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

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Finite Hermitian diagonalization

Statement

For every integer $d \geq 1$ and Hermitian matrix $A \in M_{d} (C)$ , there is an orthonormal basis of $C^{d}$ consisting of eigenvectors of $A$ , and all eigenvalues are real. If in addition $x^{*} A x \geq 0$ for every $x \in C^{d}$ , every eigenvalue is nonnegative and the matrix $A^{1 / 2}$ obtained by replacing each eigenvalue by its nonnegative square root is well defined.

Facts & Assumptions

Given: An integer $d \geq 1$ and a Hermitian matrix $A \in M_{d} (C)$ ; for the positivity conclusion, also assume $x^{*} A x \geq 0$ for every $x$ .

[F1]

Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line: every closed bounded subset of $R^{n}$ , $n \geq 1$ , is compact.

[F2]

A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value: a continuous real function on a nonempty compact metric space attains its maximum.

[F3]

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 number has a unique nonnegative square root.

Proof technique: Rayleigh maximization and induction on $d$ .

Proof

1.1givenF1algebra

Identify $C^{d}$ with $R^{2 d}$ by real and imaginary coordinates. Its unit sphere is a closed bounded subset and is nonempty because $d \geq 1$ ; it is compact by [F1].

2.1givenF2step 1.1

The Rayleigh function $f (x) = x^{*} A x$ is real on the unit sphere, since $A = A^{*}$ , and is continuous because each coordinate of $f$ is a finite polynomial in the real and imaginary coordinates of $x$ . By [F2] it has a maximizer $u$ with $u^{*} u = 1$ .

3.1step 2.1algebra

Fix $w$ with $u^{*} w = 0$ and let $g (t) = (u + t w)^{*} A (u + t w) / (u + t w)^{*} (u + t w)$ for real $t$ . The denominator is $1 + t^{2} w^{*} w > 0$ , so $g$ is differentiable near $0$ ; maximality of $u$ gives $g^{'} (0) = 0$ . Expanding numerator and denominator gives $g^{'} (0) = 2 Re (w^{*} A u)$ , hence the real part of $w^{*} A u$ is zero.

4.1step 3.1algebra

Apply step 3.1 with $i w$ in place of $w$ . Since $(i w)^{*} A u = - i w^{*} A u$ , its real part is the imaginary part of $w^{*} A u$ ; therefore $w^{*} A u = 0$ for every $w ⊥ u$ . Decompose $A u = (u^{*} A u) u + w_{0}$ with $w_{0} ⊥ u$ ; taking $w = w_{0}$ yields $w_{0}^{*} A u = ∥ w_{0} ∥^{2} = 0$ . Thus $A u = λ u$ , where $λ = u^{*} A u \in R$ .

5.1step 4.1algebra

If $x ⊥ u$ , then $u^{*} A x = (A u)^{*} x = λ u^{*} x = 0$ , so $u^{⊥}$ is invariant under $A$ and the restriction of $A$ to it is Hermitian. For $d = 1$ the standard unit vector is already an eigenbasis. For $d > 1$ , induction on the dimension supplies an orthonormal eigenbasis on $u^{⊥}$ ; adjoining $u$ gives one for $C^{d}$ .

6.1givenF3step 5.1algebra∎

For any unit eigenvector $v$ with eigenvalue $μ$ , positivity gives $μ = v^{*} A v \geq 0$ . By [F3], each such $μ$ has a unique nonnegative square root. In the eigenbasis from step 5.1, define $A^{1 / 2}$ to be diagonal with entries $\sqrt{μ}$ . This matrix is nonnegative and squares to $A$ ; it is independent of choices inside repeated eigenspaces because the same scalar is assigned to every vector in that eigenspace.

Finite Hermitian diagonalization

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources