Question

Let \( i \) be an imaginary number such that \( i = \sqrt{-1} \). Let \( a \) and \( b \) be real numbers satisfying \( a^2 + b^2 = 1 \).
Then, the eigenvalues of the matrix \[ \begin{bmatrix} -a & b \\ b & a \end{bmatrix} \] are ...........

• A. 1 and 1
• B. \( i \) and \( i \)
• C. \( i \) and \( -i \)
• D. 1 and -1

Hint:

For symmetric matrices with real entries, eigenvalues are always real. Normalization conditions like \( a^2 + b^2 = 1 \) often simplify the characteristic equation.

Answer 1

The Correct Option is D

Step 1: Let the matrix be
\[ A = \begin{bmatrix} -a & b \\ b & a \end{bmatrix} \] Step 2: Find the characteristic equation
\[ \det(A - \lambda I) = 0 \Rightarrow \begin{vmatrix} -a - \lambda & b \\ b & a - \lambda \end{vmatrix} = 0 \] \[ (-a - \lambda)(a - \lambda) - b^2 = 0 \] \[ = -a^2 + a\lambda - a\lambda + \lambda^2 - b^2 = \lambda^2 - (a^2 + b^2) \] Step 3: Use the identity \( a^2 + b^2 = 1 \)
\[ \lambda^2 - 1 = 0 \Rightarrow \lambda = \pm 1 \] Step 4: Conclusion
So, the eigenvalues are \( \boxed{1 \text{ and } -1} \).