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Positive-definite overlap whitening preserves a generalized Ritz pencil

Statement

Let $H = H^{*} \in M_{n} (C)$ be a finite Hermitian stiffness matrix and let $S = S^{*} > 0$ be a positive-definite overlap or mass matrix. By Cholesky factorization (A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique), write $S = L L^{*}$ with $L$ invertible lower triangular and positive diagonal. Define the whitened Hermitian matrix $\tilde{H} : = L^{- 1} H L^{- *}, L^{- *} : = (L^{*})^{- 1} .$ The generalized Ritz equation $H c = λ S c$ is equivalent under the invertible change of coordinates $y = L^{*} c$ to $\tilde{H} y = λ y$ . The Rayleigh quotients agree: $\frac{c^{*} H c}{c^{*} S c} = \frac{y^{*} \tilde{H} y}{y^{*} y} (c \neq 0) .$

Facts & Assumptions

Given: A Hermitian matrix $H$ , a Hermitian positive-definite matrix $S$ , its Cholesky factor $L$ , and a nonzero coefficient vector $c$ .

[F1]

A Hermitian positive-definite matrix has a unique Cholesky factorization $S = L L^{*}$ with $L$ lower triangular and positive real diagonal; such $L$ is invertible (Hermitian positive-definite matrices and Cholesky factorisation A = LL* with positive diagonal, A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique).

[F2]

Matrix products and identity matrices use the entrywise convention of Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes.

[F3]

For compatible finite-dimensional maps, $(A B)^{*} = B^{*} A^{*}$ , $A^{* *} = A$ , and $I^{*} = I$ (Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ ). In the standard orthonormal coordinate bases these adjoints are the conjugate-transpose matrices (In orthonormal bases, the matrix of the adjoint is the conjugate transpose of the matrix).

Proof

technique · direct

Given: $H = H^{*}$ , $S = L L^{*}$ , $L$ invertible, $c \neq 0$ , and $y = L^{*} c$ .

1.1F1F2F3givenalgebra

Taking adjoints of $L^{- 1} L = I$ and $L L^{- 1} = I$ by [F3] gives $L^{*} (L^{- 1})^{*} = I$ and $(L^{- 1})^{*} L^{*} = I$ . Hence $L^{*}$ is invertible with inverse $(L^{- 1})^{*} = L^{- *}$ ; taking adjoints again gives $(L^{- *})^{*} = L^{- 1}$ . The matrix $\tilde{H} = L^{- 1} H L^{- *}$ is Hermitian, since ${\tilde{H}}^{*} = (L^{- *})^{*} H^{*} (L^{- 1})^{*} = L^{- 1} H L^{- *} = \tilde{H} .$ The coordinate map $c \mapsto y = L^{*} c$ is invertible by [F1].

2.1F1F2F3step 1.1givenalgebra

Substituting $c = L^{- *} y$ into $H c = λ S c$ and multiplying by $L^{- 1}$ gives $L^{- 1} H L^{- *} y = λ L^{- 1} L L^{*} L^{- *} y = λ y .$ Every operation is reversible because $L$ is invertible, so the generalized and whitened eigenproblems have exactly the same eigenvalues, with eigenvectors related by $y = L^{*} c$ .

3.1F1F2F3step 1.1algebra∎

With $c = L^{- *} y$ , matrix multiplication gives $c^{*} H c = y^{*} L^{- 1} H L^{- *} y = y^{*} \tilde{H} y, c^{*} S c = y^{*} L^{- 1} L L^{*} L^{- *} y = y^{*} y .$ Since $c \neq 0$ iff $y \neq 0$ , and $S > 0$ makes $c^{*} S c > 0$ , both equal denominators are positive. Division proves equality of the two Rayleigh quotients.

Remarks

The equivalence is exact linear algebra. It does not by itself assert a continuum eigenvalue bound: that requires the finite basis to be a conforming subspace of the stated form domain and the matrices to be the corresponding form and inner-product matrices.

Depends on

Used by

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Sources

Jean-Luc Guermond, MATH 661 Finite Element Methods, Fall 2025, Chapter 47 (standard reference, not scraped)
Bertrand Mercier, Topics in Finite Element Solution of Elliptic Problems, Chapter 8 (standard reference, not scraped)