Let $f:V\times V\to \Bbb R $ skew-symmetric non-degenerate bilinear form in a real vector space $V$. prove that there is operator $J:V\to\ V $ such that:

  • $J^2=-I$
  • Form $\varphi:V\times V\to \Bbb R $ such that $\varphi(a,b)=f(a,J(b))$ , $a,b \in V$ is symmetric positive definite bilinear form.

I don't really know how can I start prove this, many thanks for any guidance.


1 Answer 1


Currently, I can only show this for $V = \mathbb R^n$. In this case, $f(x,y) = \langle Ax,y\rangle$ with a skew-symmetric matrix $A\in\mathbb R^{n\times n}$, that is, $A^T = -A$ (this also has to be shown). The eigenvalues of $A$ (considered as a complex matrix) lie on the imaginary axis. As $f$ is non-degenerate, $A$ is invertible. If $ia$, $a > 0$, is an eigenvalue of $A$, also $-ia$ is an eigenvalue with the same multiplicity. Therefore, $n$ is actually even, $n = 2m$. Let $ia_j$, $j=1,\ldots,m$, be the eigenvalues of $A$ in the upper halfplane (i.e., $a_j> 0$ for each $j$), counting multiplicities. Let $x_j\in\mathbb C^n$ be an eigenvector corresponding to $ia_j$ such that $\langle x_j,x_k\rangle = 0$ for $j\neq k$. Then $\overline{x_j}$ is an eigenvector of $A$ corresponding to $-ia_j$. Put $u_j := \Re x_j$ and $v_j := \Im x_j$ (real and imaginary parts), $j=1,\ldots,m$. We have $Au_j = -a_j v_j$ and $Av_j = a_ju_j$. The $u_j$'s and $v_j$'s span $\mathbb R^n$. They are even an orthogonal basis: $au_j^Tv_j = (Av_j)^Tv_j = v_j^TA^Tv_j = 0$ since $x^TA^Tx = -x^TAx = -(x^TAx)^T = -x^TA^Tx$. Similarly, one shows that $\|u_j\|^2 = \|v_j\|^2$. Hence, we can scale the $x_j$'s so that the basis $B := \{u_j : j=1,\ldots,m\}\cup\{v_j : j=1,\ldots,m\}$ is orthonormal.

Now, put $Ju_j := -v_j$ and $Jv_j := u_j$ and extend $J$ linearly. With respect to the orthonormal basis $B$ the matrices $J$ and $A$ admit the representations $$ J = \left(\begin{matrix}0 & I_m\\-I_m & 0\end{matrix}\right) \qquad\text{and}\qquad A = \left(\begin{matrix}0 & D\\-D & 0\end{matrix}\right), $$ where $D = \operatorname{diag}(a_1,\ldots,a_m)$. Therefore, $JA$ is a diagonal matrix with only negative entries on the diagonal. In particular, $-JA$ is symmetric and positive definite. Finally, define $\varphi(x,y) := -\langle JAx,y\rangle$ for $x,y\in\mathbb R^n$. This is obviously a positive definite bilinear form and $\varphi(x,y) = \langle Ax,Jy\rangle = f(x,Jy)$.

It is more or less clear that everything above can be "generalized" to arbitrary finite-dimensional real vector spaces. Unfortunately, I don't know how to prove the result for infinite-dimensional spaces.

  • $\begingroup$ Why $f(x,y) = \langle Ax,y\rangle$? If $A^T = -A$, $ f(x,y) =- \langle Ax,y\rangle$ no? $\endgroup$
    – OfirGo
    Mar 4, 2016 at 11:51
  • $\begingroup$ He? A is skew symmetric if and only if -A is so. So, what's your point? I am just stating that there exists such an A. $\endgroup$ Mar 4, 2016 at 13:51

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