TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28

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.

Courant-Fischer min-max principle for self-adjoint endomorphisms on finite-dimensional real inner product spaces

Statement

Let $T : V \to V$ be self-adjoint on an $n$ -dimensional real inner product space, and list its eigenvalues in weakly decreasing order:

$λ_{1} \geq λ_{2} \geq \dots \geq λ_{n} .$

Then for every $1 \leq k \leq n$ ,

$λ_{k} = \min_{\dim W = n - k + 1} \max_{0 \neq v \in W} R_{T} (v) = \max_{\dim U = k} \min_{0 \neq v \in U} R_{T} (v) .$

Facts & Assumptions

Given: A self-adjoint endomorphism $T : V \to V$ of an $n$ -dimensional real inner product space.

[L1]

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

[L2]

A subspace of an $n$ -dimensional vector space cannot have dimension greater than $n$ (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

Proof

technique · direct

1.1L1algebra

By [L1], choose an orthonormal eigenbasis $(e_{1}, \dots, e_{n})$ with $T e_{j} = λ_{j} e_{j}$ . For every nonzero $v = \sum_{j} c_{j} e_{j}$ , one has $R_{T} (v) = \frac{\sum_{j} λ_{j} ∣ c_{j} ∣^{2}}{\sum_{j} ∣ c_{j} ∣^{2}}$ , so on $U_{k} : = span (e_{1}, \dots, e_{k})$ every Rayleigh quotient is at least $λ_{k}$ and on $W_{k} : = span (e_{k}, \dots, e_{n})$ every Rayleigh quotient is at most $λ_{k}$ ; equality holds at $e_{k}$ in both cases.

2.1L2step 1.1

Let $W \subseteq V$ have dimension $n - k + 1$ . If $W \cap U_{k} = {0}$ , then $W \subseteq U_{k}^{⊥} = span (e_{k + 1}, \dots, e_{n})$ , whose dimension is $n - k$ , contradicting [L2]. Thus some nonzero $v \in W \cap U_{k}$ satisfies $R_{T} (v) \geq λ_{k}$ by step 1.1, so $\max_{0 \neq v \in W} R_{T} (v) \geq λ_{k}$ . Since step 1.1 also gives $\max_{0 \neq v \in W_{k}} R_{T} (v) = λ_{k}$ , the first Courant-Fischer equality follows.

3.1L2step 1.1∎

Let $U \subseteq V$ have dimension $k$ . If $U \cap W_{k} = {0}$ , then $U \subseteq W_{k}^{⊥} = span (e_{1}, \dots, e_{k - 1})$ , whose dimension is $k - 1$ , contradicting [L2]. Thus some nonzero $v \in U \cap W_{k}$ satisfies $R_{T} (v) \leq λ_{k}$ by step 1.1, so $\min_{0 \neq v \in U} R_{T} (v) \leq λ_{k}$ . Since step 1.1 also gives $\min_{0 \neq v \in U_{k}} R_{T} (v) = λ_{k}$ , the second Courant-Fischer equality follows.

Depends on

Used by

Dependency tree · two levels

27 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

MIT 18.409, Lecture 3: Courant-Fischer and Rayleigh quotients (standard reference, not scraped)