I am studying on the optimization via vector method. The reference book is Optimization by Vector Method by Luenberg.
I have trouble in understanding the following statement [p.123];
We consider the unknown $x^*$ in a dual space $X^*$ and express the contraints in the form: $$\langle y_1,x^*\rangle=c_1 , \cdots , \langle y_n,x^*\rangle=c_n$$
If $\bar{x}^*$ is any vector satisfying the contraints, we have
$$d = \min_{\langle y_i,x^* \rangle=c_i}|| x^* || = \min_{m^*\in M^+}|| \bar{x}^*-m^* ||$$
where $M$ denotes the space generated by the $y_i$'s and $M^+$ is the orthogonoal complement of $M$.
My question is because the second equility holds.
Can you explain the reason for
$$\min_{\langle y_i,x^*\rangle=c_i}|| x^* || = \min_{m^*\in M^+}|| \bar{x}^*-m^* ||$$
not for
$$\min_{\langle y_i,x^*\rangle=c_i}|| x^* || = \min_{\bar{x}^* \in X^*}|| \bar{x}^* ||$$
Thank you ahead.
This is essentially just reformulating the problem. It is an exercise in basic linear algebra to show that if $\bar x^*$ is a particular solution to the constraint equations, then the entire solution space is give by $\bar x^* + M^\perp$. That is:
$$ \langle y_i,x^*\rangle = c_i \Leftrightarrow x^*-\bar{x}^*\in M^\perp \Leftrightarrow x^*=\bar{x}^*-m^*, \ m^*\in M^\perp $$
Your alternative statement is wrong because $\bar x^*$ should not be a variable that you minimize over. Strictly speaking, $\min_{\bar{x}^* \in X^*}|| \bar{x}^* ||=0$, which is unrelated to the original problem.