# 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!

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i am not sure but seems that every endomorphism with entries in $\mathbb{C}$ is diagonalizable, since $\mathbb{C}$ is algebraically closed –  haemhweg Apr 23 '13 at 11:54
@haemhweg Thanks for the comment. Maybe if I can recall some result in Linear Algebra that states that, I am set. I'm starting to believe that there is no such example, but I could be wrong. –  Food4Thought Apr 23 '13 at 11:59
–  haemhweg Apr 23 '13 at 12:00
When we talk about skew-symmetric matrices with complex entries, by definition it is transposing and conjugating the entries, right? NOT the condition $A = -A^T$. –  Food4Thought Apr 23 '13 at 12:07
Every skew-Hermitian matrix is diagonalizable, so you would be looking for a complex matrix such that $A = -A^\top$ and not the conjugate with the transpose, since that would make it skew-Hermitian. Sorry that I can not give an example right now. Note that if you use only real entries, then it will be again diagonalizable. –  adam W Apr 23 '13 at 12:45

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 $$P = X(B\oplus B)D = \frac1{\sqrt{2}}\pmatrix{ i &-1 &-i &-1\\ 1 & i & 1 &-i\\ -1 & i & 1 & i\\ i & 1 & i &-1}.$$