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Let $A$ be a nondegenerate skew-symmetric matrix over complex field $\mathbb{C}$. Is there an invertible matrix $P$ such that \begin{align*} P^{T}AP=\begin{bmatrix} 0 & I_{\ell} \\ -I_{\ell} & 0 \end{bmatrix}, \end{align*} where $P^{T}$ is the transpose of $P$, $I_{\ell}$ is the $\ell$ by $\ell$ identity matrix?

If $A$ is an $n \times n$ nondegenerate skew-symmetric matrix over complex field $\mathbb{C}$, then $n$ is even. Since $\mathrm{det}(A)=\mathrm{det}(-A^{T})=(-1)^{n}\mathrm{det}(A)$.

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Denote by $\langle \cdot, \cdot \rangle$ the skew-symmetric bilinear form defined by $A$ and let $V = \mathbb{C}^d$. Recall the following two facts:

  1. If $W \subseteq V$ be a subspace such that the restriction $\langle \cdot, \cdot \rangle_{W \times W} : W \times W \to \mathbb{C}$ is nondegenerate, then $V$ admits an orthogonal decomposition $V = W \boxplus W^\perp$.

  2. Moreover, if the bilinear form $\langle \cdot, \cdot \rangle$ was nondegenerate on $V$ , then so is its restriction to $W^\perp$.

Note that by reordering the basis vectors, it suffices to find a basis such that the Gram matrix $A$ is block-diagonal with blocks $\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$. As you say, $V$ must have even dimension, so let $\dim(V) = d = 2n$. We proceed by induction on $n$.

Base Case: $n=1$, ($d = 2$). Then for any choice of basis $\{v_1, v_2\}$, $A$ has the form $\begin{pmatrix} 0 & -a\\ a & 0 \end{pmatrix}$. Since $\langle \cdot, \cdot \rangle$ is nondegenerate, then $a^2 = \det(A) \neq 0$, so $a \neq 0$. Letting $\widetilde{v_2} = \frac{1}{a} v_2$, then with respect to $\{v_1, \widetilde{v_2}\}$, $A = \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$.

Inductive Step: Assume the result holds for $n$ and suppose $\dim(V) = 2(n+1)$. Choose $w_1 \in V \setminus \{0\}$. Since $\langle \cdot, \cdot \rangle$ is nondegenerate, then there exists $w_2 \in V$ such that $\langle w_1, w_2 \rangle = c \neq 0$. By taking $\widetilde{w_2} = \frac{1}{c} w_2$, we may assume $c = 1$. Letting $W = \mathbb{C}\{w_1, w_2\}$, then $\langle \cdot, \cdot \rangle |_{W \times W}$ has Gram matrix $\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$. This matrix has determinant $1$, so $\langle \cdot, \cdot \rangle |_{W \times W}$ is nondegenerate. Then $V = W \boxplus W^\perp$ by the fact 1.

Note that $\dim(W^\perp) = 2n$ and $\langle \cdot, \cdot \rangle|_{W^\perp \times W^\perp}$ is nondegenerate by fact 2. By the inductive hypothesis $W^\perp$ has a basis $\{v_1, \cdots, v_{2n}\}$ with respect to which the Gram matrix of $\langle \cdot, \cdot \rangle|_{W^\perp \times W^\perp}$ has the desired form. Taking the basis $v_1, \ldots, v_{2n}, w_1, \widetilde{w_2}$ produces a block-diagonal Gram matrix with blocks $\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$. Finally, reordering the basis as $v_1, v_3, \ldots, v_{2n-1}, w_1, v_2, v_4, \ldots, v_{2n} \widetilde{w_2}$ produces $$ \begin{pmatrix} 0 & -I_n\\ I_n & 0 \end{pmatrix} \, . $$

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  • $\begingroup$ Thank you for your consideration. If $A$ is a nondegenerate skew-symmetric matrix over real field $\mathbb{R}$, then is the above result right? $\endgroup$ – bing Mar 4 '16 at 7:45
  • $\begingroup$ The above proof works for any field of characteristic zero, so in particular for $\mathbb{R}$. $\endgroup$ – André 3000 Mar 4 '16 at 14:12
  • $\begingroup$ Thank you very much for your consideration. $\endgroup$ – bing Mar 5 '16 at 2:53

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