Section to Skew-Symmetrization Map

Question

Let $A$ be an $n\times n$ matrix skew-symmetric matrix. Define the map $\mathbb{R}^{d^2}\to Skew_d$ by $$ B\mapsto B^{\top} - B. $$ Does this map have a continuous right inverse?

Answer 1

The following addresses the question as first stated, which read, if I recall correctly:

"Let $A$ be an $n\times n$ matrix skew-symmetric matrix. Does there necessarily exist a unique $n\times n$ matrix $B$ such that

$A = B^T - B." \tag 0$

To which I responded:

Pick any symmetric matrix

$C = C^T; \tag 1$

let

$B = -\dfrac{1}{2}(A + C); \tag 2$

then

$B^T = -\dfrac{1}{2}( A^T + C^T) = -\dfrac{1}{2}(-A + C), \tag 3$

since

$A^T = -A, \tag{3.5}$

and

$B^T - B = -\dfrac{1}{2}(-A + C) - (-\dfrac{1}{2}(A + C)) = A. \tag 4$

These calculations indicate the existence of an infinite number of matrices $B$ such that (4) holds; thus no such $B$ is unique.

The preceding result may be used to address the present question of the existence of a continuous right inverse to the skew-symmetrization map, which we denote by

$\Sigma(B) = B^T - B. \tag 5$

According to what has been presented in the above, every skew-symmetric matrix is of the form $\Sigma(B)$ for some $B$. Thus a right inverse $\theta$ to $\Sigma$ will assign to every skew-symmetric $A$ a matrix $\theta(A)$ such that

$\Sigma ( \theta(A)) = A. \tag 6$

Picking any symmetric matrix $C$ as in (1), we set

$\theta(A) = -\dfrac{1}{2}(A + C) \tag 7$

as in (2); it is easy to see that such a $\theta$ is continuous, since both matrix addition and division by $2$ are continuous operations. Now (6) is simply the result of our preceding calculations, and $\theta$ is a right inverse of $\Sigma$.