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Horn and Johnson's matrix analysis makes the following interesting statements about the Jordan canonical forms of symmetric and skew-symmetric matrices. Note: I am concerned here with matrices that have complex entries, and I am considering the entrywise transpose rather than the conjugate-transpose.

Regarding symmetric matrices:

Theorem 4.4.24: Each $A \in M_n$ is similar to a complex-symmetric matrix.

Regarding skew-symmetric matrices:

4.4.P34: Although a symmetric complex matrix can have any given Jordan canonical form (4.4.24), the Jordan canonical form of a skew-symmetric complex matrix has a special form. It consists of only the following three types of direct summands: (a) pairs of the form $J_k(\lambda) \oplus J_k(-\lambda)$, in which $\lambda \neq 0$; (b) pairs of the form $J_k(0) \oplus J_k(0)$, in which k is even; and (c) $J_k(0)$, in which k is odd. Explain why the Jordan canonical form of a complex skew-symmetric matrix $A$ ensures that $A$ is similar to $−A$; also deduce this fact from [similarity of a matrix to its transpose].

In the above, $J_k(\lambda)$ denotes the Jordan block of size $k$ associated with eigenvalue $\lambda$. The exercise given is easy enough, but I'd like to prove the leading assertion.

To that end, I have found a useful trick: if $A$ is skew-symmetric and $B$ is symmetric, then $A \otimes B$ is skew-symmetric (where $\otimes$ denotes a Kronecker product). With this trick together with the theorem above, it is easy to find examples of skew-symmetric matrices similar to summands (a) and (b). However, that's as far as I got, which leaves me with two questions.

Questions:

  1. How can we construct a skew-symmetric matrix that is similar to $J_k(0)$, where $k$ is odd?
  2. Why are there no skew-symmetric matrices similar to $J_k(0)$, where $k$ is even?

Thank you for your consideration.


An update: one way to answer question 2 is as follows. We have the following result:

Corollary 4.4.19: Let $A \in M_n$ be skew-symmetric. Then $r = \operatorname{rank}(A)$ is even, the non-zero singular values of $A$ occurs in pairs $\sigma_1 = \sigma_2 = s_1 \geq \sigma_3 = \sigma_4 = s_2 \geq \cdots \geq \sigma_{r-1} = \sigma_r = s_{r/2} \geq 0$, and $A$ is unitarily congruent to $$ 0_{n-r} \oplus \pmatrix{0&s_1\\-s_1 & 0} \oplus \cdots \oplus \pmatrix{0&s_{r/2}\\-s_{r/2} & 0}. $$

By the way: $A$ is unitary congruent to $B$ means that $A = UBU^T$ for some unitary matrix $U$; note that this is not necessarily a matrix similarity.

Because $A$ has singular values that occur in pairs, we can preclude the possibility that $A$ is similar to any matrix of odd rank. For even $k$, $J_k(0)$ is such a martix.

I would still be interested in an argument that doesn't use this fact though; perhaps there is an easy way to see that a skew-symmetric matrix must have even rank.


Possibly useful observations:

  • The rank of $A$ is the same as that of the Hermitian matrix $A^*A = \overline{A^T}A = - \bar A A$.
  • Due to the above corollary, we will necessarily be able to write a matrix that is similar to $J_3(0)$ in the form $$ A = U\pmatrix{0&1&0\\-1&0&0\\0&0&0}U^T = u_1u_2^T - u_2u_1^T $$ where columns $u_1,u_2$ of $U$ are orthonormal.
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A construction can be found in lemma 5.2.2, pp.36-37 of Olga Ruff's master thesis The Jordan canonical forms of complex orthogonal and skew-symmetric matrices: characterization and examples.

To summarise, let $z=\frac{1-i}{2}$. Since $\pmatrix{z&\overline{z}\\ \overline{z}&z}^2=\pmatrix{0&1\\ 1&0}$, if we set $X$ to the $(2n+1)\times(2n+1)$ matrix $$ \pmatrix{ z&&&&&&&&&&\overline{z}\\ &iz&&&&&&&&i\overline{z}\\ &&z&&&&&&\overline{z}\\ &&&iz&&&&i\overline{z}\\ &&&&\ddots&&\unicode{x22F0}\\ &&&&&\sqrt{(-1)^n}\\ &&&&\unicode{x22F0}&&\ddots\\ &&&i\overline{z}&&&&iz\\ &&\overline{z}&&&&&&z\\ &i\overline{z}&&&&&&&&iz\\ \overline{z}&&&&&&&&&&z}, $$ then \begin{aligned} X^2&=\operatorname{antidiag}(1,-1,1,-1,\ldots,1)=DR=RD,\text{ where}\\ D&=\operatorname{diag}(1,-1,1,-1,\ldots,1),\\ R&=\operatorname{antidiag}(1,1,\ldots,1). \end{aligned} Let $J=J_{2n+1}(0)$. Since $X$ is symmetric and $X^4=I$, we have $$ (XJX^{-1})^T=X(X^2J^TX^2)X^{-1} =XDRJ^TRDX^{-1}=XDJDX^{-1}=-XJX^{-1}, $$ i.e. $K=XJX^{-1}$ is skew-symmetric and similar to $J$.


We can prove by a parity argument that nilpotent Jordan blocks of even sizes are not similar to any complex skew-symmetric matrices. First, we need the following result of Horn and Merino (2009) (which is also part of lemma 5.1.2 in Olga Ruff's thesis).

Lemma. A complex square matrix $A$ is similar to a complex skew-symmetric matrix $K$ only if $SA$ is skew-symmetric for some complex symmetric matrix $S$.

Proof. If $A=P^{-1}KP$ where $K^T=-K$, then $A^T=-(P^TP)A(P^TP)^{-1}$. Hence $P^TPA$ is skew-symmetric. $\square$

Now suppose an $m\times m$ nilpotent Jordan block $J=J_m(0)$ is similar to a skew-symmetric matrix. By the above lemma, $SJ$ is skew-symmetric for some non-singular symmetric matrix $S$. Note that the first column of $SJ$ is zero. Therefore $$ S_{1j}=(SJ)_{1,j+1}=-(SJ)_{j+1,1}=0 \textrm{ for all } j<m.\tag{1} $$ Moreover, by the symmetry of $S$ and skew-symmetry of $SJ$, $$ S_{ij}=S_{ji}=(SJ)_{j,i+1}=-(SJ)_{i+1,j}=-S_{i+1,j-1}.\tag{2} $$ Equality $(1)$ means that all entries on the first row of $S$ except the rightmost one are zero. Equality $(2)$ means that if we travel down an anti-diagonal of $S$, the entries are basically constant but they have alternating signs. It follows from $(1)$ and $(2)$ that all entries of $S$ above the main anti-diagonal are zero and the main anti-diagonal of $S$ is $\left(s,-s,s,-s,\ldots,(-1)^{m-1}s\right)$ for some $s$. As $S$ is non-singular, $s$ must be nonzero. Yet, as $S$ is symmetric, the first and the last entries on the anti-diagonal must be equal. Hence $s=(-1)^{m-1}s$ and $m$ is odd.

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  • $\begingroup$ Excellent find! Any thoughts regarding question 2? $\endgroup$ Jan 8, 2020 at 7:08
  • $\begingroup$ @Omnomnomnom Can't think of anything easy but I have an elementary argument. $\endgroup$
    – user1551
    Jan 8, 2020 at 16:29
  • $\begingroup$ Fantastic! I think that's as good as I can expect to get here, thank you very much. $\endgroup$ Jan 8, 2020 at 16:32

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