Pillai Trace as Bounded Explained Variation

From Hypothesis and Error Variation to a Bounded Criterion

In a multivariate linear model, the hypothesis sum-of-products matrix $H$ records variation associated with the tested restrictions, while the error sum-of-products matrix $E$ records residual variation. Their generalized roots describe the relative size of hypothesis variation against error variation along particular response directions. Pillai’s trace converts each such root into a fraction, then adds those fractions. The conversion matters: a very large root can make a contribution close to one, but cannot make that contribution grow without bound.

Assume throughout that $H$ is a symmetric positive semidefinite $p\times p$ matrix and $E$ is a symmetric positive definite $p\times p$ matrix. Positive definiteness of $E$ ensures that the inverses and finite generalized roots below are well-defined. The hypothesis matrix may be singular, as is usual when the tested effect has rank less than the response dimension.

Definition: The Pillai trace for $H$ and $E$ is $$ V=\operatorname{tr}\!\left[H(H+E)^{-1}\right]. $$ The generalized roots $\lambda_1,\ldots,\lambda_p$ are the eigenvalues of $E^{-1/2}HE^{-1/2}$, equivalently the roots of $\det(H-\lambda E)=0$, counted with multiplicity. Under the stated assumptions, each $\lambda_i$ is nonnegative.

The order of the factors in the displayed trace is part of its definition. In general, $H$ and $E$ do not commute, so their matrix products cannot be rearranged arbitrarily. The product $H(H+E)^{-1}$ need not be symmetric; its trace is nevertheless real and has a useful spectral interpretation.

Why Generalized Roots Become Fractions

A generalized root compares hypothesis variation with error variation in a response direction. To see the associated fraction, let $v\ne 0$ satisfy $Hv=\lambda Ev$. Multiplying on the left by $v^\mathsf{T}$ gives $v^\mathsf{T}Hv=\lambda v^\mathsf{T}Ev$. Since $E$ is positive definite, $v^\mathsf{T}Ev>0$. Thus the ratio of hypothesis variation to total variation in this direction is

\frac{v^\mathsf{T}Hv}{v^\mathsf{T}(H+E)v} = \frac{\lambda v^\mathsf{T}Ev}{(1+\lambda)v^\mathsf{T}Ev} = \frac{\lambda}{1+\lambda}.

This ratio lies between zero and one. For a fuller decomposition, choose generalized eigenvectors that are orthonormal in the $E$-inner product, so $v_i^\mathsf{T}Ev_j$ is one when $i=j$ and zero otherwise. Such a choice is available because $E^{-1/2}HE^{-1/2}$ is symmetric and can be orthogonally diagonalized. In this basis, both the hypothesis and total quadratic forms are diagonal: their entries in direction $i$ are $\lambda_i$ and $1+\lambda_i$. The fraction $\lambda_i/(1+\lambda_i)$ is therefore the hypothesis share of hypothesis-plus-error variation along that generalized direction.

Theorem: Under the stated assumptions, $$ V=\operatorname{tr}\!\left[H(H+E)^{-1}\right] =\sum_{i=1}^{p}\frac{\lambda_i}{1+\lambda_i}. $$ Every summand belongs to $[0,1)$. If $r=\operatorname{rank}(H)$, then $0\le V<r$ when $r>0$, and $V=0$ when $r=0$.

Proof. Let $A=E^{-1/2}HE^{-1/2}$. This matrix is symmetric positive semidefinite, and its eigenvalues are precisely the generalized roots $\lambda_i$. The factorization

H=E^{1/2}AE^{1/2}, \qquad H+E=E^{1/2}(A+I)E^{1/2}

implies

H(H+E)^{-1} = E^{1/2}A(A+I)^{-1}E^{-1/2}.

By invariance of trace under cyclic permutation of factors,

V =\operatorname{tr}\!\left(E^{1/2}A(A+I)^{-1}E^{-1/2}\right) =\operatorname{tr}\!\left(A(A+I)^{-1}\right).

Diagonalize $A=Q\operatorname{diag}(\lambda_1,\ldots,\lambda_p)Q^\mathsf{T}$, where $Q$ is orthogonal. Then $A(A+I)^{-1}$ has eigenvalues $\lambda_i/(1+\lambda_i)$, so its trace is their sum. Since every $\lambda_i\ge 0$, each summand is nonnegative and less than one. Exactly $r$ roots are positive, while the other $p-r$ roots are zero. If $r>0$, the sum of the $r$ positive terms is strictly less than $r$; if $r=0$, all terms vanish. $\square$

Worked Calculations

Worked Example: Diagonal Hypothesis and Error Matrices

Let $H=\begin{pmatrix}4&0\\0&1\end{pmatrix}$ and $E=\begin{pmatrix}2&0\\0&3\end{pmatrix}$. Compute the generalized roots and Pillai’s trace in two ways.

Because both matrices are diagonal, the generalized roots are the ratios of corresponding diagonal entries of $H$ and $E$: $\lambda_1=4/2=2$ and $\lambda_2=1/3$. Their bounded contributions are $\lambda_1/(1+\lambda_1)=2/3$ and $\lambda_2/(1+\lambda_2)=(1/3)/(4/3)=1/4$. Thus the root sum is $2/3+1/4=8/12+3/12=11/12$.

For a direct trace calculation, $H+E=\begin{pmatrix}6&0\\0&4\end{pmatrix}$, so $H(H+E)^{-1}=\begin{pmatrix}4/6&0\\0&1/4\end{pmatrix}$. Its trace is $2/3+1/4=11/12$, as required.

Worked Example: A Nonsymmetric Matrix Product

Let $H=\begin{pmatrix}2&0\\0&0\end{pmatrix}$ and $E=\begin{pmatrix}2&1\\1&2\end{pmatrix}$. The error matrix is positive definite: its eigenvalues are $1$ and $3$. Find the roots and trace, noting that $H(H+E)^{-1}$ is not symmetric.

The generalized roots solve $\det(H-\lambda E)=0$. Directly, \[ \det\begin{pmatrix}2-2\lambda&-\lambda\\-\lambda&-2\lambda\end{pmatrix} =(2-2\lambda)(-2\lambda)-\lambda^2 =3\lambda^2-4\lambda =\lambda(3\lambda-4). \] Hence $\lambda_1=4/3$ and $\lambda_2=0$. Their contributions sum to $(4/3)/(1+4/3)+0=4/7$.

Also, $H+E=\begin{pmatrix}4&1\\1&2\end{pmatrix}$, whose determinant is $8-1=7$. Therefore $(H+E)^{-1}=\frac17\begin{pmatrix}2&-1\\-1&4\end{pmatrix}$, and \[ H(H+E)^{-1} =\frac17\begin{pmatrix}4&-2\\0&0\end{pmatrix}. \] This product is not symmetric, but its trace is $4/7$, matching the root calculation. Symmetry of the product is not required for the trace identity.

Worked Example: Two Positive Roots with Identity Error

Let $E=I$ and $H=\begin{pmatrix}3&1\\1&1\end{pmatrix}$. The eigenvalues of $H$, and thus the generalized roots, solve $\det(H-\lambda I)=\lambda^2-4\lambda+2=0$, giving $\lambda_1=2+\sqrt2$ and $\lambda_2=2-\sqrt2$. Both are positive.

The root contributions add to \[ \frac{2+\sqrt2}{3+\sqrt2}+\frac{2-\sqrt2}{3-\sqrt2} =2-\left(\frac{1}{3+\sqrt2}+\frac{1}{3-\sqrt2}\right) =2-\frac{6}{9-2} =\frac87. \] Here the middle equality uses $\lambda/(1+\lambda)=1-1/(1+\lambda)$.

For verification, $H+I=\begin{pmatrix}4&1\\1&2\end{pmatrix}$ has inverse $\frac17\begin{pmatrix}2&-1\\-1&4\end{pmatrix}$. Multiplying by $H$ gives $\frac17\begin{pmatrix}5&1\\1&3\end{pmatrix}$, whose trace is $8/7$. Each root contributes less than one, and the total is less than the rank $2$.

What the Bound Means—and What It Does Not Mean

The transformation $g(\lambda)=\lambda/(1+\lambda)$ is increasing for $\lambda\ge0$, since $g'(\lambda)=1/(1+\lambda)^2>0$. For a small root, $g(\lambda)$ is close to $\lambda$; for a large root, the contribution approaches one. Consequently, Pillai’s trace retains information from every positive root but limits the influence of any single very large root. Its possible range is governed by the number of positive roots, not by the numerical scale of the largest root alone.

The word “explained” is useful when interpreting the directional ratios, but it should not be taken to mean that $V$ is generally an ordinary scalar $R^2$ for the original responses. It is a sum of normalized hypothesis shares across generalized directions. Those directions are defined jointly by $H$ and $E$, and need not coincide with the original response coordinates.

Common pitfall: It is incorrect to call $V$ the sum of the generalized roots, or to assume $H(H+E)^{-1}$ is symmetric and read its diagonal entries as separate explained proportions. The trace is instead the sum of the transformed roots $\lambda_i/(1+\lambda_i)$. This spectral identity remains valid even when the matrix product is nonsymmetric; individual diagonal entries of that product need not be the directional fractions.

The rank bound gives a quick interpretation check. If a tested effect can contribute variation in only one independent response direction, then $r=1$ and $V<1$, regardless of how large its root becomes. With $r$ positive roots, $V<r$. In applications, this bound is also a useful arithmetic check: a computed trace outside the interval $[0,r)$ signals a problem with the matrices, the calculation, or the assumptions.

Check Your Understanding

Use the definitions and results above to answer these questions.

Why does positive definiteness of $E$ ensure that every generalized root is nonnegative?
If the generalized roots are $0$, $1$, and $4$, what is Pillai’s trace, and how does it compare with the number of positive roots?
Explain why a nonsymmetric product $H(H+E)^{-1}$ can still have a trace equal to a sum of real, nonnegative terms.
For a single generalized root that increases without bound, what value does its Pillai contribution approach?
Why should Pillai’s trace not automatically be described as an ordinary scalar $R^2$ for the original response variables?

Pillai Trace as Bounded Explained Variation

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