Question

Let Q∈Rⁿ×ⁿ

be symmetric. Show that any two eigenvectors of Q, associated with two distinct eigenvalues, are Q-conjugate.

Answer 1

For a symmetric matrix Q in ℝ^n×n, any two eigenvectors associated with distinct eigenvalues are Q-conjugate.

Step-by-step explanation:

Let Q be a symmetric matrix in ℝ^n×n and v₁, v₂ be eigenvectors of Q corresponding to distinct eigenvalues λ₁ and λ₂ respectively, i.e., Qv₁ = λ₁v₁ and Qv₂ = λ₂v₂.

To show that v₁ and v₂ are Q-conjugate, consider the inner product of these vectors: v₁^T * v₂, where ^T denotes the transpose. Since Q is symmetric, Q is equal to its transpose: Q^T = Q. Therefore, we can express the eigenvalue equations as v₁^T * Q * v₂ = λ₁ * v₁^T * v₂ and v₁^T * Q * v₂ = λ₂ * v₁^T * v₂.

Subtracting these equations results in (λ₁ - λ₂) * v₁^T * v₂ = 0. Given that λ₁ and λ₂ are distinct eigenvalues, (λ₁ - λ₂) ≠ 0. Hence, v₁^T * v₂ = 0, implying that v₁ and v₂ are orthogonal or Q-conjugate.

Additionally, it's essential to note that if Q is symmetric, it possesses a complete set of eigenvectors that are orthogonal to each other. Therefore, any two eigenvectors associated with different eigenvalues are orthogonal, fulfilling the condition of being Q-conjugate.

This property holds true for symmetric matrices, ensuring that eigenvectors corresponding to distinct eigenvalues are Q-conjugate, guaranteeing orthogonality in the vector space.