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Machine Learning: Difference between revisions
→Mercer's Theorem
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Positive Definite:<br> | Positive Definite:<br> | ||
Let <math>\mathbf{v} \in \mathbb{R}^n</math>.<br> | Let <math>\mathbf{v} \in \mathbb{R}^n</math>.<br> | ||
Then <math>\mathbf{v}^T K v | Then | ||
= v^T | <math> | ||
= \sum_i \sum_j v_{i}K_{ij}v_{j} | \begin{aligned} | ||
= \sum_i \sum_j v_{i}\phi(x^{(i)})^T\phi(x^{(j)})v_{j} | \mathbf{v}^T K v | ||
= \sum_i \sum_j v_{i} \sum_k \phi_k(x^{(i)}) \phi_k(x^{(j)})v_{j} | &= v^T [\sum_j K_{ij}v_j]\\ | ||
= \sum_k \sum_i \sum_j v_{i} \phi_k(x^{(i)}) \phi_k(x^{(j)})v_{j} | &= \sum_i \sum_j v_{i}K_{ij}v_{j}\\ | ||
= \sum_k \sum_i v_{i} \phi_k(x^{(i)}) \sum_j \phi_k(x^{(j)})v_{j} | &= \sum_i \sum_j v_{i}\phi(x^{(i)})^T\phi(x^{(j)})v_{j}\\ | ||
= \sum_k (\sum_i v_{i} \phi_k(x^{(i)}))^2 | &= \sum_i \sum_j v_{i} \sum_k \phi_k(x^{(i)}) \phi_k(x^{(j)})v_{j}\\ | ||
\geq 0 | &= \sum_k \sum_i \sum_j v_{i} \phi_k(x^{(i)}) \phi_k(x^{(j)})v_{j}\\ | ||
&= \sum_k \sum_i v_{i} \phi_k(x^{(i)}) \sum_j \phi_k(x^{(j)})v_{j}\\ | |||
&= \sum_k (\sum_i v_{i} \phi_k(x^{(i)}))^2\\ | |||
&\geq 0 | |||
\end{aligned} | |||
</math> | </math> | ||
}} | }} | ||