Solution
= Solution
Adding and subtracting the two coordinates shows that $A(1,1)^T=(c+s)(1,1)^T$ and $A(1,-1)^T=(c-s)(1,-1)^T$. The normalized <eigenvectors> can therefore be chosen as
$$
\boxed{v_+=\frac1{\sqrt2}(1,1)^T,\qquad v_-=\frac1{\sqrt2}(1,-1)^T.}
$$
Their <inner product> is $(1-1)/2=0$, and their squared <norms> are one, proving <orthonormality>. If $s=0$, every nonzero <vector> is an <eigenvector>; this chosen <orthonormal basis> still works.