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matrices – Design an orthogonal foundation for rank $2$ matrix in a scalar convex duty

Consider the next duty $$f(Q) = |Q-Q_d|_F^2,$$ the place

- $Q = qq^T$ 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 minimal.

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

Goal: Design a $f(Q)$ such that the matrix in $|cdot|_F^2$ is of

rank$2$. Meanwhile, it has anorthogonalfoundation. On the opposite hand, $f(Q)$ has to have the properties 2. and three.

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

What I strive is $$Omega Q_dQ-Q_d,$$ the place $Omega$ is skew-symmetric. it is because $$Omega Q_dQ-Q_d=Omega q_dq_d^Tqq^T-q_dq_d^T$$ So foundation ${Omega q_d, q_d}$. However, the minimal worth of $|Omega Q_dQ-Q_d|_F^2$ could not emerge when $Q=Q_d$.

Any good options? I value it!

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