LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A positive-semidefinite matrix has a unique positive-semidefinite square root

Statement

Let $d \geq 1$ and let $G \in R^{d \times d}$ be symmetric and positive semidefinite, meaning $v^{⊤} G v \geq 0$ for every $v \in R^{d}$ . Then there is a unique symmetric positive-semidefinite matrix $H$ such that $H^{2} = G$ . If $G = Q diag (λ_{1}, \dots, λ_{d}) Q^{⊤}$ is an orthogonal diagonalization, this root is $H = Q diag (\sqrt{λ_{1}}, \dots, \sqrt{λ_{d}}) Q^{⊤}$ .

Statement references

Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis: a real self-adjoint matrix has an orthonormal eigenbasis.

In orthonormal bases, the matrix of the adjoint is the conjugate transpose of the matrix: for a real Euclidean matrix, self-adjointness is symmetry.

Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes: matrix products and transposes use ordinary finite-dimensional matrix algebra.

Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication: matrix multiplication is associative.

Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ : every nonnegative real eigenvalue has a unique nonnegative scalar square root.

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : the quadratic form and orthogonality use the standard Euclidean inner product.

Facts & Assumptions

Given: A symmetric positive-semidefinite $G \in R^{d \times d}$ with $d \geq 1$ .

[F1]

Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis: a self-adjoint endomorphism of a finite-dimensional real inner-product space has an orthonormal eigenbasis.

[F2]

In orthonormal bases, the matrix of the adjoint is the conjugate transpose of the matrix: in an orthonormal real basis, the adjoint matrix is the transpose, so a symmetric matrix is self-adjoint.

[F3]

Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes: products, powers, and transposes are the usual finite-dimensional matrix operations.

[F4]

Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ : every real $λ \geq 0$ has a unique scalar $r \geq 0$ with $r^{2} = λ$ .

[F5]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : $v^{⊤} w = ⟨ v, w ⟩$ and $∥ v ∥_{2}^{2} = ⟨ v, v ⟩$ in standard orthonormal coordinates.

[F6]

Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication: matrix multiplication is associative and unital, so $S S^{2} = S^{2} S$ and $Q^{⊤} Q = I$ for an orthogonal matrix $Q$ .

[A1]

$G$ is symmetric and satisfies $v^{⊤} G v \geq 0$ for all $v \in R^{d}$ .

Proof

1.1givenA1F1F2F5choosealgebra

By [F1, F2], choose an orthogonal matrix $Q$ whose columns are an orthonormal eigenbasis of $G$ , so $G = Q diag (λ_{1}, \dots, λ_{d}) Q^{⊤}$ . For each unit eigenvector $q_{i}$ , $λ_{i} = q_{i}^{⊤} G q_{i} \geq 0$ by [A1].

1.2givenA1F1F2F3F4F6algebra

Let $S$ be any symmetric positive-semidefinite matrix with $S^{2} = G$ . By [F6], $S G = S S^{2} = S^{3} = S^{2} S = G S$ , so $S$ preserves every eigenspace $E_{λ} (G)$ : if $G v = λ v$ , then $G (S v) = S (G v) = λ S v$ . Its restriction to each $E_{λ} (G)$ is symmetric and positive semidefinite. By [F1, F2] that restriction has an orthonormal eigenbasis; each eigenvalue $μ$ satisfies $μ \geq 0$ and $μ^{2} = λ$ , hence [F4] forces $μ = \sqrt{λ}$ . Therefore $S = \sqrt{λ} I$ on each $E_{λ} (G)$ .

2.1step 1.1F3F4F5F6constructalgebra

By [F4], choose $r_{i} = \sqrt{λ_{i}} \geq 0$ with $r_{i}^{2} = λ_{i}$ , and set $H = Q diag (r_{1}, \dots, r_{d}) Q^{⊤}$ . This matrix is symmetric and, using [F6] and $Q^{⊤} Q = I$ , $H^{2} = Q diag (λ_{1}, \dots, λ_{d}) Q^{⊤} = G$ . For every $v$ , $v^{⊤} H v = \sum_{i} r_{i} (q_{i}^{⊤} v)^{2} \geq 0$ , so $H$ is positive semidefinite.

3.1step 1.1step 2.1step 1.2F1algebra∎

The eigenspaces in step 1.1 form an orthogonal basis of $R^{d}$ , and step 1.2 shows that $S$ and $H$ act identically on every vector in that basis. Hence $S = H$ , proving uniqueness.

Depends on

Used by

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Sources

Elad Hazan, Introduction to Online Convex Optimization, second edition (standard reference, not scraped)
John Duchi, Elad Hazan, and Yoram Singer, Adaptive Subgradient Methods for Online Learning and Stochastic Optimization (standard reference, not scraped)