Question

If \(P\) is an orthogonal matrix of order \(3\times3\), \(A=\begin{bmatrix}1& 2& 2\\2& 1& 2\\2& 2& 1\end{bmatrix}\) and the eigenvalues of \(P^{T}AP\) are \(\alpha,\beta,\gamma\), then \(\alpha^2+\beta^2+\gamma^2=\)

• A. 3
• B. 27
• C. 15
• D. 12

Hint:

If \(A\) has eigenvalues \(\lambda_1,\lambda_2,\ldots,\lambda_n\), then \[ \boxed{\lambda_1^2+\lambda_2^2+\cdots+\lambda_n^2=\operatorname{tr}(A^2).} \] Also, orthogonally similar matrices have identical eigenvalues.

Answer 1

The Correct Option is B

Step 1: Use the property of orthogonal similarity. Since \(P\) is an orthogonal matrix, \[ P^{T}AP \] is orthogonally similar to \(A\). Hence, \[ A \text{ and } P^{T}AP \] have the same eigenvalues. Therefore, \[ \alpha^2+\beta^2+\gamma^2 =\operatorname{tr}(A^2). \]

Step 2: Compute \(A^2\). Given, \[ A= \begin{bmatrix} 1& 2& 2\\ 2& 1& 2\\ 2& 2& 1 \end{bmatrix} \] Multiplying, \[ A^2= \begin{bmatrix} 9& 8& 8\\ 8& 9& 8\\ 8& 8& 9 \end{bmatrix} \] Hence, \[ \operatorname{tr}(A^2)=9+9+9=27. \] Therefore, \[ \alpha^2+\beta^2+\gamma^2=27. \] Thus, \[ \boxed{27} \] is the correct answer. Hence, \[ \boxed{(B)} \] is the correct answer.