LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The maximum trace over q-dimensional orthonormal frames is the sum of the top q eigenvalues

Statement

Let $A \in R^{p \times p}$ be symmetric with eigenvalues $λ_{1} \geq \dots \geq λ_{p}$ , counted with multiplicity. For $1 \leq q \leq p$ and every $Q \in R^{p \times q}$ satisfying $Q^{T} Q = I_{q}$ ,

$tr (Q^{T} A Q) \leq \sum_{j = 1}^{q} λ_{j} .$

Equality is attained by taking the columns of $Q$ to be orthonormal eigenvectors for the $q$ largest eigenvalues. Equality holds exactly when the column space of $Q$ contains every eigenspace with eigenvalue greater than $λ_{q}$ , is orthogonal to every eigenspace with eigenvalue less than $λ_{q}$ , and uses a subspace of dimension $q - # {j : λ_{j} > λ_{q}}$ from the eigenspace with eigenvalue $λ_{q}$ .

Statement references

Cameron, Linear and Nonlinear Dimensionality Reduction, §2.1, PDF p. 3

von Luxburg, A Tutorial on Spectral Clustering, §5.2, PDF pp. 11–12

Facts & Assumptions

Given: A symmetric real matrix $A \in R^{p \times p}$ and an integer $q$ with $1 \leq q \leq p$ .

[F1]

In the standard real coordinate inner product, $⟨ x, y ⟩ = \sum_{i < p} x_{i} y_{i}$ (The standard formulas $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and $\sum_{k < n} x_{k} \overline{y_{k}}$ on $C^{n}$ are inner products).

[F2]

In an orthonormal basis, a real endomorphism is self-adjoint exactly when its matrix is symmetric (In an orthonormal basis, self-adjoint means conjugate-transpose symmetry and normal means commuting with the conjugate transpose).

[F3]

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

[F4]

The trace of a square matrix is the sum of its diagonal entries (The trace $tr (A)$ as the sum of the diagonal entries).

[F5]

For conformable rectangular matrices $A, B$ , $tr (A B) = tr (B A)$ (For $A \in M_{m \times n} (F)$ and $B \in M_{n \times m} (F)$ , $tr (A B) = tr (B A)$ ).

Proof

1.1givenF1F2F3

By [F1] and [F2], the matrix $A$ represents a self-adjoint map on $R^{p}$ with its standard inner product; [F3] gives an orthonormal eigenbasis $u_{1}, \dots, u_{p}$ ordered with eigenvalues $λ_{1} \geq \dots \geq λ_{p}$ .

2.1step 1.1F4F5algebra

For any feasible $Q$ , put $P = Q Q^{T}$ . Since $Q^{T} Q = I_{q}$ , $P^{T} = P$ and $P^{2} = P$ , so $u_{j} = P u_{j} + (I - P) u_{j}$ is an orthogonal decomposition and $a_{j} : = u_{j}^{T} P u_{j} = ∥ Q^{T} u_{j} ∥_{2}^{2} = ∥ P u_{j} ∥_{2}^{2}$ lies in $[0, 1]$ . This decomposition gives $a_{j} = 0$ exactly when $P u_{j} = 0$ , and $a_{j} = 1$ exactly when $P u_{j} = u_{j}$ . Also $\sum_{j = 1}^{p} a_{j} = tr (P) = q$ by [F4]. The frame trace equals $tr (P A)$ by [F5]. If $U$ has columns $u_{j}$ and $Λ = U^{T} A U = diag (λ_{j})$ , [F5] and $U U^{T} = I$ give $tr (P A) = tr (U^{T} P A U) = tr ((U^{T} P U) Λ) = \sum_{j = 1}^{p} λ_{j} a_{j}$ .

3.1step 2.1algebra

Because $λ_{j} \geq λ_{q}$ for $j \leq q$ and $λ_{j} \leq λ_{q}$ for $j > q$ , $\sum_{j = 1}^{p} λ_{j} a_{j} \leq \sum_{j = 1}^{q} (λ_{j} a_{j} + λ_{q} (1 - a_{j})) \leq \sum_{j = 1}^{q} λ_{j}$ . The first inequality uses $\sum_{j} a_{j} = q$ ; the second uses $0 \leq a_{j} \leq 1$ and $λ_{j} \geq λ_{q}$ for $j \leq q$ .

4.1step 2.1step 3.1algebra

The first inequality in step 3.1 has nonnegative difference $\sum_{j > q} (λ_{q} - λ_{j}) a_{j}$ , so equality requires $a_{j} = 0$ whenever $λ_{j} < λ_{q}$ . The second has nonnegative difference $\sum_{j \leq q} (λ_{j} - λ_{q}) (1 - a_{j})$ , so equality requires $a_{j} = 1$ whenever $λ_{j} > λ_{q}$ . By step 2.1, these conditions say the chosen subspace contains all higher-eigenvalue eigenspaces, is orthogonal to all lower-eigenvalue eigenspaces, and takes its remaining dimension in the $λ_{q}$ eigenspace. Conversely, each such subspace makes both differences zero. If $A = 0$ , all eigenvalues are zero and both differences vanish for every feasible $Q$ , as does the stated equality condition.

5.1step 1.1step 3.1F3∎

Taking the first $q$ vectors of the eigenbasis from step 1.1 as the columns of $Q$ gives a feasible frame and makes the trace equal $\sum_{j = 1}^{q} λ_{j}$ . Thus the upper bound is the attained maximum, with the stated equality cases.

Depends on

Used by

Dependency tree · two levels

20 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Elements of Statistical Learning, second edition, corrected 12th printing (standard reference, not scraped)
Maria Cameron, Linear and Nonlinear Dimensionality Reduction, lecture notes (standard reference, not scraped)
Ulrike von Luxburg, A Tutorial on Spectral Clustering (standard reference, not scraped)