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This shows we have two vectors, say $v_1$ and $v_2$ in $ker(T-3I)$.
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This gives two vectors in $ker(T-3I)^2$ that are not in $ker(T-3I)$; call these $v_3$ and $v_4$.
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Since we are now up to dimension 5, which is the power of $(x-3)$ appearing in the characteristic polynomial, we have all the vectors.
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