Non singular matrix $M$ such that $MAM^T=F$, normal form Show that there is a non-singular matrix $M$ such that $MAM^T = F$ for any antisymmetric matrix $A$ where the normal form $F$ is a matrix with $2 \times 2$ blocks on its principal diagonal which are either zero or $$\begin{pmatrix} 0 &amp;1 \\ -1&amp;0 \end{pmatrix}$$

To do so, consider the analogue of the Gram-Schmidt method for antisymmetric matrices using the antisymmetric inner product $\langle x,y \rangle = x^TA y$

I am just really looking for some hints to start. For vectors $x$ and $y$, $$x^T A y = (x_1 \dots x_n) \begin{pmatrix} A_{11} &amp; A_{12} &amp;..&amp;A_{1n} \\ ..&amp;..  \\ ..&amp; ..&amp;..&amp;A_{nn} \end{pmatrix} \begin{pmatrix} y_1 \\ y_2\\ ..\\y_n \end{pmatrix}$$ and $A_{ii} = 0, A_{ij} = -A_{ji} $ but I am not seeing how the hint is useful. 

Thanks!
We suppose that $A$ is written in the canonical basis. Consider the product defined by $\langle x,y\rangle=x^TAy$,  it is a bilinear antisymetric product. There exists a basis $e_1,...e_{2n}$ such that this product in this basis is $e_{n+1}\wedge e_{1}+...+e_{2n}\wedge e_{n}$. The matrix of the bilinear for in this basis is $\pmatrix{0 &amp; 1 \cr -1 &amp; 0}$.

Let $M$ be the transtion matrix from the  basis to $e_1,...,e_{2n}$ to the canonical basis $M^TAM =\pmatrix{ 0 &amp;1\cr -1 &amp; 0}$
We suppose that $A$ is written in the canonical basis. Consider the product defined by $\langle x,y\rangle=x^TAy$,  it is a bilinear antisymetric product. There exists a basis $e_1,...e_{2n}$ such that this product in this basis is $e_{n+1}\wedge e_{1}+...+e_{2n}\wedge e_{n}$. The matrix of the bilinear for in this basis is $\pmatrix{0 &amp; 1 \cr -1 &amp; 0}$.

Let $M$ be the transtion matrix from the  basis to $e_1,...,e_{2n}$ to the canonical basis $M^TAM =\pmatrix{ 0 &amp;1\cr -1 &amp; 0}$