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Orthogonal rotation preserves the covariance factor model

Statement

In the covariance factor model, write $X = A S + ε$ with $A \in R^{p \times q}$ , $q < p$ , $E [S] = E [ε] = 0$ , $Cov (S) = I_{q}$ , $Cov (ε) = D$ diagonal positive, and $Cov (S, ε) = 0$ . Then

$Cov (X) = A A^{T} + D .$

For every orthogonal matrix $R \in R^{q \times q}$ , the transformed loadings and factors $A_{R} = A R^{T}$ and $S_{R} = R S$ give another representation with the same covariance matrix. Thus covariance alone does not identify a unique loading matrix or latent orientation. This is a covariance-model claim; if the factors are additionally required to be mutually independent, the proposition does not claim that arbitrary rotations preserve that extra requirement.

Statement references

Hastie, Tibshirani, and Friedman, §14.7.1

Facts & Assumptions

Given: A covariance factor model and an orthogonal matrix $R$ , so $R^{T} R = I_{q}$ .

[F1]

Covariance factor model with diagonal residual covariance: $X = A S + ε$ , with zero means, $Cov (S) = I_{q}$ , $Cov (ε) = D$ , and $Cov (S, ε) = 0$ .

Proof

1.1F1givenalgebra

By [F1], bilinearity of covariance gives $Cov (X) = A Cov (S) A^{T} + A Cov (S, ε) + Cov (ε, S) A^{T} + Cov (ε) = A A^{T} + D$ .

2.1F1step 1.1algebra∎

The transformed factors have mean zero, covariance $R I_{q} R^{T} = I_{q}$ , and cross-covariance $Cov (R S, ε) = R Cov (S, ε) = 0$ . Also $A_{R} S_{R} = A R^{T} R S = A S$ , and $A_{R} A_{R}^{T} + D = A R^{T} R A^{T} + D = A A^{T} + D$ . Hence the transformed pair is another covariance factor model with the same covariance.

Depends on

Covariance factor model with diagonal residual covariance

Used by

Two Gaussian latent factors admit a nontrivial orthogonal rotation Example

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Sources

Trevor Hastie, Robert Tibshirani, and Jerome Friedman, The Elements of Statistical Learning, second edition, corrected 12th printing (standard reference, not scraped)