Orthogonal Equivariance Of Matrix Beta Laws

Why Orthogonal Conjugation Matters

A matrix beta variable is a random positive-definite matrix, so its orientation can matter: changing coordinates can alter its entries even when it leaves its eigenvalues unchanged. Orthogonal conjugation is a special coordinate change. It rotates or reflects the axes while preserving lengths, and it preserves the identity matrix that appears in the standard matrix beta supports and densities. The resulting invariance is useful when a calculation depends on a convenient orientation rather than on a particular coordinate system.

The qualification “orthogonal” is essential. A general nonsingular linear transformation can stretch some directions, changing the beta type I support and changing the beta type II density. We will verify both the invariance and this limitation directly.

Matrix Beta Laws and Their Supports

Let $\mathbb S_p$ denote the real symmetric $p\times p$ matrices, and let $\mathbb S_p^{++}$ denote the positive-definite matrices. Write $A\prec B$ when $B-A$ is positive definite. The shape parameters $a,b$ are real and satisfy $a,b>(p-1)/2$, the conditions for the standard central matrix beta type I and type II densities to be proper.

Definition: A central matrix beta type I variable $X$ has density, with respect to Lebesgue measure on the distinct entries of $\mathbb S_p$, $$ f_{\mathrm I}(X)=\frac{1}{B_p(a,b)} |X|^{a-(p+1)/2}|I-X|^{b-(p+1)/2}, \qquad 0\prec X\prec I. $$ A central matrix beta type II variable $Y$ has density $$ f_{\mathrm {II}}(Y)=\frac{1}{B_p(a,b)} |Y|^{a-(p+1)/2}|I+Y|^{-(a+b)}, \qquad Y\in\mathbb S_p^{++}. $$ Here $|A|$ denotes the determinant, $I$ is the $p\times p$ identity matrix, and $B_p(a,b)$ is the normalizing constant.

The type I support requires both $X$ and $I-X$ to be positive definite. The type II support requires only $Y$ to be positive definite; its density also involves $I+Y$, which is positive definite whenever $Y$ is. In these central laws, the parameters $a,b$ are scalars. They affect the determinant powers but do not specify preferred directions in the coordinate space.

The Orthogonal-Conjugation Result

A real matrix $Q$ is orthogonal if $Q'Q=QQ'=I$, where $Q'$ denotes its transpose. For a symmetric matrix $A$, conjugation sends $A$ to $QAQ'$. It preserves eigenvalues and determinant. We also need that it preserves volume on the space of symmetric matrices.

Theorem: If $X$ has a central matrix beta type I distribution with parameters $a,b>(p-1)/2$, and $Q$ is a fixed orthogonal matrix, then $QXQ'$ has the same distribution as $X$. The same assertion holds for a central matrix beta type II variable $Y$. Thus both laws are invariant under orthogonal conjugation for every admissible scalar parameter pair $a,b$.

Proof. First, if $A\succ0$, then for any nonzero vector $v$, $v'QAQ'v=(Q'v)'A(Q'v)>0$, because $Q'v\ne0$. The same reasoning shows that $A\prec B$ implies $QAQ'\prec QBQ'$. Since $QIQ'=I$, conjugation maps the type I support $0\prec X\prec I$ onto itself, and maps the type II support $\mathbb S_p^{++}$ onto itself.

For any $A\in\mathbb S_p$, $|QAQ'|=|Q||A||Q'|=|A|$, because $|Q|=\pm1$. Therefore $|I-QXQ'|=|Q(I-X)Q'|=|I-X|$, and $|I+QYQ'|=|Q(I+Y)Q'|=|I+Y|$. In each density, all determinant factors are unchanged.

It remains to check the change-of-variables factor. The map $A\mapsto QAQ'$ preserves the Frobenius inner product on $\mathbb S_p$: for symmetric $A,B$, $\operatorname{tr}((QAQ')(QBQ'))=\operatorname{tr}(AB)$. Hence it is an isometry of this finite-dimensional real vector space, so its absolute Jacobian is one. The transformed density is therefore the original density at the transformed matrix, with no extra volume factor. The supports, density values, and volume element are preserved, proving both distributional claims. $\square$

This is an invariance statement for each fixed $Q$, not a claim that the entries of $X$ or $Y$ remain unchanged. The matrices usually change; their probability law does not. Since the central densities contain only scalar shape parameters and the identity reference matrix, no direction is singled out. Unequal values of $a$ and $b$ do not break the invariance.

Worked Examples

Worked Example: Conjugating a Type I Matrix

Consider the symmetric matrix $$ X=\begin{pmatrix}0.6&0.2\\0.2&0.5\end{pmatrix} \quad\text{and}\quad Q=\begin{pmatrix}0&-1\\1&0\end{pmatrix}. $$ Check that $X$ is in the type I support, and compute its conjugate.

For a symmetric $2\times2$ matrix, positive definiteness follows from a positive first diagonal entry and a positive determinant. Here the first diagonal entry of $X$ is $0.6>0$, and $|X|=(0.6)(0.5)-(0.2)^2=0.26>0$. Also, $$ I-X=\begin{pmatrix}0.4&-0.2\\-0.2&0.5\end{pmatrix}, \qquad |I-X|=(0.4)(0.5)-(-0.2)^2=0.16>0. $$ Its first diagonal entry is positive as well, so $I-X\succ0$, and $0\prec X\prec I$.

The matrix $Q$ is orthogonal. Multiplication gives $$ QXQ'=\begin{pmatrix}0.5&-0.2\\-0.2&0.6\end{pmatrix}. $$ Its determinant is $0.26$, and $I-QXQ'=\begin{pmatrix}0.5&0.2\\0.2&0.4\end{pmatrix}$ has determinant $0.16$ and positive first diagonal entry. Thus the conjugate is also in the type I support. Its two determinant factors are unchanged, as the theorem requires.

Worked Example: Type II Density Factors Under Rotation

Let $Y=\begin{pmatrix}2&1\\1&2\end{pmatrix}$, and use the same $Q$ as in the previous example. Verify that the type II density factors agree at $Y$ and $QYQ'$.

The first diagonal entry of $Y$ is 2 and $|Y|=4-1=3>0$, so $Y\succ0$. Direct multiplication gives $$ QYQ'=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}, \qquad |QYQ'|=4-1=3. $$ The other determinant in the type II density is $$ |I+Y|=\left|\begin{pmatrix}3&1\\1&3\end{pmatrix}\right|=9-1=8, \qquad |I+QYQ'|=\left|\begin{pmatrix}3&-1\\-1&3\end{pmatrix}\right|=9-1=8. $$ Consequently, for any admissible $a,b$, the unnormalized density factor at either matrix is $3^{a-3/2}8^{-(a+b)}$. Orthogonal conjugation changes the off-diagonal signs here, but not the density.

Why General Congruence Is Different

A congruence transformation replaces $A$ by $CAC'$ for a nonsingular matrix $C$. It maps positive-definite matrices to positive-definite matrices, but it need not preserve $I$, distances, or the volume element. Those differences matter for both beta laws. For type I, the upper support bound is particularly restrictive: in general $CIC'=CC'\ne I$. For type II, positive definiteness is preserved, but the determinant involving $I+Y$ generally does not retain its original value after transformation.

Result: Under an invertible congruence $A\mapsto CAC'$, the absolute Jacobian on $\mathbb S_p$ is $|C|^{p+1}$. For orthogonal $C=Q$, this equals one. Unless the congruence is orthogonal, neither the unit upper bound of the standard type I support nor the identity-based type II density is generally preserved.

To see the Jacobian formula, use a singular value decomposition of $C$. The orthogonal factors have absolute Jacobian one by the isometry argument above. For a diagonal factor with diagonal entries $c_1,\ldots,c_p$, a diagonal coordinate $A_{ii}$ is multiplied by $c_i^2$, while an off-diagonal coordinate $A_{ij}$, $i<j$, is multiplied by $c_ic_j$. The absolute product of these factors is $\prod_i |c_i|^2\prod_{i<j}|c_ic_j|=\prod_i|c_i|^{p+1}=|C|^{p+1}$. This also shows why simply checking that a transformation preserves positive definiteness is not enough to establish invariance of a density.

Worked Example: A Congruence Leaves the Type I Support

Take the type I matrix $X=\operatorname{diag}(0.8,0.4)$ and the nonsingular matrix $C=\operatorname{diag}(\sqrt{2.5},\sqrt{1.25})$. First, $X\succ0$, and $I-X=\operatorname{diag}(0.2,0.6)\succ0$, so $X$ is in the type I support.

The congruence is $$ CXC'=\operatorname{diag}\big((2.5)(0.8),(1.25)(0.4)\big) =\operatorname{diag}(2,0.5). $$ This matrix is positive definite, but it is not below $I$, since $$ I-CXC'=\operatorname{diag}(-1,0.5). $$ The first diagonal entry is $1-2=-1$, not $-1.5$; in particular, the difference is not positive definite. Thus the congruence has taken a valid type I support point outside $0\prec X\prec I$. Positive definiteness of the transformed matrix alone does not establish type I support membership.

Worked Example: A Type II Congruence Changes the Density

Consider a type II law with $p=2$, $a=b=2$, and the congruence $C=\operatorname{diag}(2,1)$. The matrix $Y=I$ is positive definite, and $D=CYC'=\operatorname{diag}(4,1)$ remains positive definite. Compare the density at $D$ with the density of the transformed random matrix.

Write the original density as $f$, omitting its common normalizing constant. Since $a-(p+1)/2=1/2$ and $a+b=4$, $$ f(I)=|I|^{1/2}|2I|^{-4}=2^{-4}=\frac{1}{16}, \qquad f(D)=|D|^{1/2}|I+D|^{-4} =4^{1/2}\,|\,\operatorname{diag}(5,2)\,|^{-4} =\frac{2}{10^4}=\frac{1}{5000}. $$ If $Z=CYC'$, its density at $D$ is $g(D)=f(C^{-1}DC^{-1})|C|^{-3}$, using the inverse congruence Jacobian for $p=2$. Here $C^{-1}DC^{-1}=I$ and $|C|=2$, so $g(D)=1/128$. This is not $f(D)=1/5000$. The transformed variable still has positive-definite support, but its density is not the original standard type II density.

Common Mistake and Practical Interpretation

A common mistake is to reason: “Congruence preserves positive definiteness, so it must preserve a matrix beta law.” The first clause is true for any invertible $C$, but the conclusion does not follow. Type I support also requires $I-CXC'\succ0$; type II invariance requires the density and the change-of-variables factor to match, not merely positive-definite support. The orthogonal case works because it preserves the identity, determinants, and volume simultaneously.

More generally, a model with a fixed directional matrix parameter can only be invariant under orthogonal transformations that preserve that parameter. For example, the transformations preserving a symmetric matrix $\Omega$ satisfy $Q\Omega Q'=\Omega$. Full invariance under every orthogonal $Q$ requires $\Omega$ to be a scalar multiple of $I$: invariance under coordinate sign changes forces off-diagonal entries to vanish, and invariance under coordinate permutations forces all diagonal entries to be equal. The standard central beta laws have precisely this direction-free form, with scalar shape parameters and the identity as their reference.

The practical test is therefore to track three things separately: the support, every determinant in the density, and the Jacobian. For orthogonal conjugation, all three are preserved, which gives a direct and dependable argument for equivariance. This conclusion concerns the distribution under a change of orientation; it does not itself provide moment formulas or a sampling algorithm.

Check Your Understanding

Use the support conditions, density forms, and change-of-variables reasoning developed above.

State the support conditions for matrix beta type I and type II variables, and identify which condition is not automatically ensured by positive definiteness of a type I matrix.
Why does conjugation by an orthogonal matrix preserve both $|X|$ and $|I-X|$?
Does choosing $a\ne b$ destroy orthogonal-conjugation invariance for a central matrix beta type I law? Explain.
For $C=\operatorname{diag}(3,1)$, what is the absolute Jacobian of $A\mapsto CAC'$ on $\mathbb S_2$?
Explain why positive-definite support alone cannot prove that a congruence-transformed type II variable has the same law.