Question

Let \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) be the two eigenvectors corresponding to distinct eigenvalues of a \( 3 \times 3 \) real symmetric matrix. Which one of the following statements is true?

• A. \( \mathbf{v}_1^T \mathbf{v}_2 \neq 0 \)
• B. \( \mathbf{v}_1^T \mathbf{v}_2 = 0 \)
• C. \( \mathbf{v}_1 + \mathbf{v}_2 = 0 \)
• D. \( \mathbf{v}_1 - \mathbf{v}_2 = 0 \)

Hint:

For real symmetric matrices, {eigenvectors corresponding to distinct eigenvalues are always orthogonal.} This property is heavily used in spectral decomposition and principal component analysis.

Answer 1

The Correct Option is B

For a real symmetric matrix, a fundamental result from linear algebra is that eigenvectors corresponding to distinct eigenvalues are orthogonal. Let \( A \) be a real symmetric matrix, and suppose: \[ A \mathbf{v}_1 = \lambda_1 \mathbf{v}_1, \quad A \mathbf{v}_2 = \lambda_2 \mathbf{v}_2, \quad {with } \lambda_1 \neq \lambda_2 \] Then, \[ \lambda_1 (\mathbf{v}_1^T \mathbf{v}_2) = \mathbf{v}_1^T A \mathbf{v}_2 = (A \mathbf{v}_1)^T \mathbf{v}_2 = \lambda_2 (\mathbf{v}_1^T \mathbf{v}_2) \] Since \( \lambda_1 \neq \lambda_2 \), it must be that: \[ \mathbf{v}_1^T \mathbf{v}_2 = 0 \] Therefore, \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are orthogonal.