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We all know that for a feature map $\Phi$, there exists a kernel function $K_1$ satisfying $\langle\Phi (x),\Phi (y)\rangle=K_1(x,y)$.
For a positive-deļ¬nite matrix $A$, the quadratic form $\langle\Phi (x),A\Phi (y)\rangle=\langle A^{1/2}\Phi (x),A^{1/2}\Phi (y)\rangle$, so there also exists a kernel function $K_2$ satisfying $\langle\Phi (x),A\Phi (y)\rangle=K_2(x,y)$.

I want to know whether there is some relationship between $K_2$ and $K_1$.

Math Attack
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vegetabletina
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1 Answers1

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If $A$ is positive definite, by the properties of its eigenvalues (see Rayleigh Quotient, for example), the values of the quadratic form are bounded. For any vector $v$ in the feature space, we have that: \begin{equation} \lambda_\min \langle v, v \rangle \leq \langle v, Av \rangle \leq \lambda_\max \langle v, v \rangle\,, \end{equation} where $\lambda_\min$ and $\lambda_\max$ denote the minimum and maximum eigenvalues of $A$, respectively. Therefore, we can say that: \begin{equation} \lambda_\min K_1(x,x) \leq K_2(x,x) \leq \lambda_\max K_1(x,x)\,, \end{equation} for all $x$.

rafaol
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