Find orthogonal matrices satisfying constraints I have come across the following problem, just wondered if there is an analytical way to solve it:
Given arbitrary $\mathbf{C} \in \mathbb{R}^{N \times N} $, Find matrices $\mathbb{R}^{N \times N} \ni \mathbf{A} \neq \mathbf{0}$ and $\mathbb{R}^{N \times N} \ni \mathbf{B} \neq \mathbf{0} $ such that:
$\mathbf{A}$+$\mathbf{B}$ = $\mathbf{C}$
$\mathbf{A}^T\mathbf{B}$ = $\mathbf{0}$
So far, I can solve this numerically (Newton-Raphson algorithm) for specific cases of $\mathbf{C}$. However, if anyone has any different methods I will be grateful.
 A: Since you did not provide more information, I assume that your matrices are quadratic, i.e., $A,B,C\in\mathbb R^{n\times n}$. Then for every subspace $M\subset\mathbb R^n$ there is a solution to your problem. By this, I mean the following:
Let $M\subset\mathbb R^n$ be a subspace. Then a solution is given by $A = P_MC$ and $B = (I-P_M)C$, where $P_M$ is the orthogonal projection onto $M$.
A: I assume that $C$ is a square matrix, $C\in M_{n\times n}(\mathbb{R})$. We need to find matrix $A$ such that $A^T(C-A)=0$, that is $$A^TA= A^TC = C^TA$$
We have then
$$ (A-\frac12 C)^T(A-\frac12 C) = \frac14 C^TC$$
Using RQ decomposition of matrices $C$ and $A-\frac12 C$ it can be proved that this means that
$$ A-\frac12 C = \frac12 R C$$ 
for some orthogonal matrix $R$, that is
$$ A =\frac12(\mathbf{1}+R)C $$
 Therefore the condition $A^T(C-A)=0$ takes form
$$ C^T(\mathbf{1}+R^{-1})(\mathbf{1}-R)C = 0 $$
$$ C^T(R^{-1}-R)C = 0$$
If matrix $C$ is nondegenerate, it's solution is any $R$ such that $R^2=\mathbf{1}$. You can find many such $R$, for example for any subspace $M\in \mathbb{R}^n$ you can take $R = 2P_M - 1$, where $P_M$ is an orthogonal projection on $M$, which gives you $A=P_M C$, $B = (1-P_M)C$.
If $C$ is degenerate, the situation is more complicated, there are additional solutions.
