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

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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

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 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 an orthogonal foundation. 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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