I'm reading some notes in multivariate data analysis, in particular factor analysis.

  • A data vector $X_{p\times 1}$, with $E(X) = \mu$

  • A vector $F_{m \times 1} $ of factors,

  • A matrix $L_{p\times m}$ of loadings, and

  • A vector $\varepsilon_{p \times 1}$ of $p$ errors with a diagonal covariance matrix $Var(\varepsilon) = \Psi$

which gives the model

$$ X-\mu = LF + \varepsilon $$

The model has $mp + p$ parameters after considering $L$ and $\Psi$.

The solutions are not unique, and are only determined up to orthogonal rotations $L^* = LT$ where $T$ is an orthogonal matrix. This is used profitably to rotate the factors in a way that provides better interpretation.


Then, the notes say:

after rotating with $T$, there are $\frac{m(m-1)}{2}$ fewer parameters.

I just can't figure out where that comes from. Can someone please explain?

  • $\begingroup$ @amoeba No, you're right, I probably don't want it there. However, I would appreciate either a hint or a full solution. Whichever gets me closer to understanding. $\endgroup$
    – RMurphy
    Oct 17, 2016 at 23:58
  • $\begingroup$ I don't have time for a full answer right now, but here is a hint: how many parameters does $T$ have? $\endgroup$
    – amoeba
    Oct 18, 2016 at 0:06
  • $\begingroup$ @amoeba Hmm... I thought it was fixed and known, based on what I've read so far. I'll continue reading, and keep that hint in mind. $\endgroup$
    – RMurphy
    Oct 18, 2016 at 1:42
  • 1
    $\begingroup$ Hi, I think I figured one thing out. Observe that, in order for a matrix $T$ to be orthogonal, its columns need to sum to one be mutually orthogonal. Hence, that gives a system of equations that the elements of the matrix need to satisfy. I found that $\frac{m(m-1)}{2}$ parameters are freely varying, in the sense that if you wanted to make an orthogonal matrix, you could start by picking $\frac{m(m-1)}{2}$ elements but then the rest would be determined. So I kind of get that. Is that the right direction? $\endgroup$
    – RMurphy
    Oct 18, 2016 at 18:10
  • 1
    $\begingroup$ @amoeba I understand that $T$ has $m(m-1)/2$ free parameters and that L is only unique to multiplication by an orthogonal matrix. However, I am unable to understand how these two facts put together reduces the number of free parameters in $L$ by $m(m-1)/2$? $\endgroup$ Jul 31, 2019 at 17:32

1 Answer 1


I found the following references - Rotation in Factor analysis by Darton and this presentation wherein it is stated that because of the rotational indeterminacy, an extra constraint is imposed - primarily for computational convenience.

The extra constraint is usually of the form: $L^{T} \Psi^{-1} L$ is a diagonal matrix. By construction $L^{T} \Psi^{-1} L$ is a symmetric matrix, and setting the non-diagonal elements to zero is equivalent to a reduction in $m(m-1)/2$ degrees of freedom.


This answer by Amoeba explains it in more detail.


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