entries in $\mathbb{C}$ that is not-diagonalizable with non-zero eigenvalues. I have tried using Wolfram-Alpha (Jordan Normal Form Calculator Online), inputting different values to make different skew symmetric matrices, but the matrix I end up concocting ends up being diagonalizable. 
First of all, I was reading the following paper:

Olga Ruff, The Jordan Canonical Forms of complex orthogonal and skew-symmetric matrices: characterization and examples, Master thesis, Iowa State University, 2007.

The fact that there are nondiagonalizable skew symmetric matrices is mentioned on page 35 under Lemma 5.2.1. 
As far as I know, if this fact is true, can someone provide me a simple example. If not, maybe a link or theorem that states such scenario is not possible. 
Thanks!
 A: If you read the thesis carefully, you will see that it has already offered a way to constuct the desired skew symmetric matrix. For instance, the Jordan form $J=J_2(1)\oplus J_2(-1)$ -- which is not diagonalisable -- is similar to the skew symmetric matrix
$$
Y = \pmatrix{
0       &\frac12+i &0     &\frac{i}2\\
-\frac12-i  &0     &-\frac{i}2  &0\\
0       &\frac{i}2   &0     &-\frac12+i\\
-\frac{i}2    &0     &\frac12-i &0
}.
$$
You may verify that the Jordan form of $Y$ is indeed $J_2(1)\oplus J_2(-1)$.

Here are the details of construction. First of all, $J_2(1)\oplus J_2(-1)$ is similar to $\widetilde{J}=J_2(1)\oplus-J_2(1)$:
$$
\underbrace{\pmatrix{1\\ &1\\ &&1\\ &&&-1}}_{D}
\pmatrix{1&1\\ &1\\ &&-1&1\\ &&&-1}
\underbrace{\pmatrix{1\\ &1\\ &&1\\ &&&-1}}_{D^{-1}}
=\pmatrix{1&1\\ &1\\ &&-1&-1\\ &&&-1}.
$$
Yet $J_2(1)$ is similar to a complex symmetric matrix (theorem 2.1.4):
$$
\underbrace{\frac1{\sqrt{2}}\pmatrix{1&i\\ i&1}}_{B}
\ \pmatrix{1&1\\ &1}
\ \underbrace{\frac1{\sqrt{2}}\pmatrix{1&-i\\ -i&1}}_{B^{-1}}
=\underbrace{\pmatrix{1-\frac{i}{2}&\frac12\\ \frac12&1+\frac{i}{2}}}_{S}.
$$
Therefore $\widetilde{J}$ is similar to $A=S\oplus -S$.

Let $H=\pmatrix{0&I_2\\ I_2&0}$. Then $HAH^{-1}=-A$. We have $H=X^TX$, where
$$
X=\frac1{\sqrt{2}}\pmatrix{
i &0 &-i &0\\
1 &0 &1  &0\\
0 &i &0  &-i\\
0 &1 &0  &1}.
$$
Therefore, by the proof of lemma 5.1.2 and by lemma 5.2.1, $Y = XAX^{-1}$ is skew symmetric.
Putting all the pieces together, we have $Y=PJP^{-1}$, where
$$
P = X(B\oplus B)D = \frac1{\sqrt{2}}\pmatrix{
 i &-1 &-i &-1\\
 1 & i & 1 &-i\\
-1 & i & 1 & i\\
 i & 1 & i &-1}.
$$
