 Solution

Call the given matrix $M$. Its middle coordinate decouples, immediately giving the eigenvalue $3$ with normalized eigenvector $(0,1,0{)}^T$. The remaining characteristic polynomial comes from the $xz$ block:For $\lambda =-4$, the first row of $(M+4I)v=0$ gives $x=\sqrt{3}z$; for $\lambda =8$, the first row of $(M-8I)v=0$ gives $z=-\sqrt{3}x$. After vector normalization, one possible complete set isEach has unit Euclidean norm. The dot products with $v_2$ vanish because the other two middle components are zero, and $v_1\cdot v_3=(-\sqrt{3}+\sqrt{3})/4=0$. This explicitly verifies orthonormality, also predicted by the spectral theorem for real symmetric matrices with distinct eigenvalues. Changing any eigenvector's sign gives an equally valid normalized choice.