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The smallest and largest eigenvalues of a self-adjoint endomorphism are the minimum and maximum Rayleigh quotients

Statement

If $T$ is self-adjoint on a nonzero finite-dimensional real inner product space and its eigenvalues are ordered as

$λ_{1} \geq \dots \geq λ_{n},$

then

$λ_{1} = \max_{v \neq 0} R_{T} (v) and λ_{n} = \min_{v \neq 0} R_{T} (v) .$

Facts & Assumptions

Given: A self-adjoint endomorphism $T$ on a nonzero finite-dimensional real inner product space.

[L1]

Courant-Fischer gives $λ_{k} = \min_{\dim W = n - k + 1} \max_{0 \neq v \in W} R_{T} (v) = \max_{\dim U = k} \min_{0 \neq v \in U} R_{T} (v)$ (Courant-Fischer min-max principle for self-adjoint endomorphisms on finite-dimensional real inner product spaces).

Proof

technique · direct

1.1L1

Taking $k = 1$ in [L1] gives $λ_{1} = \max_{\dim U = 1} \min_{0 \neq v \in U} R_{T} (v) = \max_{v \neq 0} R_{T} (v)$ , because every one-dimensional subspace consists of the nonzero scalar multiples of any one of its nonzero vectors.

2.1L1∎

Taking $k = n$ in [L1] gives $λ_{n} = \min_{\dim W = 1} \max_{0 \neq v \in W} R_{T} (v) = \min_{v \neq 0} R_{T} (v)$ for the same reason.

Depends on

Used by

Dependency tree · two levels

6 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)