Show that skew symmetric A and B are congruent $\text{Let}\ \ A,B\in M_n\ \ \text{be skew symmetric. Show that there is a nonsingular}\ \ S\in M_n\ \ \text{such that}$
$A = SBS^T\ \ \text{if and only if}\ \ rankA = rankB. $
 A: This is a partial answer which I hope will be of some use. There are two general approaches that come to mind, both rely on the idea of showing that a rank-$2k$ skew-symmetric matrix is congruent to some canonical form.
Noting that $A$ must have even rank, let $k = \frac 12 \operatorname{rank}(A)$.
Option 1:


*

*Noting that $iA$ is Hermitian (where $i^2 = -1$), apply the either spectral theorem or Sylvester's law of inertia to conclude that there exists a non-singular matrix $S_1$ with complex entries for which $S_1AS_1^*$ has the form $\operatorname{diag}(iI_k,-iI_k,0)$.

*Using this, conclude that there exists a non-singular matrix $S_2$ such that $S_2AS_2^*$ has the form
$$ 
S_2AS_2^* = \pmatrix{0&I_k&0\\-I_k&0&0\\0&0&0}
$$

*It now suffices to show that if $SAS^* = B$ has a complex, non-singular solution $S$, then it must also have a real, non-singular solution $S$.  Unfortunately, I don't see a quick approach here.


Option 2:


*

*Show that if $A$ is skew-symmetric with $A^2 = -I_{2k}$, then $A$ is orthogonally similar to $\operatorname{diag}(C_k,0)$ with
$$
C_k = \pmatrix{0&I_k\\-I_k&0}.
$$

*Show that if $A$ is skew-symmetric with $A^2 = \operatorname{diag}(0,-\lambda_1 I_{2k_1}, \dots, -\lambda_m I_{2k_m})$ with $0<\lambda_1 < \cdots < \lambda_m$, then $A = \operatorname{diag}(A_0,A_1,\dots,A_m)$ with $A_j^2 = \lambda_j I_{2k_j}$.

*Using the spectral theorem, conclude that $A$ is orthogonally similar to $\operatorname{diag}(0, \lambda_1 C_{k_1},\dots, \lambda_{m}C_{k_m})$.

*If $M = \operatorname{diag}(0, \lambda_1 C_{k_1},\dots, \lambda_{m}C_{k_m})$, show that there is a non-singular $S$ such that $SMS^T = \operatorname{diag}(C_k,0)$. Combine with what was said so far conclude that there exists such a matrix $S$ with $SAS^T = \operatorname{diag}(C_k,0)$.
