Find $A$ if $B=A-A^T$

Question

Suppose $B=A-A^T$ and I know $B\in\mathbb R^{n\times n}$. What is a simple way to get $A$? And what if I have the constraint that $A_{ij}\ge 0$ $\forall i,j$?

Clarifications:

• $B$ is skew symmetric
• $a_{ii}$ (the diagonal elements of $A$) are zero

Answer 1

If you only know $B$ then for the equation to have a solution, $B$ must be skew symmetric. $$\implies b_{ij}=a_{ij}-a_{ji} \forall i,j$$

Now you can pick many $a_{ij}, a_{ji}$ that satisfy this equation. One choice is $a_{ij}=k, a_{ji}=k-b_{ij}$. Similarly you can pick nearly anything for the non diagonal elements. For diagonals you can have any $a_{ii}=k \in \mathbb{R}$.

Answer 2

You can’t, not without further information.

Suppose that $B$ is :

$$ B= \begin{bmatrix} 0 & 1\\ -1&0\\ \end{bmatrix} $$

NB : $B$ is antisymmetric from the given condition and therefore the leading diagonal is zero.

Then all we can deduce about $A$ is that it has the form :

$$ A= \begin{bmatrix} p&q+1\\ q&r\\ \end{bmatrix} $$

Where $p$, $q$, and $r \in\mathbb{R}$

Even with the constraint, you are only requiring that $p, q, r > 0$, which doesn’t allow you to solve for them uniquely.