## Lemma for the eigenspaceEdit

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

### Proof by Shilov (1969)Edit

In fact, if *Ax*_{1} = λ*x*_{1}, and *Ax*_{2} = λ*x*_{2}, then

*A*(α*x*_{1}+ β*x*_{2}) = α*Ax*_{1}+ β*Ax*_{2}= αλ*x*_{1}+ βλ*x*_{2}= λ (α*x*_{1}+ β*x*_{2})

with which the statement in the lemma is proven.

## Lemma for linear independence of eigenvectorsEdit

Eigenvectors *x*_{1}, *x*_{2}, ... , *x*_{n} of the (linear) transformation *A* with respective pairwise distinct eigenvalues λ_{1}, λ_{2}, ... , λ_{n}, are linearly independent.

### Proof by Shilov (1969)Edit

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:

- α
_{1}*x*_{1}+ α_{2}*x*_{2}+ ... + α_{n}*x*_{n}= 0.

Applying transformation *A* to this identity, one has

- α
_{1}λ_{1}*x*_{1}+ α_{2}λ_{2}*x*_{2}+ ... + α_{n}λ_{n}*x*_{n}= 0.

Multiply the first equation by λ_{n} and subtract from the second one; one obtains

- α
_{1}(λ_{1}– λ_{n})*x*_{1}+ α_{2}(λ_{2}– λ_{n})*x*_{2}+ ... + 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
_{n}*x*_{n}= 0

which means α_{n} = 0, too. Consequently, all coefficients α_{i} are 0. Therefore, the vectors *x*_{1}, *x*_{2}, ..., *x _{n}* are linearly independent.