1
$\begingroup$

I was just watching this tutorial about Support Vector Machines, and I came to a halt because of the following problem.

Given that $\vec{w}$ is a vector perpendicular to a hyperplane separating two classes, does the equation $\vec{u} \cdot\vec{w}+b=1$ or ($\vec{u}\cdot\vec{w}+b=-1$) represent another hyperplane parallel to this one? Where $\vec{u}$ is a some vector.

That is, in the case of a 2D plane, does $\vec{u}\cdot\vec{w}+b=1$ represent a line, that is parallel to the original hyperplane?

$\endgroup$
0
$\begingroup$

Yes it does.

We can directly see the relation to the equation of a line segment $mx+b=y$.

Anyways, let's get the equation of the hyperplane given here. We're given that $\vec{w}$ is a perpendicular vector to it (from the origin). Let's assume that $\vec{x}$ is a vector that represents a point on the hyperplane given. Therefore, since $\vec{x}-\vec{w}$ is vector perpendicular to $\vec{w}$ therefore, it can be represented by $$(\vec{x}-\vec{w})\cdot\vec{w}=0$$

Now, consider a hyperplane that is parallel to the original one. Let that be represented by $\lambda\vec{w}$. So, it's equation would be $(\vec{x}-\lambda\vec{w})\cdot\lambda\vec{w}=0$. That is $$\vec{x}\cdot\vec{w}-\lambda|w|^2=0$$

As we can see, that if we change $b$ to $1-\lambda|w|^2$ (since $b$ is an arbitrary constant), then $\vec{x}\cdot\vec{w}+b=1$ changes to $\vec{x}\cdot\vec{w}-\lambda|w|^2=0$, which is an equation of an hyperplane parallel to the original one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.