Are eigenvectors corresponding to distinct eigenvalues always linearly independent?

• A. Not always; they can be dependent.
• B. Yes, always independent only if the matrix is symmetric.
• C. Yes, eigenvectors corresponding to distinct eigenvalues are always linearly independent.
• D. They are always orthogonal.

Answer: (C)

The key idea is that eigenvectors from different eigenvalues form a linearly independent set. Suppose you have eigenvectors v1, v2, ..., vm with corresponding eigenvalues λ1, λ2, ..., λm, all distinct, and a linear combination that sums to zero: c1 v1 + c2 v2 + ... + cm vm = 0. Apply the matrix product P(A) = ∏{j≠1} (A − λ_j I) to both sides. Each (A − λ_j I) acting on vi gives (λi − λj) vi. After applying the product, you get c1 ∏{j≠1} (λ1 − λj) v1 = 0, because for i ≠ 1 one of the factors becomes zero (since λi = λj for some j in the product when i ≠ 1). The scalar ∏_{j≠1} (λ1 − λj) is nonzero because all λs are distinct, and v1 ≠ 0, so c1 must be 0. Repeating this argument for the remaining vectors shows all coefficients are zero, so the vectors are linearly independent.

This result does not require symmetry of the matrix. Symmetry (or normality) would guarantee orthogonality of the eigenvectors, which is a stronger condition, but independence holds for eigenvectors from distinct eigenvalues regardless.