C.2 The eigenvalue equation in kernel PCA

The equivalence of the equations

$$\lambda \bf K \alpha \bf K \bf K \alpha$$ (C.8)

and

$$\lambda \alpha \bf K \alpha$$ (C.9)

is shown.

Equation (C.8) follows immediately from (C.9). To prove the opposite direction, we assume that a vector $\beta$ exists that is not an eigenvector of $\bf K$, while $\bf K$$\beta$ is an eigenvector of $\bf K$. This assumption infers that (C.8) is fulfilled and (C.9) is not. Thus, we need to show that the assumption leads to a contradiction.

We only consider the case that $\beta$ is orthogonal to the subspace ker $\bf K$ (the space of vectors $\alpha$ fulfilling $\bf K$$\alpha$ = 0) because the elements of ker $\bf K$--if they exist--solve already both (C.8) and (C.9). Since $\bf K$ is symmetric, $\beta$ can be written as a linear combination of the pairwise orthogonal eigenvectors $\alpha$^l of $\bf K$, $\beta$ = $\sum_{l}^{}$u_l$\alpha$^l. At least, two u_l must differ from zero because $\beta$ itself is not an eigenvector. It follows that $\bf K$$\beta$ = $\sum_{l}^{}$u_l$\lambda{^\prime}_{l}$$\alpha$^l with $\lambda{^\prime}_{l}$ being the eigenvalues corresponding to $\alpha$^l. All eigenvalues are non-zero because $\beta$ is orthogonal to ker $\bf K$. Thus, $\bf K$$\beta$ can be also not an eigenvector of $\bf K$. This contradicts our first assumption. Therefore, (C.9) follows from (C.8).