Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 2 / e / Solution

True. Orthogonality gives $A_{ki}{A}_{kj}={\delta }_{ij}$ and $A_{ik}{A}_{jk}={\delta }_{ij}$, so $A^T=A^{-1}$. Thus $C_{ij}=A_{ki}{B}_{k\ell }{A}_{\ell j}$ is a similar matrix to $B$. If $B_{ij}{v}_j=\lambda v_i$, set $w_i=A_{ji}{v}_j$. ThenThe transformed eigenvector is nonzero because $A$ is invertible. Applying the inverse transformation gives the reverse correspondence, and similar matrices also have the same characteristic polynomial, including algebraic multiplicities of all eigenvalues.