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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.

Sebastiano
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user142054
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1 Answers1

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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.

p.s.
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