Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 8

An orthogonal matrix satisfies $A^{-1}=A^T$, while a symmetric matrix satisfies $A^T=A$. Combining them gives $\boxed{A^{-1}=A}$. For the product with the antisymmetric matrix $B$, the rule for the transpose of a matrix product givesThus $AB$ is an antisymmetric matrix exactly when $AB=BA$. The hypotheses do not force this commutativity. For a counterexample, takeHere $A$ is a symmetric matrix and an orthogonal matrix, and $B$ is an antisymmetric matrix and an orthogonal matrix, but $AB$ is a nonzero symmetric matrix, so it is not an antisymmetric matrix. On the other hand, the product of two orthogonal matrices is always an orthogonal matrix:Therefore the answers are $\boxed{\text{(i) no; (ii) yes.}}$

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.