= Solution
If $B=S^{-1}AS$ with $S$ invertible, then $tI-B=S^{-1}(tI-A)S$. Multiplicativity of the <determinant> gives
$$
\chi_B(t)=\det(S^{-1})\det(tI-A)\det(S)=\chi_A(t).
$$
The roots of the <characteristic polynomial>, counted with their multiplicities, are the <eigenvalues> with their <algebraic multiplicities>. Thus \b[<similar matrices> have the same <eigenvalues> with the same <algebraic multiplicities>].
\b[The converse is false.] The matrices
$$
I_2\quad\text{and}\quad J=\begin{pmatrix}1&1\\0&1\end{pmatrix}
$$
both have <characteristic polynomial> $(t-1)^2$, but every matrix similar to $I_2$ is $S^{-1}I_2S=I_2$, whereas $J\ne I_2$. Alternatively, their <geometric multiplicities> for $1$ differ.
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