Proving Linear Independence of Eigenvectors with Distinct Eigenvalues via Mathematical Induction

Eigenvectors and linear independence are fundamental concepts in linear algebra. In this article, we explore a method to prove that eigenvectors corresponding to distinct eigenvalues are linearly independent using mathematical induction.

Introduction to Mathematical Induction

Mathematical induction is a powerful proof technique used to establish that a statement is true for all natural numbers. It involves two main steps: the base case and the inductive step. We apply this technique to prove the linear independence of eigenvectors associated with distinct eigenvalues of a square matrix.

Theorem Statement

Given a square matrix A, if λ?, λ?, ..., λ? are distinct eigenvalues of A with corresponding eigenvectors v?, v?, ..., v?, then the set of eigenvectors {v?, v?, ..., v?} forms a linearly independent set.

Proof via Induction

Base Case (n 1)

For n 1: Consider a matrix A with a single distinct eigenvalue λ? and its corresponding eigenvector v?. The set {v?} is trivially linearly independent. A single vector is linearly independent by definition since it cannot be expressed as a non-trivial linear combination of other vectors.

Inductive Step

Assume the statement holds for n k, i.e., any set of k eigenvectors corresponding to k distinct eigenvalues of A is linearly independent. We aim to prove the statement for n k 1, meaning any set of k 1 eigenvectors corresponding to k 1 distinct eigenvalues is also linearly independent.

Assume for Contradiction

Suppose there exist k 1 distinct eigenvalues λ?, λ?, ..., λ_{k 1} and their corresponding eigenvectors v?, v?, ..., v_{k 1} that are linearly dependent. This implies the existence of scalars c?, c?, ..., c_{k 1}, not all zero, such that:

c?v?   c?v?   ...   c_{k 1}v_{k 1}  0

Isolating v_{k 1} in this equation gives us:

v_{k 1}  -frac{c?}{c_{k 1}}v? - frac{c?}{c_{k 1}}v? - ... - frac{c_k}{c_{k 1}}v_k

This implies that v_{k 1} can be expressed as a linear combination of v?, v?, ..., v_k. This contradicts our inductive hypothesis, which states that the set of eigenvectors {v?, v?, ..., v_k} is linearly independent for any k distinct eigenvalues.

Conclusion

Since assuming the linear dependence of v?, v?, ..., v_{k 1} leads to a contradiction, we conclude that the set of eigenvectors corresponding to k 1 distinct eigenvalues must be linearly independent.

Thus, by induction, we have shown that eigenvectors corresponding to different eigenvalues are linearly independent for any finite number of distinct eigenvalues.