Let p and q be real numbers such that p² + q² = 1. The eigenvalues of the matrix $\left[\begin{array}{cc}p& q\\ q& -p\end{array}\right]$ are

The correct option is 1 and -1.

To find the eigenvalues of the given 2 × 2 matrix, let us denote the matrix as $A$:

$$A=\left[\begin{array}{cc}p& q\\ q& -p\end{array}\right]$$

The eigenvalues $\lambda$ of matrix $A$ are the roots of the characteristic equation:
$\det (A-\lambda I)=0$

Substituting the values of $A$ into the characteristic equation, we get:

$$\det \left[\begin{array}{cc}p-\lambda & q\\ q& -p-\lambda \end{array}\right]=0$$

Expanding the determinant, we have:
$(p-\lambda )(-p-\lambda )-q^2=0$

Let us simplify the terms inside the product:
$-(p-\lambda )(p+\lambda )-q^2=0$
Using the difference of squares identity, $(p-\lambda )(p+\lambda )=p^2-{\lambda }^2$, we obtain:
$-(p^2-{\lambda }^2)-q^2=0$
$-p^2+{\lambda }^2-q^2=0$
Rearranging the terms:
${\lambda }^2-(p^2+q^2)=0$

We are given that $p^2+q^2=1$. Substituting this relationship into our equation yields:
${\lambda }^2-1=0$
${\lambda }^2=1$
Taking the square root of both sides gives the eigenvalues:
$\lambda =\pm 1$

Thus, the eigenvalues of the matrix are 1 and -1.