Question:
Let \( Q, A, B \) be matrices of order \( n \times n \) with real entries such that \( Q \) is orthogonal and \( A \) is invertible. Then the eigenvalues of \( Q^T A^{-1} B Q \) are always the same as those of
Let \( Q, A, B \) be matrices of order \( n \times n \) with real entries such that \( Q \) is orthogonal and \( A \) is invertible. Then the eigenvalues of \( Q^T A^{-1} B Q \) are always the same as those of
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Orthogonal transformations preserve eigenvalues, and for matrices of the form \( Q^T A^{-1} B Q \), the eigenvalues are the same as those of \( BA^{-1} \).
Updated On: Dec 15, 2025
- \( AB \)
- \( Q^T A^{-1} B Q \)
- \( A^{-1} B Q^T \)
- \( BA^{-1} \)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the matrices.
Since \( Q \) is orthogonal, \( Q^T = Q^{-1} \). The transformation \( Q^T A^{-1} B Q \) preserves the eigenvalues of \( AB \) due to the property of orthogonal matrices. However, the eigenvalues of \( Q^T A^{-1} B Q \) will match those of \( BA^{-1} \) because of the way matrix multiplication and inversion work. Therefore, the eigenvalues of \( Q^T A^{-1} B Q \) are the same as those of \( BA^{-1} \).
Step 2: Conclusion.
Therefore, the correct answer is \( BA^{-1} \), option (D).
Since \( Q \) is orthogonal, \( Q^T = Q^{-1} \). The transformation \( Q^T A^{-1} B Q \) preserves the eigenvalues of \( AB \) due to the property of orthogonal matrices. However, the eigenvalues of \( Q^T A^{-1} B Q \) will match those of \( BA^{-1} \) because of the way matrix multiplication and inversion work. Therefore, the eigenvalues of \( Q^T A^{-1} B Q \) are the same as those of \( BA^{-1} \).
Step 2: Conclusion.
Therefore, the correct answer is \( BA^{-1} \), option (D).
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