Eigenfunctions and eigenvalues of the product of two exponential kernels - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T16:11:39Z http://mathoverflow.net/feeds/question/118785 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118785/eigenfunctions-and-eigenvalues-of-the-product-of-two-exponential-kernels Eigenfunctions and eigenvalues of the product of two exponential kernels Ivan 2013-01-13T07:25:25Z 2013-01-13T15:32:52Z <p>Consider the following exponential kernel:</p> <p><code>$k(x_1, x_2) = \exp\left(\frac{|x_1 - x_2|}{L}\right)$</code>,</p> <p>which is symmetric and non-negative definite. By virtue of Mercer's theorem, we have</p> <p><code>$k(x_1, x_2) = \sum_{i = 1}^\infty \lambda_i \phi_i(x_1) \phi_i(x_2)$</code></p> <p>where $\lambda_i$ and $\phi_i$ are the eigenvalues and eigenfunctions of $k$, respectively. Now, consider the following product:</p> <p><code>$K((x_1, y_1), (x_2, y_2)) := k(x_1, x_2) k(y_1, y_2) = \exp\left( -\frac{|x_1 - x_2|}{L} - \frac{|y_1 - y_2|}{L}\right)$</code>.</p> <p>Since the product of two symmetric, non-negative definite kernels is another kernel with the same properties, Mercer's theorem still applies.</p> <p>The question is: Having computed $\lambda_i$ and $\phi_i$ of $k$, what can we say about the eigenfunctions and eigenvalues of $K$?</p> <p>Thank you.</p> <p>Regards, Ivan</p> http://mathoverflow.net/questions/118785/eigenfunctions-and-eigenvalues-of-the-product-of-two-exponential-kernels/118817#118817 Answer by Uwe Franz for Eigenfunctions and eigenvalues of the product of two exponential kernels Uwe Franz 2013-01-13T15:32:52Z 2013-01-13T15:32:52Z <p>You have two independent sets of variables, so it is a tensor product, not a Hadamard product, no? I am not sure I understand Suvrit's comment...</p> <p>I think you just get</p> <p>$K((x_1,y_1),(x_2,y_2)) = k(x_1,x_2)k(y_1,y_2) = \sum_{i=1}^\infty \lambda_i \phi_i(x_1)\phi_i(x_2)\sum_{j=1}^\infty \lambda_j \phi_j(y_1)\phi_j(y_2)$</p> <p>$=\sum_{i,j=1}^\infty \lambda_i \lambda_j \phi_i(x_1)\phi_j(y_1) \phi_i(x_2)\phi_j(y_2)$</p> <p>so that the eigenvalues of $K$ are just products of the eigenvalues of $k$,</p> <p>$\Lambda_{i,j} = \lambda_i \lambda_j$.</p> <p>And the eigenfunctions are simply products, too, i.e.</p> <p>$\Phi_{i,j}(x,y) = \phi_i(x)\phi_j(y)$.</p> <p>If you want sums over one index, you can now use your favorite enumeration of $N\times$,</p> <p>$\Lambda_1 = \lambda_1 \lambda_1$,</p> <p>$\Lambda_2 = \lambda_1 \lambda_2$, $\Lambda_3 = \lambda_2 \lambda_1$,</p> <p>$\Lambda_4 = \lambda_1 \lambda_3$, $\Lambda_5 = \lambda_2 \lambda_2$, $\Lambda_6 = \lambda_3 \lambda_1$,</p> <p>etc. (and similarly for the $\Phi$'s).</p>