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Machine Learning: Difference between revisions
→Mercer's Theorem
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\mathbf{v}^T \mathbf{K} \mathbf{v}= \mathbf{v}^T [\sum_j K_{ij}v_j] | \mathbf{v}^T \mathbf{K} \mathbf{v}= \mathbf{v}^T [\sum_j K_{ij}v_j] | ||
= \sum_i \sum_j v_{i}K_{ij}v_{j} | = \sum_i \sum_j v_{i}K_{ij}v_{j} | ||
</math><br> | |||
<math> | |||
= \sum_i \sum_j v_{i}\phi(\mathbf{x}^{(i)})^T\phi(\mathbf{x}^{(j)})v_{j} | = \sum_i \sum_j v_{i}\phi(\mathbf{x}^{(i)})^T\phi(\mathbf{x}^{(j)})v_{j} | ||
</math><br> | |||
<math> | |||
= \sum_i \sum_j v_{i} \sum_k \phi_k(\mathbf{x}^{(i)}) \phi_k(\mathbf{x}^{(j)})v_{j} | = \sum_i \sum_j v_{i} \sum_k \phi_k(\mathbf{x}^{(i)}) \phi_k(\mathbf{x}^{(j)})v_{j} | ||
</math><br> | |||
<math> | |||
= \sum_k \sum_i \sum_j v_{i} \phi_k(\mathbf{x}^{(i)}) \phi_k(\mathbf{x}^{(j)})v_{j} | = \sum_k \sum_i \sum_j v_{i} \phi_k(\mathbf{x}^{(i)}) \phi_k(\mathbf{x}^{(j)})v_{j} | ||
</math><br> | |||
<math> | |||
= \sum_k \sum_i v_{i} \phi_k(\mathbf{x}^{(i)}) \sum_j \phi_k(\mathbf{x}^{(j)})v_{j} | = \sum_k \sum_i v_{i} \phi_k(\mathbf{x}^{(i)}) \sum_j \phi_k(\mathbf{x}^{(j)})v_{j} | ||
</math><br> | |||
<math> | |||
= \sum_k (\sum_i v_{i} \phi_k(\mathbf{x}^{(i)}))^2 | = \sum_k (\sum_i v_{i} \phi_k(\mathbf{x}^{(i)}))^2 | ||
\geq 0 | \geq 0 | ||