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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.

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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).

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  • $\begingroup$ 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: $0 \neq \det(A) = \det(A^T) = \det(-A) = (-1)^n \det(A)$. $\endgroup$
    – Chase Bender
    Commented Jan 21, 2022 at 14:57
  • $\begingroup$ @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. $\endgroup$
    – user1551
    Commented Jan 21, 2022 at 15:12
  • $\begingroup$ 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. $\endgroup$
    – Chase Bender
    Commented Jan 21, 2022 at 15:38

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