Does every gutter point in SVM have positive multiplier? I understand that SVM is about solving the constrained optimization such that 
$$\min_{\mathbf{w}} \dfrac{1}{2} \mathbf{w}^T\mathbf{w}$$
subject to 
$$y_i(\mathbf{w}^T\mathbf{x_i}+b)\geq{1}, i=1, 2, ...,n$$

And this is handled using nonlinear optimization method Karush–Kuhn–Tucker approach where one step is based on the necessary complementary slackness condition such that 
$${\alpha}_i\left(y_i(\mathbf{w}^T\mathbf{x_i}+b)-1\right)=0, i=1, 2, ...,n$$ has to be satified.

Because for non-gutter dots (i.e., the points not on the edge of the separating hyperplane), we have
$$y_i(\mathbf{w}^T\mathbf{x_i}+b)-1 > 0$$
the corresponding multiplier $\alpha_i$ then must be $0$. But my question is for gutter points, because 
$$y_i(\mathbf{w}^T\mathbf{x_i}+b)-1 = 0$$
we know that the corresponding multiplier $\alpha_i$ should be non-negative, but are they necessarily positive? In other words, if I define support vector as any $\mathbf{x_i}$ on the gutter, then is this the same as if I define support vector as any $\mathbf{x_i}$ whose multiplier is positive?
 A: It's mathematically possible to have cases where the constraint $y_i(\mathbf{w}^T\mathbf{x_i}+b)\geq{1}$ is satisfied with equality and where Lagrange multiplier $a_i = 0$ if the constraint isn't active in the sense that without the constraint, you would still have the same solution $\mathbf{w}$. In practice, this is probably a knife-edge case.
Intuitively, multipliers are penalties for violating a constraint
The intuitive interpretation of Lagrangian multipliers is that they are the price of violating a constraint.
Imagine that we have the problem:
\begin{equation}
 \begin{array}{*2{>{\displaystyle}r}}
 \mbox{minimize (over $\mathbf{x}$)} & f(\mathbf{x}) \\
 \mbox{subject to} & g(\mathbf{x}) \leq 0
 \end{array}
\end{equation}
The Lagrangian is:
$$ \mathcal{L}\left(\mathbf{x}, \lambda \right) = f(\mathbf{x}) + \lambda g(\mathbf{x}) $$
$\lambda$ is a penalty for having a positive $g$. The intuitive idea is that there exists a price $\lambda^*$ for violating the constraint such that the optimizer is better off not violating the constraint.
(Side note: it's worth the time to understand KKT sufficient vs. necessary conditions, the primal vs. dual problem, weak duality, strong duality, and the saddle point property of the Lagrangian.)
Simple example:
\begin{equation}
 \begin{array}{*2{>{\displaystyle}r}}
 \mbox{minimize (over $x$)} & x^2 \\
 \mbox{subject to} & -x \leq 0
 \end{array}
\end{equation}
Lagrangian is:
$$ \mathcal{L} = x^2 - \lambda x $$
First order condition implies:
$$ 2x = \lambda $$
Solution is $x=0$, $\lambda = 0$. The solution to the unconstrained problem $\min x^2$ is $x = 0$. Basically, the point is on the boundary, but the constraint isn't doing any work.
If the unconstrained solution to the problem would have placed the point $x$ on the boundary of the constraint anyway, the penalty for violating the constraint may be zero. In practice, for many problems, this will be a knife-edge case.
