Congruence of invertible skew symmetric matrices
The key fact is that every real and normal matrix is orthogonally similar to its real Jordan form. However, as real Jordan forms are seldom taught in universities, we may need to start from scratch here.
- As $A$ is skew symmetric and invertible, its eigenvalues are conjugate pairs of nonzero pure imaginary numbers $\pm b_1i,\,\pm b_2i,\,\ldots,\pm b_ni$ (where the size of $A$ is $2n\times2n$).
- Furthermore, as $A$ is skew symmetric, it is a normal matrix. Hence it possesses an orthonormal eigenbasis. (In some lecture notes, this fact is sometimes stated as "if $A$ is normal, $A=UDU^\ast$ for some unitary matrix $U$ and diagonal matrix $D$".) Let $u_1+iv_1,\,u_2+iv_2,\,\ldots,\,u_n+iv_n$ be $n$ orthogonal eigenvectors corresponding the eigenvalues $b_1i,\,b_2i,\,\ldots,\,b_ni$.
- Note that $\left\{\frac{u_1}{\|u_1\|},\frac{u_2}{\|u_2\|},\ldots,\frac{u_n}{\|u_n\|},\frac{v_1}{\|v_1\|},\frac{v_2}{\|v_2\|},\ldots,\frac{v_n}{\|v_n\|}\right\}$ is an orthonormal basis of $\mathbb{R}^{2n}$ (why?). So, these $2n$ vectors form a real orthogonal matrix $Q$ and $A=Q\pmatrix{0&B\\ -B&0}Q^T$ (again, why?), where $B=\operatorname{diag}(b_1,b_2,\ldots,b_n)$. Now the rest should be straightforward.
Hint: a skew-symmetric matrix commutes with its transpose, and so it is diagonalizable. Your block matrix on your right hand side is also skew-symmetric, and so it is also diagonalizable.