by AlphaOmega
Last Updated August 29, 2017 11:19 AM

Suppose I have n data points lying in $\mathbf{R}^3$. Then, after defining my Gaussian diffusion kernel $k(x,y)$ and computing matrix $K$, I obtain $P$, whose entries $P_{ij}=p(x_i,y_i)$ is the transition probability of jumping from point i to point j in a random walk process. Now, $P\in \mathbf{R}^{n\times n}$, thus its ith eigenvector $\Phi_i$ with eigenvalue $\lambda_i$ lives in $\mathbf{R}n^{\times 1}$. The papers treating diffusion maps now say that the t-th diffusion map of data point $x$ is defined to be

$\Psi_t(x)=(\lambda_1^m\Phi_1(x),\lambda_2^m\Phi_2(x),...,\lambda_d^m\Phi_d(x))^T$, where $d$ is no. of dimensions I chose to retain.

Now I am at a loss here: If $\Phi_1$ in the discrete case is a vector, what is $\Phi_1(x)$? Normally, I would do $\Phi_1^T\cdot x$, however, this is clearly not possible here, as the dimensions don't check out. So, what can possibly be meant by $\Phi_1(x)$?

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