# Devision by norm vector to maximize margin in SVMs

Could someone give me an intuition why we devide the difference of the two contraints (1) $w^tx_++b=1$ and (2) $w^tx_-+b=-1$ by the length or the norm of the vector $w$ to get the margin of the hyperplane $w^tx+b=0$?

As far as I understand the concept, subtract the constraint from (2) from te constaint (1):

$$w^tx_++b-w^tx_-+b=1-(-1)$$

which gives us

$$w^t(x_+-x_-) =2$$

Now comes the part that I don't clearly understand. We devide the $w^t(x_+-x_-) =2$ by $\lVert w\rVert$ to get:

$$\cfrac{w^t(x_+-x_-)}{\lVert w\rVert} =\cfrac{2}{\lVert w\rVert}$$

Which tells us that the margin from hyperplane $w^tx+b=0$ to the constraints (1) and (2) is $\cfrac{2}{\lVert w\rVert}$.

What exactly does the norm of the the vector $w$ in this case tell us? $w$ is the slope of the hyperplane but what does the length of the slope express?

You're trying to maximize the distance between two parallel hyperplanes. To find the distance between two hyperplanes, you take a point $y$ from one hyperplane and find the shortest distance to the other hyperplane, using any point $x$ that is in the latter hyperplane. So we have $$d=||y-x||=\left|\left | \frac{(y-x)\cdot \omega}{\omega\cdot\omega} \omega \right| \right|=\frac{|y\cdot\omega-x\cdot\omega|}{||\omega||}$$ where we're taking the projection of the vector connecting $x$ and $y$ onto the $\omega$ direction and measuring the projection's magnitude. Using the formulas we have for each hyperplane, the above formula simplifies to the aforementioned $\frac{2}{||\omega||}$.
What is weird here, is when we do projections, we divide out by the size of $\omega$ to compensate for $\omega$'s contribution to the inner product in the numerator. That is, when we measure the component of a vector in the direction of a second vector, we don't want the size of the second vector affecting the measurement. However, in this particular case, the inner product in the numerator simplifies to a constant so when we change the size of $\omega$ we actually affect the distance between the two planes. Namely, because the two planes are offset from the origin by $b\pm 1$, the only way to affect the distance between the hyperplanes is if you minimize $||\omega||$ with the constraints that $$class(x_i)(\omega\cdot x_i+b)\geq 1\mbox{ for } 1\leq i\leq n,$$ where $class(x_i)\in \{-1,1\}$ denotes to which class $x_i$ belongs.