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Factor Rotation and Assessment · Tutorial 555 of 975

Factor Rotation as a Likelihood-Preserving Reparameterization

Learn how to transform factor loadings and factor covariance together, and distinguish covariance and likelihood invariants from quantities that depend on the chosen factor coordinates.

Advanced 10 min read

What You'll Learn

  • Derive the loading and factor-covariance transformation for an invertible change of factor coordinates
  • Verify that the common covariance and full model-implied covariance are preserved
  • Explain why identical covariance matrices give identical Gaussian observed-data likelihoods
  • Identify the quantities preserved by orthogonal rotations, including communalities and total common variance
  • Distinguish orthogonal rotations from general invertible transformations when factor covariance is constrained
  • Recognize how loading constraints can prevent a rotated representation from belonging to the same restricted model

Why Factor Coordinates Can Change Without Changing the Model

A factor model describes the covariance among observed variables through a smaller collection of latent factors. But the coordinates used to describe those factors are not always uniquely determined by the covariance model. If we change the factor coordinates and adjust the loadings and factor covariance consistently, the same observed covariance can have a different-looking parameter representation. This matters because a change in the displayed loadings does not necessarily mean the fitted model has changed.

Let \(X\) be a \(p\)-dimensional observed random vector and \(F\) a \(q\)-dimensional factor vector, with \(q\leq p\). The common-factor model is

$$ X=\mu+\Lambda F+\varepsilon, \qquad E[F]=0,\quad E[\varepsilon]=0,\quad \operatorname{Cov}(F)=\Phi,\quad \operatorname{Cov}(\varepsilon)=\Psi,\quad \operatorname{Cov}(F,\varepsilon)=0. $$

Here \(\Lambda\) is a \(p\times q\) loading matrix, \(\Phi\) is a positive-definite \(q\times q\) factor-covariance matrix, and \(\Psi\) is typically a positive diagonal \(p\times p\) uniqueness matrix. The zero cross-covariance assumption is what removes cross terms when calculating the covariance of \(X\).

Definition: The common covariance is \(\Lambda\Phi\Lambda'\), and the model-implied covariance is \(\Sigma=\Lambda\Phi\Lambda'+\Psi\). The diagonal entries of the common covariance are the communalities: \(h_i^2=(\Lambda\Phi\Lambda')_{ii}\). The diagonal entries of \(\Psi\) are the uniquenesses.

The covariance formula follows by expanding \(\operatorname{Cov}(\Lambda F+\varepsilon)\); the cross-covariance terms vanish because \(\operatorname{Cov}(F,\varepsilon)=0\). The central question is how this representation changes when the latent coordinates are replaced by another invertible system.

Invertible Changes of Factor Coordinates

Let \(A\) be any invertible \(q\times q\) matrix. Define new factors by \(F^*=A^{-1}F\). Then \(F=AF^*\), so the same observed vector can be written using new parameters \(\Lambda^*=\Lambda A\) and \(\Phi^*=A^{-1}\Phi A^{-T}\). The uniqueness matrix and mean are left unchanged.

Theorem: If \(A\) is invertible and the model parameters are transformed as \(\Lambda^*=\Lambda A\), \(\Phi^*=A^{-1}\Phi A^{-T}\), \(\Psi^*=\Psi\), and \(\mu^*=\mu\), then \(\Lambda^*\Phi^*\Lambda^{*'}+\Psi^*=\Lambda\Phi\Lambda'+\Psi\). Thus the transformed representation has the same model-implied covariance.

Proof. Since \(A^{-T}=(A^{-1})'\), substitution gives

$$ \begin{aligned} \Lambda^*\Phi^*\Lambda^{*'}+\Psi^* &=(\Lambda A)(A^{-1}\Phi A^{-T})(\Lambda A)'+\Psi\\ &=\Lambda A A^{-1}\Phi A^{-T}A'\Lambda'+\Psi\\ &=\Lambda\Phi\Lambda'+\Psi, \end{aligned} $$

because \(AA^{-1}=I_q\) and \(A^{-T}A'=I_q\). Also, \(F^*\) has mean zero and covariance \(A^{-1}\Phi A^{-T}\); it remains uncorrelated with \(\varepsilon\). Since \(\Phi\) is positive definite and \(A\) is invertible, \(\Phi^*\) is positive definite as well: for nonzero \(v\), \(v'\Phi^*v=(A^{-T}v)'\Phi(A^{-T}v)>0\). The transformed parameters therefore define a valid covariance representation. \(\square\)

If the factor model is Gaussian, the observed distribution is determined by \(\mu\) and \(\Sigma\). More specifically, for \(n\) independent observations \(x_1,\ldots,x_n\), its observed-data log-likelihood is

$$ \ell(\mu,\Sigma) =-\frac{np}{2}\log(2\pi)-\frac{n}{2}\log|\Sigma| -\frac{1}{2}\sum_{r=1}^n(x_r-\mu)'\Sigma^{-1}(x_r-\mu). $$

When the transformation leaves \(\mu\) and \(\Sigma\) unchanged, it leaves every term in this expression unchanged. Thus the observed-data likelihood is identical at the two representations. This conclusion concerns the likelihood of the observations, not a density for factor coordinates treated as if they were observed.

Worked Example: An Invertible Transformation That Preserves Covariance

Take \(q=2\), \(\Phi=I_2\), \(\Psi=\operatorname{diag}(0.3,0.4,0.2)\), and

$$ \Lambda= \begin{pmatrix} 1&0.5\\ 0.2&0.8\\ 0.7&-0.1 \end{pmatrix}. $$

The original common covariance is obtained by taking row-wise inner products:

$$ \Lambda\Lambda'= \begin{pmatrix} 1.25&0.60&0.65\\ 0.60&0.68&0.06\\ 0.65&0.06&0.50 \end{pmatrix}. $$

Choose \(A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\), whose inverse is \(A^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}\). The transformed parameters are

$$ \Lambda^*=\Lambda A= \begin{pmatrix} 1&1.5\\ 0.2&1.0\\ 0.7&0.6 \end{pmatrix}, \qquad \Phi^*=A^{-1}A^{-T}= \begin{pmatrix} 2&-1\\ -1&1 \end{pmatrix}. $$

For example, the first transformed communality is \((1,1.5)\Phi^*(1,1.5)'=2-3+2.25=1.25\). The covariance between the first and second variables is \((1,1.5)\Phi^*(0.2,1)'=0.4-1-0.3+1.5=0.60\). The remaining entries likewise give the same common covariance matrix above, as the theorem guarantees. Adding \(\Psi\) gives the model-implied covariance

$$ \Sigma= \begin{pmatrix} 1.55&0.60&0.65\\ 0.60&1.08&0.06\\ 0.65&0.06&0.70 \end{pmatrix}. $$

Thus the loadings have changed, and the transformed factor covariance is no longer the identity, but the observed covariance is exactly the same. In the Gaussian model, the observed-data likelihood is consequently unchanged.

What Orthogonal Rotations Preserve

An orthogonal matrix \(Q\) satisfies \(Q'Q=QQ'=I_q\). The corresponding rotation has \(\Lambda^*=\Lambda Q\) and \(\Phi^*=Q'\Phi Q\), which is the general transformation above with \(A=Q\). In particular, if factors are standardized and uncorrelated so that \(\Phi=I_q\), then \(\Phi^*=Q'I_qQ=I_q\). Orthogonal rotations therefore change the loading coordinates while keeping the factor covariance equal to the identity.

Theorem: Under an orthogonal rotation with the factor covariance transformed as \(\Phi^*=Q'\Phi Q\), the model-implied covariance, common covariance, communalities, uniquenesses, and total common variance are unchanged. When \(\Phi=I_q\), the communality of variable \(i\) is the sum of its squared loadings, \(\sum_{j=1}^q\lambda_{ij}^2\), and this sum is unchanged by the rotation.

Proof. The common covariance after rotation is

$$ \Lambda^*\Phi^*\Lambda^{*'} =(\Lambda Q)(Q'\Phi Q)(\Lambda Q)' =\Lambda QQ'\Phi QQ'\Lambda' =\Lambda\Phi\Lambda', $$

using \(QQ'=I_q\) twice. The uniqueness matrix is not altered, so the full covariance and each diagonal common-covariance entry are unchanged. If \(\Phi=I_q\), the communality is \((\Lambda\Lambda')_{ii}\), which equals the squared Euclidean length of row \(i\) of \(\Lambda\). Finally, the total common variance is the trace of the common covariance, \(\operatorname{tr}(\Lambda\Phi\Lambda')\), so it too is unchanged. \(\square\)

These invariants describe the fitted covariance structure, not every feature of the loading table. In particular, the loading of a given variable on a particular rotated factor can change, as can the squared norm of an individual loading column. The total common variance is preserved, but its allocation across columns generally is not.

Worked Example: Communalities Stay Fixed While Loadings Change

Consider standardized factors and

$$ \Lambda= \begin{pmatrix} 0.8&0.1\\ 0.3&0.7\\ 0.5&0.4 \end{pmatrix}, \qquad Q=\frac{1}{\sqrt{2}} \begin{pmatrix} 1&-1\\ 1&1 \end{pmatrix}. $$

The columns of \(Q\) are orthonormal, so \(Q'Q=I_2\). The original communalities are \(0.8^2+0.1^2=0.65\), \(0.3^2+0.7^2=0.58\), and \(0.5^2+0.4^2=0.41\). Rotating gives

$$ \Lambda Q= \begin{pmatrix} 0.9/\sqrt{2}&-0.7/\sqrt{2}\\ 1.0/\sqrt{2}&0.4/\sqrt{2}\\ 0.9/\sqrt{2}&-0.1/\sqrt{2} \end{pmatrix}. $$

The rotated row-wise squared lengths are \((0.81+0.49)/2=0.65\), \((1+0.16)/2=0.58\), and \((0.81+0.01)/2=0.41\), respectively. Yet the first column’s squared length changes from \(0.8^2+0.3^2+0.5^2=0.98\) to \((0.81+1+0.81)/2=1.31\). The second column’s squared length changes from \(0.01+0.49+0.16=0.66\) to \((0.49+0.16+0.01)/2=0.33\). Their total remains \(1.31+0.33=1.64=0.98+0.66\), consistent with preservation of total common variance.

When the Reparameterization Is—and Is Not—Allowed

The scope of the invariance depends on the model’s parameter constraints. If \(\Phi\) is allowed to be any positive-definite matrix, the general invertible transformation preserves the covariance representation. If the model instead fixes \(\Phi=I_q\), a general invertible transformation will usually produce \(\Phi^*\ne I_q\). To retain the identity constraint, it must satisfy \(A^{-1}A^{-T}=I_q\), equivalently \(AA'=I_q\); for a square matrix this means \(A\) is orthogonal.

Other restrictions can also matter. For example, a loading pattern that requires specified entries of \(\Lambda\) to be zero may not retain those zeros after an arbitrary rotation. The unrestricted covariance representations can be equivalent even when one of them fails to satisfy a particular restricted loading pattern. Equivalence of covariances does not automatically establish equivalence within every constrained parameter space.

Worked Example: A Scaling Transformation and an Identity Constraint

Let \(q=1\), \(\Lambda=(1,2)'\), \(\Phi=1\), and \(\Psi=I_2\). The model-implied covariance is

$$ \Sigma= \begin{pmatrix}1\\2\end{pmatrix} \begin{pmatrix}1&2\end{pmatrix}+I_2 = \begin{pmatrix}2&2\\2&5\end{pmatrix}. $$

Take the invertible one-dimensional transformation \(A=2\). Then \(\Lambda^*=(2,4)'\) and \(\Phi^*=1/4\), so

$$ \Lambda^*\Phi^*\Lambda^{*'}+\Psi =\frac14 \begin{pmatrix}2\\4\end{pmatrix} \begin{pmatrix}2&4\end{pmatrix}+I_2 = \begin{pmatrix}2&2\\2&5\end{pmatrix}. $$

The covariance is preserved, but \(\Phi^*=1/4\), not \(1\). If the model requires unit factor variance, this scaled representation is outside that constrained model; in one dimension the allowed orthogonal transformations are only \(A=1\) and \(A=-1\). Under the unrestricted factor-covariance representation, however, the two parameterizations give the same covariance and Gaussian observed-data likelihood.

Common Mistake and Practical Significance

Common mistake: “The loading matrix changed, so the fitted model and likelihood must have changed.” This reasoning confuses parameter coordinates with the observed distribution. The correction is to calculate the transformed factor covariance as well: compare \(\Lambda\Phi\Lambda'\), not \(\Lambda\) alone. If the mean and the full covariance remain the same, the Gaussian observed-data likelihood remains the same too.

This distinction is useful when interpreting fitted factors. Different loading matrices can express the same covariance structure, so a particular loading pattern is not automatically a unique discovery about the data. At the same time, not every numerical transformation is admissible under every identification or loading constraint. Always check both the covariance identity and the parameter-space restrictions before calling two representations equivalent.

Key takeaway: An invertible change of factor coordinates preserves the common covariance when loadings and factor covariance are transformed together. Orthogonal rotations are the special case that also preserves an identity factor covariance; covariance features such as communalities remain fixed even though individual loadings can change.

Check Your Understanding

Use the transformation formulas and invariance results to answer the following.

  1. For an invertible \(A\), write the transformations of \(\Lambda\) and \(\Phi\) that preserve \(\Lambda\Phi\Lambda'\).
  2. Why does preserving \(\Sigma\) preserve the observed-data likelihood in the Gaussian factor model when the mean is also unchanged?
  3. When \(\Phi=I_q\), what condition on \(A\) ensures that the transformed factor covariance remains \(I_q\)?
  4. Which quantity is preserved by an orthogonal rotation: each loading-column squared norm, or the total common variance? Explain.
  5. Can a covariance-preserving invertible transformation necessarily be used if the model requires certain loading entries to remain zero? Why or why not?
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