# Covariance matrix using squared exponential function

I'm writing down the covariance matrix $K$ of a vector X using squared exponential covariance function, and then evaluating the determinant of the matrix $K$. Let's say i add a new point to $X$ , and re-evaluate $det(K)$. Does the value of the determinant of this new matrix has to be larger/smaller than the one computed with the smaller vector? In other words, is the determinant of the covariance matrix somehow affected by the dimension of $X$?

N.B: Squared Exponential covariance function: $$k(x_{i},x_{j}) = \sigma * exp(-(|x_{i}-x_{j}|)^2)/2 l)$$ where $\sigma$ and $l$ are the parameters of the covariance function.

• Have you tried to perform some numerical experiments with some random dataset ? If so, please include that in the question. From what is discussed in earler questions here, I can say that the determinant represents the volume of spread of the data points. So, if we have an additional coordinate, it adds another dimension which I feel will increase the value of determinant.math.stackexchange.com/questions/889425/… – Srinivas K Mar 12 '15 at 21:55
• @SrinivasK Now that i've read the other post. I think i kinda get it. determinant of the covariance matrix is called "generalized variance" because i think it does the same role of variance in one dimensional data. the more the data is spread, the more this value increases. so, it's according to the location of the added point with respect to the already existing coordinates that det(K) will increase or decrease. do you agree with that logic? – rodrigo Mar 13 '15 at 17:08