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As explained in intuition of determinant : we know that determinant is the how the volume is scaled when the matrix is regarded as a projection.

Now, if we narrow down the definition of the matrix here: assuming that it is a covariance matrix, i.e., positive semi definitive and symmetric, will there be any new characteristics/intuition of determinant.

As per Ben's comments:

In a 2D space, covariance matrix represents an ellipse, which captures the distribution of a vector of 2 random variables. In such a context, what does determinant say about these 2 variables?

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  • $\begingroup$ Is there a reason that you would expect that there is a useful alternative interpretation for this narrower context? What is it that you want to do with the determinant of a covariance matrix that you couldn't do in general? $\endgroup$
    – Ben Grossmann
    Commented Dec 31, 2020 at 14:22
  • $\begingroup$ @BenGrossmann, the covariance matrix has a geometric explanation: it captures the span of an ellipstical distribution. I am wondering if we can interpret determinant in such a context $\endgroup$
    – user152503
    Commented Dec 31, 2020 at 14:29
  • $\begingroup$ You could say that the determinant is simply the volume of this ellipse (divided by the volume of the unit circle) $\endgroup$
    – Ben Grossmann
    Commented Dec 31, 2020 at 15:26
  • $\begingroup$ perhaps not what you want, but comparing the determinant of your matrix with the product of the diagonals (Hadamard Inequality) does have some nice geometric insights $\endgroup$
    – user8675309
    Commented Dec 31, 2020 at 19:51

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If $\rho_{ij}=\operatorname{Cov}(X_i,\,X_j)$ and $Y_i:=R_{ij}X_j$ for a rotation matrix $R$ then $\operatorname{Cov}(Y_i,\,Y_j)=R_{ik}\rho_{kl}R^T_{lj}$, which for a suitable choice of $R$ is diagonal. So $\det\rho$ is the determinant of this diagonal matrix, i.e. the product of variances of the $Y_i$. This rotation places Cartesian axes along the axes of the ellipse associated with $\rho$, so the rotation-invariant determinant, being a product of eigenvalues, is proportional to the ellipse's area. (Needless to say, in higher dimensions I'd be talking about a hyperellipsoid's measure, hypervolume, volume, or whatever term you prefer.)

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