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Lagrange polynomials give the three eigenspace projections of a diagonalisable endomorphism

Example

Let $T$ be diagonalisable with distinct eigenvalues $λ_{0}, λ_{1}, λ_{2}$ . Put

$e_{i} (x) = \prod_{j \neq i} \frac{x - λ_{j}}{λ_{i} - λ_{j}} .$

Then $E_{i} = e_{i} (T)$ is projection onto $E_{λ_{i}} (T)$ along the other eigenspaces; the $E_{i}$ are pairwise orthogonal idempotents and $E_{0} + E_{1} + E_{2} = I$ .

Facts & Assumptions

Given: A diagonalisable $T$ with the three displayed distinct eigenvalues.

[L1]

Distinct eigenspaces give a direct sum of the whole space (An endomorphism is diagonalisable exactly when $V = ⨁_{i < r} E_{λ_{i}} (T)$ for some finite list of distinct scalars $λ_{i}$ ).

[L2]

Primary projections are polynomial expressions in $T$ obtained from the relevant congruences (Each projection in the primary decomposition is a polynomial in the endomorphism).

Verification

technique · direct

1.1L1L2algebra

The denominators are nonzero, and substitution gives $e_{i} (λ_{j}) = 1$ for $i = j$ and $0$ otherwise. Hence $e_{i} (T)$ is identity on $E_{λ_{i}} (T)$ and zero on the other eigenspaces.

2.1step 1.1L1∎

The action on every summand in [L1] now gives $E_{i}^{2} = E_{i}$ , $E_{i} E_{j} = 0$ for $i \neq j$ , and $\sum_{i} E_{i} = I$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

Keith Conrad, The Minimal Polynomial and Some Applications, Remark 4.12 (standard reference, not scraped)