Famous Theorems of Mathematics/Algebra/Linear Transformations
Lemma for the eigenspace
[edit | edit source]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 | edit source]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
[edit | edit source]Eigenvectors x1, x2, ... , xn of the (linear) transformation A with respective pairwise distinct eigenvalues λ1, λ2, ... , λn, are linearly independent.
Proof by Shilov (1969)
[edit | edit source]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 + ... + αnxn = 0.
Applying transformation A to this identity, one has
- α1λ1x1 + α2λ2x2 + ... + αnλnxn = 0.
Multiply the first equation by λn and subtract from the second one; one obtains
- α1(λ1 – λn)x1 + α2(λ2 – λn)x2 + ... + αn – 1(λn – 1 – λn)xn – 1 = 0,
from where by induction all coefficients must be zero. Distinct eigenvalues have nonzero difference, so each αi = 0 for i < n; the first equation reduces to
- αnxn = 0
which means αn = 0, too. Consequently, all coefficients αi are 0. Therefore, the vectors x1, x2, ..., xn are linearly independent.