If $A$ is a skew symmetric matrix, is it possible to always find a vector $x$ such that $x - x^\top = A$

Question

If $A \in \mathbb{R}^{d \times d}$ is a skew symmetric matrix, is it possible to always find a vector $x \in \mathbb{R}^d$ such that $x1^\top - 1x^\top = A$, where $1$ is a vector of all ones $\in \mathbb{R}^d$?

Answer 1

Here I calculate this for dimension 2 only.

Since A is a skew-symmetric matrix,let $$A=\begin{pmatrix} 0 & a \\ -a & 0\end{pmatrix}$$

If possible,let we can always find a vector $$x=\begin{pmatrix} b \\ c\end{pmatrix}$$ s.t. $$x1^t-1x^t=A$$

Now , $$x^t=\begin{pmatrix} b &c\end{pmatrix}$$

Then,$$\begin{pmatrix} b \\ c\end{pmatrix}\begin{pmatrix} 1&1\end{pmatrix}-\begin{pmatrix} 1\\1\end{pmatrix}\begin{pmatrix} b &c \end{pmatrix}=\begin{pmatrix} 0 & a \\ -a & 0\end{pmatrix}$$

Thus, $$\begin{pmatrix} b & b \\ c&c\end{pmatrix}-\begin{pmatrix} b &c\\b&c \end{pmatrix}=\begin{pmatrix} 0 & a \\ -a & 0\end{pmatrix}$$

$$\therefore \begin{pmatrix} 0 & b-c\\ c-b & 0\end{pmatrix}=\begin{pmatrix} 0 & a \\ -a & 0\end{pmatrix}$$ Thus,if $b-c=a$ holds, only then we get such vector $x$.So,for given $A$,we know $a$.Then we have to express $a$ as the difference of two numbers,then using them, we can make the vector $x$.So,we can always find $x$.