TITLE: Symmetric Matrices with trace zero
QUESTION [21 upvotes]: Let $M_n$ denote the set of complex matrices of order $n$.
It is well known that if $A\in M_n$ has trace zero then $A$ can be written as $A=BC-CB$, where $B,C\in M_n$.
Is it true that every symmetric matrix $S\in M_n$ with trace zero can be written as $S=RR^t-R^tR$, for some $R\in M_n$?
This theorem is true for real matrices. If $S$ is a real symmetric matrix with trace zero and order $n$ then exists a real matrix $R$ of order $n$ such that $S=RR^t-R^tR$. 
Is it true for complex matrices?

REPLY [6 votes]: Here you can find an alternative proof for the real symmetric case. I am writting to answer Clin's question. Notice that the approach used by user1551 at his post on May 13 gives a different solution. 
Let $S$ be a real symmetric matrix of order k and trace zero. Let $S=\sum_{i=1}^n\lambda_iv_iv_i^t$ be a spectral decomposition of $S$ with $\lambda_1\geq\lambda_2\geq\ldots\geq\lambda_n$ 
$(\lambda_i\neq 0)$(Edit: and $v_1,\dots,v_n$ orthonormal).
Define $\beta_i=\lambda_1+\dots+\lambda_i$. Notice that $\beta_n=0$. 
Now, $\beta_i>0$ for $i<n$, otherwise $\beta_{i+1}<0$ and $\beta_n<0$.
Define the matrices $V_{k\times n-1}, W_{k\times n-1}$ using the vectors $v_i$ as columns: $V=(v_1,\dots,v_{n-1})$ and $W=(v_2,\dots,v_{n})$. Notice that $V^tV=W^tW=Id_{n-1}$.
Define the diagonal matrix $D_{n-1\times n-1}$ with $\beta_1,\dots,\beta_{n-1}$ in the diagonal.
Notice that $S=\sum_{i=1}^n\lambda_iv_iv_i^t=\sum_{i=1}^{n-1}\beta_iv_iv_i^t-\sum_{i=1}^{n-1}\beta_iv_{i+1}v_{i+1}^t$. Therefore, $S=VDV^t-WDW^t$.
Finally $S=VD^{\frac{1}{2}}W^tWD^{\frac{1}{2}}V^t-WD^{\frac{1}{2}}V^tVD^{\frac{1}{2}}W^t=RR^t-R^tR$, where $R=VD^{\frac{1}{2}}W^t$.