# Positive semidefinite vector $\bar{x}$ as $\bar{x}>0 :=\bar{x} \lambda \bar{x}^{T}>0$?

$A \lambda A^{T}$ (quadratic form?) is used with matrices to check definiteness. What about with vectors? If I see conditions such as $\bar{x} > 0$, how can I know whether it means $\bar{x}_{i} > 0 \quad \forall i \in I$ or $\bar{x}:=\bar{x} \lambda \bar{x}^{T}>0$?

I am trying to understand the conditions in the standard form integer optimization problem where I see such cases:

$$\text{minimize }\ \ \bar{c}^{T}\bar{x} + \bar{d}^{T}\bar{y}$$

so that $$A\bar{x} + B\bar{y} = b$$ $$\bar{x}, \bar{y} \geq 0$$ and $\bar{y}$ is continuous.

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I've honestly never seen the notation $\vec{x}>0$ before. What kind of problem are you seeing it in? –  anon Sep 22 '11 at 10:15
I'm guessing they mean $x_i>0$ for each $i$, as that would be a standard optimization constraint while the quadratic constraint seems unnatural. But like I said, I'm guessing. –  anon Sep 22 '11 at 10:29
@anon: I am trying to translate the lecture slide 3 into English here. Originally, they apparently mean that $x \in \mathbb Z$ but I see no point otherwise to transpose the other vars such as $\bar{c}$. So I think they mean $\bar{x} \in \mathbb Z^{n}$. Anyway this is apparently copycatted from Introduction to Linear Optimization -book, have to get it to understand the definition here. I think it should be some standard def for ILO problem. –  hhh Sep 22 '11 at 10:30

In the context of linear (or mixed integer) programming the inequality $x\geq 0$ is always meant in the "componentwise" sense, i.e. $x_i\geq 0$ for all $i$.
Note that for matrices one has to be even more cautious as there are the notions of "positive definite" (of denoted by $A\succ 0$), "positive" or even "total positive", which are all different.
In the context of semidefinite programming one often has both positivity constraints for vectors (something denoted by $Ax\geq 0$) and constraints on positive semidefiniteness (something denoted by $B\succcurlyeq 0$).