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Positive definite symmetric matrices have positive definite square roots, and the congruence M^{-1/2} K M^{-1/2} preserves symmetry and positivity

Statement

Let $n \geq 1$ and let $M \in M_{n} (R)$ be symmetric and positive definite, that is $v^{T} M v > 0$ for every nonzero $v \in R^{n}$ ; identify $R^{n}$ with its standard real inner product space (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ). Then:

$M$ is invertible and $M^{- 1}$ is symmetric and positive definite;
there is a unique symmetric positive definite matrix $M^{1 / 2}$ with $(M^{1 / 2})^{2} = M$ ;
$M^{- 1 / 2} : = (M^{1 / 2})^{- 1}$ is symmetric and positive definite, satisfies $(M^{- 1 / 2})^{2} = M^{- 1}$ , and $M^{- 1 / 2} M M^{- 1 / 2} = I_{n}$ ;
for every symmetric matrix $K$ that is non-negative, that is $v^{T} K v \geq 0$ for all $v$ , the matrix $S : = M^{- 1 / 2} K M^{- 1 / 2}$ is symmetric and non-negative, and $S$ is positive definite if and only if $K$ is.

Facts & Assumptions

Given: The symmetric positive definite matrix $M$ and, in clause 4, a symmetric non-negative matrix $K$ .

[F1]

A real matrix is symmetric if and only if the associated endomorphism of $R^{n}$ is self-adjoint; positive definite (respectively non-negative) matrices in the form sense correspond exactly to positive (respectively non-negative) operators, and a positive operator is non-negative (In an orthonormal basis, self-adjoint means conjugate-transpose symmetry and normal means commuting with the conjugate transpose, Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space, Over the reals, non-negative and positive operators correspond exactly to positive semidefinite and positive definite symmetric forms, Non-negative and positive operators).

[F2]

A non-negative operator on a finite-dimensional real inner product space has a unique non-negative square root, and a non-negative operator admits an orthonormal basis of eigenvectors with non-negative eigenvalues (A non-negative operator has a unique non-negative square root, A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S).

[F3]

Matrix algebra: composition and multiplication of matrices are associative, the identity acts as $I_{n} x = x$ , and $(A B)^{T} = B^{T} A^{T}$ ; the inverse of an invertible square matrix is unique and two-sided (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, Invertible matrices and the general linear group ${GL}_{n} (F)$ , In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided).

Proof

technique · direct; transfer the positivity statement to operators and use the unique non-negative square root

1.1givenF1F2algebra

By [F1] $M$ corresponds to a positive operator on $R^{n}$ , hence to a non-negative one. By [F2] there is a unique non-negative operator $R$ with $R^{2} = M$ ; write $M^{1 / 2}$ for its matrix, which is symmetric because $R$ is self-adjoint [F1]. If $R v = 0$ then $M v = R^{2} v = 0$ , so $v = 0$ because $M$ is positive definite; hence $R$ is injective, and by [F2] its eigenvalues are non-negative and none is zero, so all are positive: $R$ is positive definite. This proves clause 2 except for uniqueness among all symmetric positive definite square roots, which follows because a symmetric positive definite square root is in particular a non-negative square root, and [F2] gives uniqueness among non-negative square roots.

2.1step 1.1F2F3algebra

In the orthonormal eigenbasis of $R = M^{1 / 2}$ from step 1.1, write $R e_{j} = r_{j} e_{j}$ with $r_{j} > 0$ . Define $B e_{j} = r_{j}^{- 1} e_{j}$ and extend linearly. Then $B R = R B = I_{n}$ on this basis, and $B^{2} M = M B^{2} = I_{n}$ since $M = R^{2}$ . Thus $R$ and $M$ are invertible by [F3]. Since $M = M^{1 / 2} M^{1 / 2}$ [F3], we have $M^{- 1} = (M^{1 / 2})^{- 1} (M^{1 / 2})^{- 1}$ , and $M^{- 1 / 2} : = (M^{1 / 2})^{- 1}$ is well defined and satisfies $(M^{- 1 / 2})^{2} = M^{- 1}$ and $M^{- 1 / 2} M M^{- 1 / 2} = (M^{1 / 2})^{- 1} M^{1 / 2} M^{1 / 2} (M^{1 / 2})^{- 1} = I_{n}$ by associativity and [F3].

2.2step 1.12.1F1F3algebra

$M^{- 1}$ and $M^{- 1 / 2}$ are symmetric and positive definite. For $M^{- 1}$ : for $u, u^{'} \in R^{n}$ write $u = M x$ , $u^{'} = M x^{'}$ ; then $⟨ M^{- 1} u, u^{'} ⟩ = ⟨ x, M x^{'} ⟩ = ⟨ M x, x^{'} ⟩ = ⟨ u, M^{- 1} u^{'} ⟩$ by symmetry of $M$ , so $M^{- 1}$ is self-adjoint, hence symmetric [F1]; and for $u \neq 0$ , $⟨ M^{- 1} u, u ⟩ = ⟨ x, M x ⟩ > 0$ , so $M^{- 1}$ is positive definite. For $M^{- 1 / 2}$ : it is the inverse of the self-adjoint invertible $M^{1 / 2}$ , and the same computation with $M^{1 / 2}$ in place of $M$ (which is positive definite by step 1.1) shows that its inverse is self-adjoint and positive definite. This proves clauses 1 and 3.

3.1step 2.2F1F3algebra∎

Clause 4: for symmetric $K$ , $S^{T} = (M^{- 1 / 2})^{T} K^{T} (M^{- 1 / 2})^{T} = M^{- 1 / 2} K M^{- 1 / 2} = S$ by symmetry of $M^{- 1 / 2}$ and $K$ , so $S$ is symmetric [F3]. For $x \in R^{n}$ put $y : = M^{- 1 / 2} x$ ; then $x^{T} S x = y^{T} K y$ , so non-negativity of $K$ implies non-negativity of $S$ . If $K$ is positive definite and $x \neq 0$ , then $y = M^{- 1 / 2} x \neq 0$ because $M^{- 1 / 2}$ is invertible, and $x^{T} S x = y^{T} K y > 0$ . Thus $S$ is positive definite. Conversely, if $S$ is positive definite and $x \neq 0$ , then with $y : = M^{1 / 2} x \neq 0$ one has $x^{T} K x = y^{T} M^{- 1 / 2} K M^{- 1 / 2} y = y^{T} S y > 0$ , so $K$ is positive definite.

Remarks

The matrices $M^{1 / 2}$ , $M^{- 1 / 2}$ are the standard square roots of a positive definite symmetric matrix; they are used in Decoupling of the constant-coefficient quadratic system $M q^{''} + K q = 0$ into normal modes, with the zero-mode and energy identities to reduce a coupled quadratic system to decoupled scalar oscillators, and in Linearisation of the mechanical system at an equilibrium: M(q_0) xi'' + D²U(q_0) xi = 0, and the small-oscillation quadratic model to identify the linearised stiffness matrix.
Clause 4 shows that the eigenvalues of $S$ are non-negative whenever $K$ is non-negative, and positive exactly when $K$ is positive definite. This is what makes the diagonal matrix $Λ$ of Decoupling of the constant-coefficient quadratic system $M q^{''} + K q = 0$ into normal modes, with the zero-mode and energy identities have non-negative diagonal entries, with the zero entries corresponding to free (non-oscillating) modes.

Positive definite symmetric matrices have positive definite square roots, and the congruence M^{-1/2} K M^{-1/2} preserves symmetry and positivity

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

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