Algebraic Relations Among the Four MANOVA Criteria

One Root Spectrum, Four Summaries

The four classical MANOVA criteria are different summaries of the same generalized-root spectrum. This common basis makes their algebraic relationships visible: each criterion transforms or aggregates the roots, but each does so in a different way. That distinction matters most when several roots are nonzero. In that setting, the criteria can rank the same hypothesis effects differently.

This tutorial concerns the criteria as functions of the roots. It does not derive their null distributions or calibration approximations. Those are separate questions: algebraic comparisons alone do not determine a test’s p-value or power.

Definitions in Terms of Generalized Roots

Let $H$ and $E$ be the hypothesis and error sum-of-squares-and-cross-products matrices for a MANOVA problem. Assume they are symmetric $p\times p$ matrices, $H$ is positive semidefinite, and $E$ is positive definite. Write $\lambda_1,\ldots,\lambda_s$ for the generalized roots of the pair $(H,E)$, including zeros as needed; the number $s$ is at most $p$, and any further roots up to $p$ are zero. Equivalently, these are the eigenvalues of $E^{-1/2}HE^{-1/2}$. Thus every $\lambda_i\geq 0$.

Definition: The four MANOVA criteria, expressed in terms of the generalized roots, are $$ \Lambda=\prod_{i=1}^{s}\frac{1}{1+\lambda_i},\qquad V=\sum_{i=1}^{s}\frac{\lambda_i}{1+\lambda_i},\qquad U=\sum_{i=1}^{s}\lambda_i,\qquad \Theta=\max_{1\leq i\leq s}\lambda_i. $$ Here $\Lambda$ is Wilks’ lambda, $V$ is Pillai’s trace, $U$ is the Hotelling–Lawley trace, and $\Theta$ is Roy’s largest root.

The definitions highlight the different operations. Wilks’ lambda multiplies one contribution from each root. Pillai’s trace sums roots after compressing each to a value below one. Hotelling–Lawley’s trace adds the uncompressed roots. Roy’s criterion retains only the largest root.

Worked Example: Computing All Four Criteria

Suppose the nonzero generalized roots are $3$ and $1$. Compute each criterion directly.

For Wilks’ lambda, the two factors are $1/(1+3)=1/4$ and $1/(1+1)=1/2$, so $$ \Lambda=\frac14\cdot\frac12=\frac18. $$

For Pillai’s trace, $$ V=\frac{3}{1+3}+\frac{1}{1+1}=\frac34+\frac12=\frac54. $$ For Hotelling–Lawley’s trace, $U=3+1=4$. For Roy’s criterion, $\Theta=\max(3,1)=3$. Thus the four summaries are $\Lambda=1/8$, $V=5/4$, $U=4$, and $\Theta=3$.

Why the Root Formulas Hold

Theorem: Under the stated assumptions, the determinant ratio for Wilks’ lambda is $\lvert E\rvert/\lvert E+H\rvert$, Pillai’s trace is $\operatorname{tr}\{H(H+E)^{-1}\}$, and Hotelling–Lawley’s trace is $\operatorname{tr}(E^{-1}H)$. These expressions equal the root formulas in the definition above; Roy’s criterion is the largest generalized root.

Proof. Put $A=E^{-1/2}HE^{-1/2}$. This matrix is symmetric and positive semidefinite, so it has eigenvalues $\lambda_1,\ldots,\lambda_p$, all nonnegative. The determinant factorization $E+H=E^{1/2}(I+A)E^{1/2}$ gives $$ \frac{|E|}{|E+H|} =\frac{|E|}{|E|\,|I+A|} =\frac{1}{\prod_{i=1}^{p}(1+\lambda_i)}. $$ Any roots beyond the $s$ listed roots are zero, so their factors equal one. This proves the Wilks formula.

The matrix $H(H+E)^{-1}$ is similar to $A(I+A)^{-1}$: conjugating the former by $E^{-1/2}$ on the left and $E^{1/2}$ on the right gives the latter. Therefore its eigenvalues are $\lambda_i/(1+\lambda_i)$, and its trace is their sum. Likewise, $E^{-1}H$ is similar to $A$, so its trace is $\sum_i\lambda_i$. By definition, Roy’s criterion is the largest eigenvalue of the generalized-root problem. The zero roots again do not affect these sums. $\square$

Worked Example: The Single-Root Case

Suppose there is exactly one nonzero root, $\lambda=2$. Then $$ \Lambda=\frac13,\qquad V=\frac23,\qquad U=2,\qquad \Theta=2. $$

Each criterion is a one-to-one function of this root on its allowed range: Wilks’ lambda decreases as the root increases, while the other three increase. For instance, $V=2/3$ corresponds to $\lambda=V/(1-V)=2$, and $\Lambda=1/3$ corresponds to $\lambda=1/\Lambda-1=2$. Hence any threshold on one criterion can be converted exactly into a threshold on another, with the direction reversed for Wilks’ lambda. This is an algebraic equivalence of rankings in the single-root case, not a claim that their usual calibrations are identical in every setting.

Basic Bounds and Their Proofs

Several useful inequalities follow from nonnegativity of the roots. They describe the range of the criteria and show how much information the largest root alone can provide. Let $s$ be the number of roots in the displayed list, including zeros, and let $\Theta=\max_i\lambda_i$.

Theorem: For nonnegative roots, $$ \frac{\Theta}{1+\Theta}\leq V\leq \frac{s\Theta}{1+\Theta}, \qquad \Theta\leq U\leq s\Theta, \qquad \frac{U}{1+U}\leq V\leq U. $$ In addition, $$ 1-V\leq\Lambda\leq\left(1-\frac{V}{s}\right)^s, \qquad 0<\Lambda\leq 1. $$

Proof. The function $g(x)=x/(1+x)$ is increasing for $x\geq0$, since $g'(x)=1/(1+x)^2>0$. At least one root equals $\Theta$, and all $s$ roots are at most $\Theta$. Summing their transformed values therefore gives $$ g(\Theta)\leq\sum_{i=1}^s g(\lambda_i)\leq s g(\Theta), $$ which is the stated bound for $V$. The bounds for $U$ follow by summing $\lambda_i\leq\Theta$, while the lower bound follows because one root equals $\Theta$.

Since $\lambda_i\leq U$, each denominator $1+\lambda_i\leq1+U$, and hence $$ V=\sum_{i=1}^s\frac{\lambda_i}{1+\lambda_i} \geq\sum_{i=1}^s\frac{\lambda_i}{1+U} =\frac{U}{1+U}. $$ Also, $\lambda_i/(1+\lambda_i)\leq\lambda_i$, which proves $V\leq U$.

Set $v_i=\lambda_i/(1+\lambda_i)$. Then $0\leq v_i<1$, $V=\sum_i v_i$, and $\Lambda=\prod_i(1-v_i)$. Repeatedly applying $(1-a)(1-b)\geq1-a-b$ for $a,b\geq0$ with $a,b\leq1$ gives $\prod_i(1-v_i)\geq1-\sum_i v_i=1-V$. For the upper bound, the arithmetic-geometric mean inequality gives $$ \prod_{i=1}^s(1-v_i) \leq\left(\frac{\sum_i(1-v_i)}{s}\right)^s =\left(1-\frac{V}{s}\right)^s. $$ Finally, each factor $1/(1+\lambda_i)$ is positive and at most one, proving $0<\Lambda\leq1$. $\square$

Worked Example: Checking Bounds From a Root Spectrum

For roots $3$ and $1$, the previous calculation gave $V=5/4$, $U=4$, $\Theta=3$, and $\Lambda=1/8$. The largest-root bounds for $V$ are $$ \frac{3}{4}\leq\frac54\leq\frac{2(3)}{4}, $$ or $0.75\leq1.25\leq1.5$. The bounds involving $U$ give $4/5\leq5/4\leq4$. For Wilks’ lambda, the lower bound is $1-V=-1/4$, so $1-V\leq1/8$; the upper bound is $$ \left(1-\frac{(5/4)}{2}\right)^2 =\left(\frac38\right)^2 =\frac{9}{64}, $$ and $1/8=8/64\leq9/64$. Each inequality holds, although the lower Wilks bound is not informative in this example.

Why Multiple Roots Can Change the Ranking

With several nonzero roots, the criteria reward different features of the spectrum. Roy’s criterion responds to the strongest direction, while the Hotelling–Lawley trace responds to total uncompressed root strength. Pillai’s trace sums bounded contributions, and Wilks’ lambda responds to their product. A concentrated spectrum can therefore look strong to Roy even when a more diffuse spectrum has a larger aggregate summary.

Worked Example: Concentrated and Diffuse Effects Rank Differently

Compare spectrum A, with roots $(9,0)$, to spectrum B, with roots $(1,1)$. For A, $$ \Lambda_A=\frac{1}{10},\qquad V_A=\frac{9}{10},\qquad U_A=9,\qquad \Theta_A=9. $$ For B, $$ \Lambda_B=\frac{1}{2}\cdot\frac{1}{2}=\frac14,\qquad V_B=\frac12+\frac12=1,\qquad U_B=2,\qquad \Theta_B=1. $$

Roy’s criterion and Hotelling–Lawley’s trace are larger for A, so they rank A as stronger by their own summaries. Pillai’s trace is larger for B, while Wilks’ lambda is smaller for A, the direction conventionally associated with a stronger effect. In particular, Pillai and Wilks do not give the same ordering here: Pillai favors B, whereas Wilks favors A. The example shows why a general ranking of all four criteria cannot be inferred from their agreement in the single-root case.

Common Pitfall and Practical Meaning

A common mistake is to treat the four criteria as interchangeable names for the sum of the roots. Only $U$ is the untransformed sum. Pillai’s trace sums $\lambda_i/(1+\lambda_i)$, Wilks’ lambda multiplies $1/(1+\lambda_i)$, and Roy’s criterion uses only the maximum. For example, with roots $(9,0)$, $U=9$ but $V=9/10$; replacing Pillai’s trace by the raw sum would change both the value and its interpretation.

The root formulas are useful when checking calculations, comparing criteria, and understanding why a test may emphasize a particular pattern of multivariate effects. They also clarify the limits of purely algebraic comparisons: with multiple roots, no criterion universally dominates the others in ranking, and a bound does not replace the distributional calibration required for inference.

Key takeaway: The four criteria are distinct functions of one nonnegative root spectrum. With one nonzero root their rankings are equivalent after accounting for Wilks’ reversed direction; with several roots, concentration versus dispersion can make their rankings disagree.

Check Your Understanding

Use the root formulas and inequalities to answer these questions.

For roots $(2,0)$, calculate $\Lambda$, $V$, $U$, and $\Theta$.
Why does the single-root case make all four criteria monotone summaries of the same quantity, despite Wilks’ lambda decreasing rather than increasing?
For any nonnegative root spectrum, explain why $V\leq U$.
Give a reason Roy’s criterion can favor a concentrated spectrum while Pillai’s trace favors a more diffuse one.
Which criteria use only the largest root, a sum of raw roots, a sum of transformed roots, and a product of transformed roots, respectively?