1
$\begingroup$

I'm currently following this material

Optimization Theory: Chapter 2 Theory of Constrained Optimization

And I can't understand why the following statement is true, between the equations (2.9) and (2.10):

The only way that such a direction does not exist is that $∇f(x)$ and $∇h(x)$ are parallel

I can grasp the idea intuitively but not in the mathematical sense

Edit: the problem actually reduces to:

Let $a$, $b$ be two vectors belonging to $R^2$ and $c$ a scalar. There is an equivalence:

There is no $d$ such that $a^Td<0$ and $b^Td=0$ $\iff$ $a = cb$

$\endgroup$
2
$\begingroup$

Regarding your equivalence,

The direction $\impliedby$ is easy. Suppose $a$ and $b$ are parallel with $a=cb$ and $a'd<0$, where $c$ is a scalar and $a,b,d$ are vectors. Then $a'd=cb'd=0$, but this contradicts $a'd<0$, so there is no $d$ satisfying the conditions.

The direction $\implies$ can by using the fact that $a'd = ||a||\: ||d||\cos\theta$ where $\theta$ is the angle between the vectors. If $a,b$ are not parallel then we can find a $d$ such that $b$ and $\pm d$ are orthogonal and $a$ and $d$ or $-d$ form an obtuse angle (can't be orthogonal or else the original vectors were parallel).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.