I'm reading a chapter on linear classifiers, and the authors define the decision hyperplane as:
$$ g(\mathbf{x}) = \mathbf{w}^T\mathbf{x} + w_0 = 0 $$ They say that if $\mathbf{x}_1$ and $\mathbf{x}_2$ are points on the decision hyperplane, then: $$ \mathbf{w}^T\mathbf{x}_1 + w_0 = 0 \\ \mathbf{w}^T\mathbf{x}_2 + w_0 = 0 $$ and: $$ \mathbf{w}^T(\mathbf{x}_1 - \mathbf{x}_2) = 0 $$ They then write the difference vector $\mathbf{x}_1 - \mathbf{x}_2$ lies on the decision hyperplane. Here I am lost. If it were on the decision hyperplane, wouldn't it need to satisfy the original equation? Specifically: $$ \mathbf{w}^T(\mathbf{x}_1 - \mathbf{x}_2) + w_0 = 0 $$
What am I missing?
The statement as written is false for arbitrary $w_0$. Given $$ \mathbf{w}^T\mathbf{x}_1 + w_0 = 0 \\ \mathbf{w}^T\mathbf{x}_2 + w_0 = 0 $$ it follows that $\mathbf{w}^T\mathbf{x}_1=\mathbf{w}^T\mathbf{x}_2=-w_0$, and hence that $$\mathbf{w}^T(\mathbf{x}_1 - \mathbf{x}_2) =\mathbf{w}^T\mathbf{x}_1-\mathbf{w}^T\mathbf{x}_2 =(-w_0)-(-w_0)= 0. $$ So It is "obviously" true if and only if $w_0=0$. I would say you are not missing anything.
You'd be surprised how many errors slip through dozens of editions.