k(x,y)=x⋅yProof of valid kernel
Given the kernel
k(xi,xj)=xi⋅xj, we need to prove
∑i=1n∑j=1ncicjk(xi,xj)≥0∑i=1n∑j=1ncicj(xi⋅xj)Rearrange terms using the bilinearity of the dot product:
∑i=1n∑j=1ncicj(xi⋅xj)=(∑i=1ncixi)⋅(∑j=1ncjxj)Let v=∑i=1ncixi:(∑i=1ncixi)⋅(∑j=1ncjxj)=v⋅vSimplify using the definition of the Euclidean norm:
v⋅v=∥v∥2≥0Therefore, for any
c and
x ∑i=1n∑j=1ncicjk(xi,xj)≥0.