Complex skew-symmetric matrices

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?

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.

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 $$S_{1j}=(SJ)_{1,j+1}=-(SJ)_{j+1,1}=0 \textrm{ for all } j Moreover, by the symmetry of $$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.

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