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The real positive-semidefinite cone is closed in the Frobenius metric

Statement

For an integer $d \geq 1$ , let $S^{d}$ be the real symmetric $d \times d$ matrices and let $S_{+}^{d} = {M \in S^{d} : v^{T} M v \geq 0 for every v \in R^{d}}$ . With the Frobenius norm $∥ A ∥_{F} = (\sum_{i, j} A_{i j}^{2})^{1 / 2}$ , the set $S_{+}^{d}$ is closed in $S^{d}$ .

Statement references

Entry-sum Frobenius norm

Euclidean inner product on coordinate space

Cauchy–Schwarz for the Euclidean inner product

Closed sets have open complements

Open balls in the Frobenius metric

Higham, §2, Theorem 2.1, article pp. 105–106

Bindel, §6, PDF p. 5

Facts & Assumptions

Given: An integer $d \geq 1$ and the real vector space $S^{d}$ with the Frobenius metric.

[F1]

The Frobenius norm is the Euclidean norm of the matrix-entry list (The Frobenius norm $∥ A ∥_{F} = (\sum_{i, j} ∣ a_{i j} ∣^{2})^{1 / 2}$ on real or complex matrices).

[F2]

The entry inner product is the Euclidean coordinate inner product, so it satisfies Cauchy–Schwarz; expanding entries gives $∥ v v^{T} ∥_{F} = ∥ v ∥_{2}^{2}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

[F3]

A set is closed when its complement is open, and an open set contains a positive-radius ball around each of its points (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space).

[F4]

The Frobenius norm is an inner-product norm, so its restriction to $S^{d}$ is a norm; the induced metric is $d_{F} (A, B) : = ∥ A - B ∥_{F}$ (The Frobenius norm $∥ A ∥_{F} = (\sum_{i, j} ∣ a_{i j} ∣^{2})^{1 / 2}$ on real or complex matrices, The inner-product norm is definite, homogeneous, and satisfies the triangle inequality, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Proof

1.1givenchoose

Let $A \in S^{d} ∖ S_{+}^{d}$ . By the definition of $S_{+}^{d}$ , there is $v \in R^{d}$ with $v^{T} A v < 0$ ; necessarily $v \neq 0$ . Set $ρ = - v^{T} A v / (2 ∥ v ∥_{2}^{2}) > 0$ .

2.1step 1.1F1F2F4algebra

For any $Z \in S^{d}$ with $∥ Z - A ∥_{F} < ρ$ , the entry norm [F1], entry inner product, and Cauchy–Schwarz [F2] give $∣ v^{T} (Z - A) v ∣ = ∣ ⟨ Z - A, v v^{T} ⟩_{F} ∣ \leq ∥ Z - A ∥_{F} ∥ v v^{T} ∥_{F} < ρ ∥ v ∥_{2}^{2} = - \frac{1}{2} v^{T} A v$ . Hence $v^{T} Z v < v^{T} A v - \frac{1}{2} v^{T} A v = \frac{1}{2} v^{T} A v < 0$ , so $Z \notin S_{+}^{d}$ .

3.1step 1.1step 2.1F3F4∎

Step 2.1 shows that the open ball $B_{F} (A, ρ)$ in the metric $d_{F}$ is contained in the complement of $S_{+}^{d}$ . Since $A$ was arbitrary in that complement, every point of it has an open ball contained in it; the complement is open by [F3], and therefore $S_{+}^{d}$ is closed.

Depends on

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Sources

Nicholas J. Higham, Computing a Nearest Symmetric Positive Semidefinite Matrix, Linear Algebra and its Applications 103 (1988), 103–118 (standard reference, not scraped)
David Bindel, Matrix Computations, CS 6210, Matrix Nearness Problems, Fall 2022 (standard reference, not scraped)