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

- $Q = qq^T$ and $Q_d = q_dq_d^T$ with $q, q_dinmathbb{R}^4$.
- Obviously, $f(Q)$ is convex in $Q$.
- 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 anorthogonalbasis. 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!