= Same eigenvalues do not imply similarity
Consider the <diagonal matrix> $D=2I_2$ and a <Jordan block>
$$
J=\begin{pmatrix}2&1\\0&2\end{pmatrix}.
$$
Both have <characteristic polynomial> $(t-2)^2$, <trace> $4$, <determinant> $4$, and <eigenvalue> $2$ with <algebraic multiplicity> two. But $\ker(D-2I)$ is the whole plane, whereas $(J-2I)(a,b)^T=(b,0)^T$ vanishes only when $b=0$. Their <eigenspaces> have different <dimensions>, so <eigenvectors under similarity> rules out <matrix similarity>. There is an even shorter proof: $S^{-1}(2I)S=2I$ for every <invertible matrix> $S$.
The missing information is how the <linear operator> acts beyond its <eigenvectors>. Here $D-2I=0$, while $J-2I\ne0$ but $(J-2I)^2=0$. Their <minimal polynomials> are respectively $t-2$ and $(t-2)^2$. Over a <field> where the <characteristic polynomial> splits, the full sizes of the <Jordan blocks>, encoded by the <Jordan normal form>, determine the <matrix similarity> class.
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