When the STA is 0, iSTAC recovers the same subspace as the significant eigenvectors of the STC. To prove this, let
A =
B T Λ B, and the KL divergence reduces to Tr[
A] − log|
A| plus a constant. Note that the first term of this expression is the sum of the eigenvalues of
A and that the second is the negative sum of the log eigenvalues of
A. These eigenvalues, in turn, represent the variance of the STE along each major axis preserved by the basis
B. If we diagonalize
Λ using its eigenvectors, it is easy to show that the function is maximized by setting
B to contain the eigenvectors of
Λ for which the corresponding eigenvalues
σ i are greater than or less than 1. The information contributed by each eigenvector is equal to
σ i − log(
σ i) − 1, which is the function plotted in
Figure 2B, and monotonically increases as
σ i moves away from a value of 1. This means that extrema (high or low eigenvalues) of the STC will also be maxima in the iSTAC analysis, and thus, the iSTAC basis will be the same as the STC basis. Moreover, if we wish to preserve only
j axes of the stimulus space, the most informative
j-dimensional subspace is generated by the eigenvectors whose corresponding eigenvalues
σ i give the
j largest values of
σ i − log(
σ i).