What’s wrong with the following proof? If \(A\) is a complex \(n \times n\) matrix such that \(A^t = -A\), then \(A\) is 0. (Proof: Let \(J\) be the Jordan form of \(A\). Since \(A^t = -A\), \(J^t = -J\). But \(J\) is triangular so that \(J^t = -J\) implies that every entry of \(J\) is zero. Since \(J = 0\) and \(A\) is similar to \(J\), we see that \(A = 0\).) (Give an example of a non-zero \(A\) such that \(A^t = -A\).)
Solution
The error is the assertion \[ J^t=-J. \] If \(J=P^{-1}AP\), then \[ J^t=P^tA^t(P^{-1})^t=-P^tA(P^{-1})^t, \] which need not equal \[ -P^{-1}AP=-J. \] Transpose does not preserve the same conjugating matrix. What follows from \(A^t=-A\) is only that \(J^t\) is similar to \(-J\), not that the two matrices are equal [@HK71]. Therefore the triangularity argument cannot be applied.
A concrete counterexample to the claimed theorem is \[ \boxed{ A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.} \] It is nonzero and \[ A^t=\begin{pmatrix}0&-1\\1&0\end{pmatrix}=-A. \] Over \(\mathbb C\) its eigenvalues are \(i\) and \(-i\), so its Jordan form is \(\operatorname{diag}(i,-i)\), which indeed is not skew-symmetric even though it is similar to the skew-symmetric matrix \(A\).