Hotelling–Lawley Trace as Aggregate Root Strength

From Generalized Roots to an Aggregate Measure

A MANOVA hypothesis can affect several response directions at once. The generalized roots describe the strengths of those directions relative to error; the Hotelling–Lawley trace provides a single summary by adding their strengths. This makes it an aggregate criterion: a collection of nonzero roots all contributes, rather than only one direction determining the summary.

The error matrix matters. A hypothesis sum-of-products matrix \(H\) alone does not say how large an effect is relative to the variation left unexplained. The Hotelling–Lawley trace compares \(H\) with the error sum-of-products matrix \(E\), and its generalized roots are dimensionless measures on that error-scaled comparison.

Definition: Let \(H\) be a \(p\times p\) symmetric positive semidefinite hypothesis sum-of-products matrix, and let \(E\) be a \(p\times p\) symmetric positive definite error sum-of-products matrix. The generalized MANOVA roots \(\lambda_1,\ldots,\lambda_p\) are the eigenvalues of \(E^{-1}H\), equivalently the solutions of \(\det(H-\lambda E)=0\), counted with multiplicity. The Hotelling–Lawley trace is \(U=\operatorname{tr}(E^{-1}H)\).

The positive-definiteness condition on \(E\) ensures that \(E^{-1}\) exists and that the comparison is well-defined. In a standard MANOVA analysis, \(H\) and \(E\) are positive semidefinite sums-of-products matrices; \(E\) must be positive definite for the classical criterion in this form. If \(s=\operatorname{rank}(H)\), then there are \(s\) positive roots and \(p-s\) zero roots. In particular, a low-rank hypothesis matrix can have fewer than \(p\) nonzero roots.

Why the Trace Is the Sum of the Roots

The matrix \(E^{-1}H\) need not be symmetric, so it is useful to connect it to a symmetric matrix before interpreting its eigenvalues. Because \(E\) is symmetric positive definite, its symmetric square root and inverse square root exist. Define the whitened hypothesis matrix \(A=E^{-1/2}HE^{-1/2}\). This matrix is symmetric positive semidefinite.

Theorem: Under the conditions in the definition, the generalized roots \(\lambda_1,\ldots,\lambda_p\) are real and nonnegative, and \(U=\operatorname{tr}(E^{-1}H)=\sum_{i=1}^{p}\lambda_i\). Consequently, the trace is the sum of all nonzero roots, with zero roots contributing nothing.

Proof. The matrix \(A=E^{-1/2}HE^{-1/2}\) is symmetric because \(H\) and \(E^{-1/2}\) are symmetric. For every vector \(x\), \(x^{\mathsf T}Ax=(E^{-1/2}x)^{\mathsf T}H(E^{-1/2}x)\geq 0\), so \(A\) is positive semidefinite. By the spectral theorem, its eigenvalues are real and nonnegative.

Moreover, \(E^{-1}H=E^{-1/2}AE^{1/2}\). Thus \(E^{-1}H\) is similar to \(A\), so they have the same eigenvalues, including multiplicities. These are precisely the generalized roots. The spectral theorem gives \(\operatorname{tr}(A)=\sum_{i=1}^{p}\lambda_i\). By cyclic invariance of the trace, \(\operatorname{tr}(A)=\operatorname{tr}(E^{-1/2}HE^{-1/2})=\operatorname{tr}(E^{-1}H)\). Combining these equalities proves the result. Since \(H\) has rank \(s\) and multiplication by an invertible matrix preserves rank, \(A\) has rank \(s\); positive semidefiniteness then implies exactly \(s\) of its roots are positive. \(\square\)

The equality is more than a convenient computational identity. It says that the Hotelling–Lawley criterion collects the strength represented by every nonzero generalized root. It does not first select one root or convert each root to a bounded quantity; each root enters the sum in its original scale.

Worked Example: A Diagonal Hypothesis and Error Matrix

Let \(E=\begin{pmatrix}2&0\\0&1\end{pmatrix}\) and \(H=\begin{pmatrix}4&0\\0&3\end{pmatrix}\). Both matrices are positive definite, so the required conditions hold. Find the generalized roots and the Hotelling–Lawley trace.

First, \(E^{-1}=\begin{pmatrix}1/2&0\\0&1\end{pmatrix}\), and hence \(E^{-1}H=\begin{pmatrix}2&0\\0&3\end{pmatrix}\). Its eigenvalues are \(\lambda_1=2\) and \(\lambda_2=3\). Therefore the sum of the roots is \(2+3=5\).

Directly from the trace definition, \(\operatorname{tr}(E^{-1}H)=2+3=5\). The calculation confirms that both response directions contribute, with their strengths measured relative to the corresponding error entries.

Calculating Roots When the Matrices Do Not Share Axes

When \(E\) and \(H\) are not both diagonal in the same coordinates, one can find the roots from the generalized characteristic equation \(\det(H-\lambda E)=0\). The trace identity remains valid, but the eigenvalues of \(H\) alone are not the relevant quantities: the error matrix changes the comparison.

Worked Example: A Non-Diagonal Hypothesis Matrix

Take \(E=\begin{pmatrix}2&0\\0&1\end{pmatrix}\) and \(H=\begin{pmatrix}2&1\\1&2\end{pmatrix}\). The matrix \(E\) is positive definite. The eigenvalues of \(H\) are \(1\) and \(3\), so \(H\) is positive definite as well. Calculate the roots using the generalized characteristic equation, then verify their sum using the trace.

The determinant equation is \(\det(H-\lambda E)=\det\begin{pmatrix}2-2\lambda&1\\1&2-\lambda\end{pmatrix}=0\). Expanding gives \((2-2\lambda)(2-\lambda)-1=2\lambda^2-6\lambda+3=0\).

The quadratic formula yields \(\lambda_1=(3+\sqrt{3})/2\) and \(\lambda_2=(3-\sqrt{3})/2\). Both are positive because \(3>\sqrt{3}\). Their sum is \(\big((3+\sqrt{3})+(3-\sqrt{3})\big)/2=3\).

For a direct trace check, \(E^{-1}H=\begin{pmatrix}1&1/2\\1&2\end{pmatrix}\), so \(\operatorname{tr}(E^{-1}H)=1+2=3\), agreeing with the sum of the generalized roots. Although this product is not symmetric, the proof above shows why its roots are nevertheless real and nonnegative in this setting.

How Contributions Accumulate

Linearity of the trace gives a useful consequence when a hypothesis matrix is expressed as a sum of positive semidefinite components. The next result states this precisely, and also shows that adding a positive semidefinite component cannot decrease the aggregate trace.

Theorem: Fix a symmetric positive definite \(E\). If \(H_1\) and \(H_2\) are symmetric positive semidefinite matrices, then \(U(E,H_1+H_2)=U(E,H_1)+U(E,H_2)\). In particular, \(U(E,H_1+H_2)\geq U(E,H_1)\).

Proof. By the trace definition and linearity, \(\operatorname{tr}(E^{-1}(H_1+H_2))=\operatorname{tr}(E^{-1}H_1)+\operatorname{tr}(E^{-1}H_2)\), which proves the additivity equality. To prove the inequality, set \(D=H_2\). The matrix \(E^{-1/2}DE^{-1/2}\) is positive semidefinite, since for every \(x\), \(x^{\mathsf T}E^{-1/2}DE^{-1/2}x=(E^{-1/2}x)^{\mathsf T}D(E^{-1/2}x)\geq 0\). Its eigenvalues are nonnegative, so its trace is nonnegative. Cyclic invariance gives \(\operatorname{tr}(E^{-1}D)=\operatorname{tr}(E^{-1/2}DE^{-1/2})\geq0\). Therefore \(U(E,H_1+H_2)-U(E,H_1)=\operatorname{tr}(E^{-1}H_2)\geq0\), as required. \(\square\)

Worked Example: Two Components Contribute to the Total

Let \(E=\operatorname{diag}(2,1,2)\), \(H_1=\operatorname{diag}(2,0,0)\), and \(H_2=\operatorname{diag}(0,1,2)\). Each hypothesis component is positive semidefinite. Calculate their separate trace contributions and the trace for their sum.

Since \(E^{-1}=\operatorname{diag}(1/2,1,1/2)\), \(E^{-1}H_1=\operatorname{diag}(1,0,0)\), giving \(U(E,H_1)=1\). Likewise, \(E^{-1}H_2=\operatorname{diag}(0,1,1)\), giving \(U(E,H_2)=2\).

Their sum is \(H_1+H_2=\operatorname{diag}(2,1,2)\), so \(E^{-1}(H_1+H_2)=\operatorname{diag}(1,1,1)\), and \(U(E,H_1+H_2)=3\). The combined value equals \(1+2=3\): the roots from each contribution accumulate in the total.

Interpretation and a Common Pitfall

The Hotelling–Lawley trace is an unbounded aggregate of error-scaled root strengths. A zero root contributes nothing; every positive root contributes its full value. Thus, several moderate roots can produce a substantial total even when no single root accounts for all of it. This is the sense in which the criterion summarizes evidence across all nonzero MANOVA roots.

The criterion is a matrix summary, not by itself a probability, a percentage of variation explained, or a decision threshold. Whether a particular observed value constitutes statistically significant evidence requires an appropriate null distribution or approximation under the model; the trace identity alone does not provide that calibration.

Wrong reasoning: “The trace is the sum of the diagonal entries of \(H\), so the error matrix can be ignored.”
Correction: The definition is \(\operatorname{tr}(E^{-1}H)\). The inverse error matrix is essential to the scaling. For example, in the first worked example, \(\operatorname{tr}(H)=7\), whereas the Hotelling–Lawley trace is \(5\).
Wrong reasoning: “Because \(E^{-1}H\) is not necessarily symmetric, its roots might be complex.”
Correction: With \(E\) positive definite and \(H\) positive semidefinite, \(E^{-1}H\) is similar to the symmetric positive semidefinite matrix \(E^{-1/2}HE^{-1/2}\); its roots are therefore real and nonnegative.

Key takeaway: When \(E\) is positive definite and \(H\) is positive semidefinite, the Hotelling–Lawley trace is \(\operatorname{tr}(E^{-1}H)=\sum_{i=1}^{p}\lambda_i\). It measures aggregate root strength by adding all nonzero generalized MANOVA roots.

Check Your Understanding

Use the definitions and results above to answer the following questions.

Why does the condition that \(E\) is positive definite matter for defining the classical Hotelling–Lawley trace?
If \(H\) has rank \(s<p\), how many generalized roots are zero under the stated assumptions, and how do those roots affect the trace?
For \(E=\operatorname{diag}(1,2)\) and \(H=\operatorname{diag}(3,4)\), calculate the two roots and the Hotelling–Lawley trace.
Explain why the generalized roots remain real even if \(E^{-1}H\) is not symmetric.
If a positive semidefinite matrix \(H_2\) is added to \(H_1\), what can be said about the change in the Hotelling–Lawley trace, and why?