## introduction

Recall Harman’s or Bartholomew’s model

Z = A F

with $Z = X^T\$, X standardized, and A a $\sqrt{\text{eigenvalue}}\$-weighted eigenvector matrix, with eigenvalues from the correlation matrix.

We saw how to compute the scores $F^T$ in the case that A was invertible (here). If, however, any eigenvalues are zero then A will have that many columns of zeroes and will not be invertible.

What to do?

One possibility – shown in at least one of the references, and, quite honestly, one of the first things I considered – is to use a particular example of a pseudo-inverse. I must tell you up front that this is not what I would recommend, but since you will see it out there, you should see why I don’t recommend it.

(Answer: it works, it gets the same answer, but computing the pseudo-inverse explicitly is unnecessary. In fact, it’s unnecessary even if we don’t have the Singular Value Decomposition (SVD) available to us.)
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