Cholesky-like decomposition of indefinite matrix

Question

Given a trace-$0$ matrix $W \in \mathbb{R}^{n \times n}$ with an equal number of positive and negative eigenvalues, I want to find matrix $L$ such that $L^T L = W$. Can one do such a factorisation?Right now I am able to factorise it as $$D=A\otimes A^{T}-C\otimes C^{T}$$ The main goal is to prove that the matrix has got equal number of positive and negative Eigen values.

Answer 1

If a matrix $W$ has any negative eigenvalues at all, then it can't be written as $L^tL$ (for $L$ real). For let $v$ be an eigenvector of $W$, with eigenvalue $\lambda$. Then $v^tv>0$, and $$\lambda v^tv=v^tWv=v^tL^tLv=(Lv)^t(Lv)\ge0$$ so $\lambda\ge0$.