# Why does the supporting vector satisfy $y_i(\mathbf{w}^T\mathbf{x_i}+b) = 1$ instead of $> 1$ or $= 2$

The SVM is about solving the constrained optimization such that

$$\min_{\mathbf{w}} \dfrac{1}{2} \mathbf{w}^T\mathbf{w}$$ subject to $$y_i(\mathbf{w}^T\mathbf{x_i}+b)\geq{1}, i=1, 2, ...,n$$

Suppose that $$(\mathbf{w}, b)$$ is the optimal hyperplane that we get by solving the SVM problem, then it must be true that the supporting vectors satisfy the following relationship:

$$y_i(\mathbf{w}^T\mathbf{x_i}+b) = 1$$

My question is why this must be the case?

Our constraint says that our hyperplane is the optimal one as long as

$$y_i(\mathbf{w}^T\mathbf{x_i}+b)\geq{1}, i=1, 2, ...,n$$

is satisfied.

So why couldn't the supporting vectors satisfy $$y_i(\mathbf{w}^T\mathbf{x_i}+b) > 1?$$ or $$y_i(\mathbf{w}^T\mathbf{x_i}+b) = 2?$$

or $$y_i(\mathbf{w}^T\mathbf{x_i}+b) = 100?$$

I think the confusion comes from the definition of support vectors : a point $$i$$ is defined as a support vector precisely if $$y_i(w^Tx_i+b) = 1$$. The inequality $$y_i(w^Tx_i+b) \geq 1$$ must hold for all data points, but the support vectors are the data points for which this constraint is attained the least.
Support vectors satisfy $$y_i(w^Tx_i+b) = 1$$... because that's how they are defined !
Indeed, if you add training data to an SVM, any point that does not change the support vectors (i.e. any new point for which $$y_i(w^Tx_i+b)>1$$) does not entail any computation or additional training. New training is needed only if you add a training point for which $$y_i(w^Tx_i+b)<1$$, and in that case the margins will need to be recomputed and this new point will become most likely a support vector, so $$y_i(w^Tx_i+b)=1$$ (here $$w$$ and $$b$$ are changed)