Question

Let $p$ and $q$ be real numbers such that $p^2 + q^2 = 1$. The eigenvalues of the matrix $\begin{bmatrix} p & q \\ q & -p \end{bmatrix}$ are

• A. 1 and 1
• B. 1 and -1
• C. $j$ and $-j$
• D. $pq$ and $-pq$

Hint:

For 2×2 symmetric matrices, the characteristic polynomial often simplifies using $p^2 + q^2$ identities.

Answer 1

The Correct Option is B

Consider the matrix \[ A = \begin{bmatrix} p & q \\ q & -p \end{bmatrix} \]
The characteristic equation is \[ \det(A - \lambda I) = \begin{vmatrix} p - \lambda & q \\ q & -p - \lambda \end{vmatrix} = (p - \lambda)(-p - \lambda) - q^2 \] \\ Simplify: \[ = -(p^2 - \lambda^2) - q^2 = \lambda^2 - (p^2 + q^2) \] \\ Given $p^2 + q^2 = 1$, so \[ \lambda^2 - 1 = 0 \] \\ Thus, \[ \lambda = \pm 1 \]
Therefore, the eigenvalues are 1 and −1.
Final Answer: 1 and -1 \\