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

• A. \( BA^{-1} \)
• B. \( Q^T A^{-1} B Q \)
• C. \( A^{-1} B Q^T \)
• D. \( AB \)

Hint:

Orthogonal transformations preserve eigenvalues, so for matrices of the form \( Q^T A^{-1} B Q \), the eigenvalues are the same as those of \( AB \).

Answer 1

The Correct Option is A

Step 1: Recall the property of orthogonal similarity.
If Q is an orthogonal matrix, then for any matrix M, the matrices M and Q^T M Q are similar.
Similar matrices have the same eigenvalues.

Step 2: Apply the property to the given matrix.
The matrix in question is Q^T A^{-1} B Q.
Let M = A^{-1} B. Then Q^T M Q is similar to M.
Thus, Q^T A^{-1} B Q has the same eigenvalues as A^{-1} B.

Step 3: Eigenvalues of A^{-1} B.
Recall that the eigenvalues of A^{-1} B are the same as those of B A^{-1} because if λ is an eigenvalue of A^{-1} B with eigenvector v:
A^{-1} B v = λ v → B v = λ A v → B A^{-1} (A v) = λ (A v).
Hence, the eigenvalues of A^{-1} B and B A^{-1} are the same.

Final Answer: BA^{-1}