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Lemma for the eigenspace

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All eigenvectors of the linear transformation A that correspond to the eigenvalue λ form a subspace L^(λ) in L.

Proof by Shilov (1969)

In fact, if Ax₁ = λx₁, and Ax₂ = λx₂, then

A(αx₁ + βx₂) = αAx₁ + βAx₂ = αλx₁ + βλx₂ = λ (αx₁ + βx₂)

with which the statement in the lemma is proven.

Lemma for linear independence of eigenvectors

Eigenvectors x₁, x₂, ... , x_n of the (linear) transformation A with respective pairwise distinct eigenvalues λ₁, λ₂, ... , λ_n, are linearly independent.

Proof by Shilov (1969)

This statement is proved by induction to number n. It is obvious that for n = 1 the lemma is true. Suppose that the lemma is true for all n – 1 eigenvalues of the transformation A; it remains to show that it is true for all n eigenvectors of the transformation A. Suppose a linear combination of n eigenvectors of the transformation A is 0:

α₁x₁ + α₂x₂ + ... + α_nx_n = 0.

Applying transformation A to this identity, one has

α₁λ₁x₁ + α₂λ₂x₂ + ... + α_nλ_nx_n = 0.

Multiply the first equation by λ_n and subtract from the second one; one obtains

α₁(λ₁ – λ_n)x₁ + α₂(λ₂ – λ_n)x₂ + ... + α_{n – 1}(λ_{n – 1} – λ_n)x_{n – 1} = 0,

from where by induction all coefficients must be zero. Distinct eigenvalues have nonzero difference, so each α_i = 0 for i < n; the first equation reduces to

α_nx_n = 0

which means α_n = 0, too. Consequently, all coefficients α_i are 0. Therefore, the vectors x₁, x₂, ..., x_n are linearly independent.