For an eigenvalue $\lambda$, $1\le\dim\ker(A-\lambda I)\le m_\lambda$, where $m_\lambda$ is its algebraic multiplicity. Equality for one eigenvalue alone does not imply the whole matrix is diagonalizable. Diagonalizability over the chosen field requires the characteristic polynomial to split and equality for every eigenvalue.
For example, $\operatorname{diag}(2,J_2(1))$ has equality at eigenvalue $2$ but is not diagonalizable because of the Jordan block at $1$.
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