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C.2 The eigenvalue equation in kernel PCA

The equivalence of the equations

n$\displaystyle \lambda$$\displaystyle \bf K$$\displaystyle \alpha$ = $\displaystyle \bf K$$\displaystyle \bf K$$\displaystyle \alpha$ (C.8)

and

n$\displaystyle \lambda$ $\displaystyle \alpha$ = $\displaystyle \bf K$$\displaystyle \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$$ \alpha$ = 0) because the elements of ker $ \bf K$--if they exist--solve already both (C.8) and (C.9). Since $ \bf K$ is symmetric, $ \beta$ can be written as a linear combination of the pairwise orthogonal eigenvectors $ \alpha$l of $ \bf K$, $ \beta$ = $ \sum_{l}^{}$ul$ \alpha$l. At least, two ul must differ from zero because $ \beta$ itself is not an eigenvector. It follows that $ \bf K$$ \beta$ = $ \sum_{l}^{}$ul$ \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).


next up previous contents
Next: C.3 Estimate of error Up: C. Proofs Previous: C.1 Probabilistic PCA and
Heiko Hoffmann
2005-03-22