= Solution
\b[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 $w_i=A_{ji}v_j$. Then
$$
C_{ij}w_j=A_{ki}B_{k\ell}A_{\ell j}A_{mj}v_m=A_{ki}B_{km}v_m=\lambda A_{ki}v_k=\lambda w_i.
$$
The 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>.
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