Square Root of a Complex, Anti-Symmetric Matrix

Suppose $$A$$ is an invertible matrix with complex entries and satisfies $$A^T = -A$$. By the Jordan decomposition of $$A$$ it is seen that there exist at least one (and hence several) square roots of $$A$$, and by Hermite interpolation of the square root function there exists a distinguished square root of $$A$$ that is a polynomial in $$A$$. Up to this point, these results are irrespective of the anti-symmetric condition. Once we impose anti-symmetry, does there exist a square root $$\sqrt{A}$$ such that $$\sqrt{A}^{T} = i \sqrt{A}$$?

P.S. The existence of such a square root would yield a proof of a technical detail in my master's thesis, but there is another (would-be) result that would suffice, so I ask it here as well.

Given a complex-symmetric (invertible) matrix $$A$$ a result known as the Autonne-Takagi factorization gives the existence of a unitary $$U$$ such that $$U^TAU$$ is a positive diagonal. If instead $$A^T = -A$$, does there exist a unitary $$U$$ such that $$U^T A U$$ is a positive diagonal times $$J$$, where $$J$$ is a block diagonal with blocks $$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$

I have tried to prove this by simply adapting the proof of Autonne-Takagi that I know, but I have come up short on the last few steps. Any information will be greatly appreciated.

The answer to your first question is negative, unless $$A=0$$. For any matrix $$B$$, if $$B^T=iB$$, then $$B=(B^T)^T=(iB)^T=iB^T=i(iB)=-B$$ and hence $$B$$ is necessarily zero.
For your second question, we have $$A=U(s_1J\oplus\cdots\oplus s_kJ\oplus0)U^T$$ for some unitary matrix $$U$$ and some positive real numbers $$s_1,\ldots,s_k$$, where the zero diagonal sub-block is possibly empty when the size of $$A$$ is even. This is a well-known result, first proved by the number theorist L.K. Hua in the article On the theory of automorphic functions of a matrix variable $$I$$-geometrical basis, Amer. J. Math., 66(1944).
• Thank you for the answer to the second question. This should still suffice. But isn't the decomposition you give just a more specific version of the one I claim? I said that it should be a positve diagonal, but if it is of the form $\diag(s_1, s_1, \dots , s_m, s_m)$ where $n=2m$, then these decompositions seem to coincide. Also, the even dimension is forced, by a similar argument that there is no square root with $\sqrt{A}^T = i\sqrt{A}$ since $A$ is invertible: $0 \neq \det(A) = \det(A^T) = \det(-A) = (-1)^n \det(A)$. Commented Jan 21, 2022 at 14:57
• @ChaseBender I've misread your question, but the statement in your question is still incorrect. As it stands, it states that $U^TAU=D(J\oplus\cdots\oplus J)$ for some positive diagonal matrix $D$. But then $A$ must be nonsingular, which we know isn't the case when the size of $A$ is odd. Even if you correct it by adding a zero sub-block to the direct sum, it is still assuming a less precise form, because the nonzero singular values of $D(J\oplus\cdots\oplus J\oplus0)$ in general can all be distinct. but those nonzero singular values of $s_1J\oplus\cdots\oplus s_kJ\oplus0$ must occur in pairs. Commented Jan 21, 2022 at 15:12
• It is assumed that $A$ is nonsingular, which in turn forces the even dimension. Further, I am not claiming that the diagonal is arbitrary, i.e. that it is sufficient for $A$ to have such a decomposition in order to be antisymmetric. Commented Jan 21, 2022 at 15:38