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

Hello pricey customer to our community We will proffer you an answer to this query matrices – Design an orthogonal foundation for rank \$2\$ matrix in a scalar convex duty ,and the respond will breathe typical by means of documented data sources, We welcome you and proffer you fresh questions and solutions, Many customer are questioning in regards to the respond to this query.

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!

we are going to proffer you the answer to matrices – Design an orthogonal foundation for rank \$2\$ matrix in a scalar convex duty query through our community which brings all of the solutions from a number of dependable sources.