How to minimize a linear function over a halfspace efficiently and intuitively

Question

Consider the following fundamental problem:

Two methods:

1. By duality: ($\lambda, b \in R$)
   $L(x,\lambda)=c^Tx+\lambda(a^Tx-b)=x^T(c+\lambda a)-\lambda b \ \ $. Therefore,
   $g(\lambda)=-\lambda b, \ \ c+\lambda a=0 \ \ \& \ \ \lambda >0$. Otherwise, $g(\lambda)=-\infty$.
2. By the following:

My question is:(yellow portion)

1. I have no idea why the solution has to decompose $c$ in that way (no other choice? why not decompose $a$?)
2. And the last part:
   $p^* = c^Tx = (\lambda a)^Tx = \lambda a^Tx \leq \lambda b$ (by the constraint).
   How do we know the optimal value for this part is not unbounded below?

Should we do all effort to try every possibility to show it is not bounded below like the previous a few cases?

Answer 1

First of all, in the first solution, the correct condition on $\lambda$ is $\lambda \ge 0$, not $\lambda > 0$.

1. I have no idea why the solution has to decompose $c$ in that way (no other choice? why not decompose $a$?)

Not sure how the author came up with the second solution, but when reading it, I immediately saw that it just follows from the first solution ($\lambda$ in the first is equivalent to $-\lambda$ in the second).

2. And the last part: $p^* = c^Tx = (\lambda a)^Tx = \lambda a^Tx \leq \lambda b$ (by the constraint). How do we know the optimal value for this part is not unbounded below?

The actually it is bounded below. The inequality should be reserved since $\lambda \le 0$.