3
$\begingroup$

I am trying to understand the geometry for linear classification models. A linear model, according to Bishop's books, is defined as: $\mathbf{y} = \mathbf{w}^T \mathbf{x} + w_0$.

For instace we have have the following example with a dataset with two features (feature A and feature B) and two classes (class A and class B):

enter image description here

The decision boundary here is the: $\mathbf{y} = \mathbf{w}^T \mathbf{x} + w_0$. The input $\mathbf{x} \in \mathbb{R}^{2}$. Here the decision boundary is a line that can seperate these two classes, however, I am bit puzzled on what $\mathbf{w}$ is. Shouldnt be also $\mathbf{w} \in \mathbb{R}^{2}$ and if yes what is it exactly if not just $w = -1$?

How is this $\mathbf{w}$ perpendidular to the decision boundary?

$\endgroup$
1
  • $\begingroup$ If $x \in \mathbb{R}^2$ then so is $-x$ but then $-x+9$ doesn't make sense since $9 \notin \mathbb{R}^2$. $\endgroup$ Commented Sep 5 at 15:30

1 Answer 1

Reset to default
3
$\begingroup$

Write $z=-x-y+9=(-1, -1)\cdot (x,y)+9$

Given $\begin{bmatrix} x \\ y\end{bmatrix} \in \mathbb{R}^2$, let $w=\begin{bmatrix} -1 \\ -1 \end{bmatrix}$. we can evaluate the dot product and then add $w_0=9$. If the sign is positive, it is class $B$.

The decision boundary is parallel to the vector $\begin{bmatrix} 1 \\ -1\end{bmatrix}$, which is perpendicular to $w$.

$\endgroup$
0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.