# matrices – Design an orthogonal basis for rank \$2\$ matrix in a scalar convex function

Consider the following function $$f(Q) = |Q-Q_d|_F^2,$$ where

1. $$Q = qq^T$$ and $$Q_d = q_dq_d^T$$ with $$q, q_dinmathbb{R}^4$$.
2. Obviously, $$f(Q)$$ is convex in $$Q$$.
3. So $$f(Q_d) =0$$, which is minimum.

We know, $$Q-Q_d$$ is of rank $$2$$ with the basis for its column space $${q,, q_d}$$.

Goal: Design a $$f(Q)$$ such that the matrix in $$|cdot|_F^2$$ is of rank $$2$$. Meanwhile, it has an orthogonal basis. On the other hand, $$f(Q)$$ has to have the properties 2. and 3.

For example, $$f(Q)=|Omega Q+QOmega^T|_F^2$$ where $$Omega$$ is a skew symmetric matrix. The above has an orthogonal basis $${Omega q,, q}$$, since $$q^TOmega q=0$$.

What I try is $$Omega Q_dQ-Q_d,$$ where $$Omega$$ is skew-symmetric. this is because $$Omega Q_dQ-Q_d=Omega q_dq_d^Tqq^T-q_dq_d^T$$ So basis $${Omega q_d, q_d}$$. However, the minimal value of $$|Omega Q_dQ-Q_d|_F^2$$ may not appear when $$Q=Q_d$$.

Any nice suggestions? I appreciate it!