Find orthogonal matrices satisfying constraints

Question

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.

Answer 1

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.

Answer 2

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 $M$.