Question

If \( \overline{X}_1 \) and \( \overline{X}_2 \) are any two eigenvectors corresponding to eigenvalues \( \lambda_1 \) and \( \lambda_2 \) respectively, then

• A. \( \exists \) a constant \( K \) such that \( \overline{X}_1 = K \overline{X}_2 \), whenever \( \lambda_1 \ne \lambda_2 \)
• B. \( \nexists \) constant \( K \) such that \( \overline{X}_1 = K \overline{X}_2 \), whenever \( \lambda_1 \ne \lambda_2 \)
• C. \( \nexists \) constant \( K \) such that \( \overline{X}_1 = K \overline{X}_2 \), whenever \( \lambda_1 = \lambda_2 \)
• D. \( \overline{X}_1 = \overline{X}_2 \) whenever \( \lambda_1 = \lambda_2 \)

Hint:

Eigenvectors corresponding to distinct eigenvalues are always linearly independent and hence cannot be scalar multiples of each other.

Answer 1

The Correct Option is B

Eigenvectors corresponding to distinct eigenvalues are linearly independent. That means if \( \lambda_1 \ne \lambda_2 \), the corresponding eigenvectors \( \overline{X}_1 \) and \( \overline{X}_2 \) cannot be scalar multiples of each other.
So, there does not exist any constant \( K \) such that \( \overline{X}_1 = K \overline{X}_2 \) when \( \lambda_1 \ne \lambda_2 \).
This is a fundamental property in linear algebra for eigenvectors of a matrix.