Famous Theorems of Mathematics/Algebra/Linear Transformations

      Lemma for the eigenspace

      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

      Ax1 + 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, ... , xn of the (linear) transformation A with respective pairwise distinct eigenvalues λ1, λ2, ... , λn, 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 the opposite, that there is a linear dependence between n eigenvectors of the transformation A:

      α1x1 + α2x2 + ... + αnxn = 0,

      where, i. e., α1 ≠ 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

      α11 – λn)x1 + α22 – λn)x2 + ... + αn – 1n – 1 – λn)xn – 1 = 0,

      from where by induction all coefficients must be zero. In particular, α11 – λn)x1 = 0, which contradicts the conditions α1 ≠ 0, λ1 ≠ λn. Therefore, our supposition is false and the vectors x1, x2, ..., xn are linearly independent.

      Last modified on 17 April 2010, at 01:24