The orthogonal standard form of antisymmetric matrix Matrix $A=\left( a_{ij} \right) \in M_n\left( \mathbb{R} \right) 
$ ,if $A$ is an antisymmetric matrix,then $a_{ij}=-a_{ji}$.
If $B$ is a real symmetric matrix,then exist $P\in O_n\left( \mathbb{R} \right) $,$P^TBP=\mathrm{diag}\left\{ \lambda _1,...,\lambda _n \right\} 
$,$\mathrm{diag}\left\{ \lambda _1,...,\lambda _n \right\} $ is called the orthogonal standard form of $B$
I want to prove that if $A$ is a antisymmetric matrix, then exist $P\in O_n\left( \mathbb{R} \right) 
$,$P^TAP=\mathrm{diag}\left\{ \left( \begin{matrix}
 0&  a_1\\
 -a_1&  0\\
\end{matrix} \right) ,...,\left( \begin{matrix}
 0&  a_m\\
 -a_m&  0\\
\end{matrix} \right) ,0,..,.0 \right\} 
$.the block diagonal matrix called the orthogonal standard form of $A$.
There is my method:
I want to prove it by induction,$A$ is an antisymmetric matrix,suppose
$A=\left( \begin{matrix}
 A_1&  \alpha _1\\
 -{\alpha _1}^T&  0\\
\end{matrix} \right)$
,where $A_1\in \,\,M_{n-1}\left( \mathbb{R} \right) 
$,$A_1$ is an antisymmetric matrix,by induction,there exist $P=\left( \begin{matrix}
 P_1&  \\
 &  1\\
\end{matrix} \right) \in O_n\left( \mathbb{R} \right) 
$ $s.t.$ $P^TAP=\left( \begin{matrix}
 {P_1}^TA_1P_1&  {P_1}^T\alpha _1\\
 -{\alpha _1}^TP_1&  0\\
\end{matrix} \right) 
$.
where ${P_1}^TA_1P_1
$ become the standard form,
but I don't know how to deal with ${P_1}^T\alpha _1
$ and $-{\alpha _1}^TP_1
$.
Thank you for sharing your mind.
 A: This can be derived as a consequence of (i) the nice case when your skew-symmetric matrix is also orthogonal and (ii) commutativity properties for normal matrices.
As is often the case, we need only prove the claim when $A\in GL_n(\mathbb R)$.  In particular the singular case is easily dealt with by finding a length 1 vector in the kernel, extending to orthogonal $Q$ and considering skew-symmetric $Q^TAQ$ (and in effect calling on an induction hypothesis).
(i.)
First prove the desired claim for the special case when $A\in O_n(\mathbb R)$, I gave a proof here:
Orthogonal Skew Symmetric matrices are orthogonally conjugate
(ii.)
Using polar form write $A=VP$ where $V\in O_n(\mathbb R)$ and $P$ is real symmetric PD.  This is unique since $A$ is invertible.  Confirm that $VP=PV$ (or ref Prove that the polar decomposition of normal matrices, $A=SU$, is such that $SU=US$) and that $PV^T = A^T=-A = P(-V)\implies V$ is orthogonal and skew-symmetric, i.e. it follows case (i.).
$VP=PV\implies V$ respects $P$'s eigenspaces, $\ker \big(P-\lambda_k I\big)$ for $1\leq k\leq m$ (where $P$ has $m$ distinct eigenvalues).  That is, for $\mathbf w\in \ker\big(P-\lambda_k I\big)$ we have $\big(P-\lambda_k I\big)(V\mathbf w)=V\big(P-\lambda_k I\big)\mathbf w = V\mathbf 0 = \mathbf 0\implies (V\mathbf w) \in \ker \big(P-\lambda_k I\big)$. Then spectral thereom for real symmetric matrices says there is a $Q\in \mathbb O_n(\mathbb R)$ such that
$Q^TPQ=\begin{bmatrix}  \lambda_1 I_{r_1} &\mathbf 0 &\dots &\mathbf 0\\\mathbf 0& \lambda_2 I_{r_2} &\dots &\mathbf 0\\  \vdots  &  \vdots&\ddots&\vdots\\ \mathbf 0&\mathbf 0&\dots & \lambda_m I_{r_m}\end{bmatrix}\implies Q^TV Q = \begin{bmatrix}  V_{r_1} &\mathbf 0 &\dots &\mathbf 0\\\mathbf 0& V_{r_2} &\dots &\mathbf 0\\  \vdots  &  \vdots&\ddots&\vdots\\ \mathbf 0&\mathbf 0&\dots & V_{r_m}\end{bmatrix}$
where $r_k = \dim \ker \big(P-\lambda_k I\big)$
(The block diagonal structure on the right is another way of saying $V$ respects $P$'s eigenspaces.  This also implies each $r_k$ is even since $Q^TVQ$ is skew-symmetric and invertible hence each $V_{r_k}$ is skew symmetric and invertible.)
By (i.) we know for each $k$ that there is an orthogonal $U_{r_k}$ such that $U_{r_k}^TV_{r_k}U_{r_k}$ is block diagonal with repeated $J=\begin{bmatrix}  0 &-1 \\  1 &0 \end{bmatrix}$ along the diagonal.
$U := \begin{bmatrix}  U_{r_1} &\mathbf 0 &\dots &\mathbf 0\\\mathbf 0& U_{r_2} &\dots &\mathbf 0\\  \vdots  &  \vdots&\ddots&\vdots\\ \mathbf 0&\mathbf 0&\dots & U_{r_m}\end{bmatrix}$, and conclude $\big(QU\big)\in O_n(\mathbb R)$ gives desired result.  Explicitly this reads
$\big(QU\big)^T A\big(QU\big)=\big(QU\big)^T VP\big(QU\big)=\Big(\big(QU\big)^T V\big(QU\big)\Big)\Big(\big(QU\big)^T P\big(QU\big)\Big)=\begin{bmatrix}  J &\mathbf 0 &\dots &\mathbf 0\\\mathbf 0&J &\dots &\mathbf 0\\  \vdots  &  \vdots&\ddots&\vdots\\ \mathbf 0&\mathbf 0&\dots &J\end{bmatrix}\begin{bmatrix}\lambda_1 I_{r_1} &\mathbf 0 &\dots &\mathbf 0\\\mathbf 0& \lambda_2 I_{r_2} &\dots &\mathbf 0\\  \vdots  &  \vdots&\ddots&\vdots\\ \mathbf 0&\mathbf 0&\dots & \lambda_m I_{r_m}\end{bmatrix}$

optional extension:
the above techniques and result allow us to conclude that a real normal matrix is orthogonally similar to a block diagonal matrix with each block either $2\times 2$ or $1\times 1$.
for normal $A$, write
$A=\big(\frac{A+A^T}{2}\big)+\big(\frac{A-A^T}{2}\big)=H+S$
(where $H$ is a nod to Hermitian and $S$ indicates skew)
$H^2 -S^2 -SH +HS =A^TA=AA^T=H^2-S^2 +SH-HS$
$\implies HS=SH$ and as before $S$ respects $H$'s eigenspaces so by real symmetric spectral theorem there exists $Q\in O_n(\mathbb R)$
$Q^THQ=\begin{bmatrix}  \lambda_1 I_{r_1} &\mathbf 0 &\dots &\mathbf 0\\\mathbf 0& \lambda_2 I_{r_2} &\dots &\mathbf 0\\  \vdots  &  \vdots&\ddots&\vdots\\ \mathbf 0&\mathbf 0&\dots & \lambda_m I_{r_m}\end{bmatrix}\implies S':= Q^TS Q = \begin{bmatrix}  S_{r_1} &\mathbf 0 &\dots &\mathbf 0\\\mathbf 0& S_{r_2} &\dots &\mathbf 0\\  \vdots  &  \vdots&\ddots&\vdots\\ \mathbf 0&\mathbf 0&\dots & S_{r_m}\end{bmatrix}$
where $r_k = \dim \ker \big(H-\lambda_k I\big)$
and by the result of this post, each $S_{r_k}$ is orthogonally similar to a block diagonal skew symmetric matrix having only $2\times 2$ and $1\times 1$ blocks, which implies $S'$ itself is orthogonally similar a block diagonal skew symmetric matrix having only $2\times 2$ and $1\times 1$ blocks via a matrix  $U:= \begin{bmatrix}  U_{r_1} &\mathbf 0 &\dots &\mathbf 0\\\mathbf 0& U_{r_2} &\dots &\mathbf 0\\  \vdots  &  \vdots&\ddots&\vdots\\ \mathbf 0&\mathbf 0&\dots & U_{r_m}\end{bmatrix}$ and
$\big(QU\big)^TA\big(QU\big)=\big(QU\big)^T\big(H+S\big)\big(QU\big)$ gives the result.
