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.


  • $\begingroup$ i am not sure but seems that every endomorphism with entries in $\mathbb{C}$ is diagonalizable, since $\mathbb{C}$ is algebraically closed $\endgroup$ – nakajuice Apr 23 '13 at 11:54
  • $\begingroup$ @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. $\endgroup$ – Food4Thought Apr 23 '13 at 11:59
  • $\begingroup$ here you go: en.wikipedia.org/wiki/Diagonalizable_matrix#Examples $\endgroup$ – nakajuice Apr 23 '13 at 12:00
  • $\begingroup$ When we talk about skew-symmetric matrices with complex entries, by definition it is transposing and conjugating the entries, right? NOT the condition $A = -A^T$. $\endgroup$ – Food4Thought Apr 23 '13 at 12:07
  • 2
    $\begingroup$ 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. $\endgroup$ – 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 $$ P = X(B\oplus B)D = \frac1{\sqrt{2}}\pmatrix{ i &-1 &-i &-1\\ 1 & i & 1 &-i\\ -1 & i & 1 & i\\ i & 1 & i &-1}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.