It appears that the matrix from rbf is a correlation matrix, with ones on the diagonal. It's also a real-square symmetric matrix, since the upper triangular (off-diagonal elements) are a mirror image of the lower triangular. The cov matrix looks like a covariance matrix, with diagonal elements that are the variance of each column, and off-diagonals are covariances. You can check if the following are true: $\sigma_{1}^2 = \frac{\sum_{i=1}^3 (x_{i1} - \bar{x}_1 )^2}{n-1} = 0.1803 \rightarrow\textrm{variance of feature 1}$ $\sigma_{12} = \frac{\sum_{i=1}^3 (x_{i1} - \bar{x}_1 )(x_{i2} - \bar{x}_2 )}{n-1} = 0.4715 \rightarrow\textrm{covariance between feature 1 and 2}$ If so, the cov is the covariance matrix. I would not commingle kernel methods with the covariance matrix, mostly because kernel tricks could throw you off. If you stick with statistical notation and calculations for obtaining the covariance matrix: \begin{equation} \boldsymbol{\sigma} = \begin{bmatrix} \sigma_1^2 & \sigma_{12} & \cdots & \sigma_{14} \\ \sigma_{21} & \sigma_2^2 & \cdots & \sigma_{24} \\ \vdots & \vdots & \ddots & \vdots \\ \sigma_{41} & \sigma_{42} & \cdots & \sigma_4^2 \\ \end{bmatrix}, \end{equation} then you can't go wrong. The correlation matrix is simply: \begin{equation} \boldsymbol{\rho} = \begin{bmatrix} 1 & \rho_{12} & \cdots & \rho_{14} \\ \rho_{21} & 1 & \cdots & \rho_{24} \\ \vdots & \vdots & \ddots & \vdots \\ \rho_{41} & \rho_{42} & \cdots & 1 \\ \end{bmatrix}, \end{equation} where all diagonal elements are ones, and elements in the off-diagonal are calculated as $\rho_{jk} = \frac{\sigma_{jk}}{\sigma_j \sigma_j}$, where $\sigma_j $ = $\sqrt{\sigma_j^2}$, which are on the diagonal of the covariance matrix.