AIDAID202 · Lectures 9–10

Lecture 10 · Linear discriminant functions

Geometry of y(x)=wᵀx+b

The weight vector is normal to the decision boundary. The score becomes a geometric signed distance after normalization by ||w||.

Predict first: If we double w and b together, does the geometric boundary move? Does y(x) change? Does y(x)/||w|| change?

2-D geometry

y(x)=wᵀx+b—
||w||—
signed distance—
class sign—
Why is the boundary perpendicular to w?

Any displacement Δ along the boundary leaves the score unchanged, so wᵀΔ=0. Thus w is orthogonal to directions within the boundary.