Question:

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

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Eigenvectors corresponding to distinct eigenvalues are always linearly independent and hence cannot be scalar multiples of each other.
Updated On: Jun 12, 2025
  • \( \exists \) a constant \( K \) such that \( \overline{X}_1 = K \overline{X}_2 \), whenever \( \lambda_1 \ne \lambda_2 \)
  • \( \nexists \) constant \( K \) such that \( \overline{X}_1 = K \overline{X}_2 \), whenever \( \lambda_1 \ne \lambda_2 \)
  • \( \nexists \) constant \( K \) such that \( \overline{X}_1 = K \overline{X}_2 \), whenever \( \lambda_1 = \lambda_2 \)
  • \( \overline{X}_1 = \overline{X}_2 \) whenever \( \lambda_1 = \lambda_2 \)
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The Correct Option is B

Solution and Explanation

Eigenvectors corresponding to distinct eigenvalues are linearly independent. This means if λ₁ ≠ λ₂, then X₁ cannot be a scalar multiple of X₂.

Option 1 Analysis

∃ a constant K such that \(\overline{X_1} = \overline{KX_2}\), whenever \(\lambda_1 \neq \lambda_2\)

This is false. Eigenvectors for different eigenvalues cannot be scalar multiples of each other.

Option 2 Analysis

∃ no constant K such that \(\overline{X_1} = \overline{KX_2}\), whenever \(\lambda_1 \neq \lambda_2\)

This is true. This directly follows from the linear independence of eigenvectors for distinct eigenvalues.

Option 3 Analysis

∃ no constant K such that \(\overline{X_1} = \overline{KX_2}\), whenever \(\lambda_1 = \lambda_2\)

This is false. When eigenvalues are equal, eigenvectors may be scalar multiples (though not necessarily).

Option 4 Analysis

\(\overline{X_1} = \overline{X_2}\) whenever \(\lambda_1 = \lambda_2\)

This is false. Equal eigenvalues don't imply identical eigenvectors, just that they belong to the same eigenspace.

Final Answer

The correct option is: Option 2

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