I am trying to get some intuition behind SVM algorithm by reading slides Support Vector Machines by Andrew W. Moore.
Let's consider the case of Linear SVM.
In the slides it's descrives that two planes can be defined.
plus-plane:$x:w.x+b=+1$
minus-plane:$x:w.x+b=-1$
Q: I know the plane can be defined as $n(r-r_0)=0$, however I don't understand why I can ensure that there is possible to define two planes $x:w.x+b=+1$ and $x:w.x+b=-1$, as I understood it's possible to define the separating plane as $x:w.x+b=0$ which is indeed in the middle of the plus plane and the minus plane.
In addition as I see from the formula $w.x+b=0$ is exactly $nr-nr_0=0$, where $w=n$, $b=-nr_0$, $x=r$, so obviously $b$ is a function of $w$, however we missed this information is $w.x+b=0$. It might be better to save this information, what the reason we don't use it?
On the page 7, tehre is a claim
The vector w is perpendicular to the Plus Plane. Why?
Initially I thought it's because the construction of the formula of $w.x+b=0$ and it follows from the definition of the plane, $w=n$ and $n$ is perpendicular to the plane, so $w$ is perpendicular too.
But, there is a hint in the slides.
Let u and v be two vectors on the Plus Plane. What is w . ( u – v ) ?
How I am supposed to use this hint?
For your first question, $w.x+b=\pm1$ are two general planes parallel to $w.x+b=0$. So, there is loss of generality in supposing them to be the canonical planes.
As for your second question, for two vectors $u$ and $v$, in the plus plane, with end points as $u_{1}, u_{2}$ and $v_{1}, v_{2}$, we have \begin{eqnarray} u=u_{1}-u_{2}\\ v=v_{1}-v_{2}\\ \end{eqnarray} Hence, from the eqn of the canonical plane \begin{eqnarray} w.u_{1,2}=w.v_{1,2}&=&1-b\\ w.(u-v)=w.(u_{1}-u_{2}-v_{1}+v_{2})&=&0 \end{eqnarray} More generally, $w.(nu+nv)=0\ \ \forall m,n\in R$. Thus any vector in the plus plane represented by $nu+nv$ is orthogonal to w.
So, $w$ is orthogonal to the canonical plane. (and also the separating plane)