System of equations involving rank-1 matrices

Question

Let $u,v \in \mathbb{R}^n$. Then I want to find all matrices $X \in \mathbb{R}^{n \times n}$ such that: \begin{align} X^T u u^T X &= u u^T \\ X^T v v^T X &= v v^T \end{align} The answer to this question analyzes the first equation thoroughly, and finds that the most general solution is $$ X = \pmatrix{ P & 0 \\ R & S } $$ where $P \in O(1,\mathbb{R})$, $R\in\text{Mat}_{(n-1)\times 1}(\mathbb{R})$ and $S\in\text{Mat}_{(n-1)\times (n-1)}(\mathbb{R})$. Applying this solution to the second equation gives \begin{align} X^T v v^T X &= \pmatrix{ P^T & R^T \\ 0 & S^T } v v^T \pmatrix{ P & 0 \\ R & S } &= \pmatrix{ (P^T v_1 + R^T v_2)(v_1^T P + v_2^T R) & (P^T v_1 + R^T v_2)v_2^T S \\ S^T v_2 (v_1^T P + v_2^T R) & S^T v_2 v_2^T S } \end{align} where I've set $v = \pmatrix{v_1 \\ v_2}$. Equating this matrix to $v v^T$ gives the following conditions: \begin{align} 0 &= R^T v_2 v_1^T P + P^T v_1 v_2^T R + R^T v_2 v_2^T R \\ v_1 v_2^T &= P^T v_1 v_2^T S + R^T v_2 v_2^T S \\ v_2 v_2^T &= S^T v_2 v_2^T S \end{align} I don't see how to find the most general forms of $P,R,S$ which satisfy these equations. Ultimately I would like to generalize this to the case of $m$ equations $X^T u_i u_i^T X = u_i u_i^T$, $1 \le i \le m$, but right now I'm trying to solve the simpler $m = 2$ case. Any insights would be appreciated.

Answer 1

The equation $X^Tuu^TX=uu^T$ implies that $X^Tu=\pm u$. So, if $U$ denotes the augmented matrix $[u_1,\ldots,u_m]$, then the system of equations $$ X^Tu_iu_i^TX=u_iu_i^T\ (i=1,2,\ldots,m)\tag{1} $$ is equivalent to the matrix equation $$ X^TU=UD\tag{2} $$ where $D$ is a diagonal matrix whose diagonal entries are $\pm1$ and $X,D$ are the unknowns.

Since $(X,D)$ is a solution to $(2)$ iff $(-X,-D)$ is a solution, we may assume that the first diagonal entry of $D$ is $1$. Therefore there are $2^{m-1}$ choices of $D$. For every fixed $D$, the least-norm matrix $X$ that minimises $\|X^TU-UD\|_F$ is $X^T=UDU^+$. It follows that $X^TU=UD$ is solvable if and only if $UD(I_m-U^+U)=0$. And if it is solvable (such as when $D=\pm I$), the general solution $X$ with this $D$ is given by $X^T=UDU^++M(I_n-UU^+)$, where $M$ is arbitrary.