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Let the repeated eigenvalue be $a$, and choose an eigenvector $v_1\ne 0$. Complete it to a basis $(v_1,v_2)$ of ${\mathbb{C}}^2$. In this basis the matrix is an upper triangular matrix, with first column $(a,0{)}^T$, say $T=\left(\begin{array}{cc}a& c\\ 0& d\end{array}\right)$. Its characteristic polynomial is $(t-a)(t-d)$. By part (a)(ii), this must equal $(t-a{)}^2$, so $d=a$.

If $c=0$, the matrix is already $aI$. If $c\ne 0$, rescale the first basis vector by $k=c$ and apply part (a)(iii); the upper-right entry becomes one. ThusThe second possibility is a size-two Jordan block. It has only one independent eigenvector, whereas $aI$ has a two-dimensional eigenspace, so the two types cannot be related by matrix similarity. This repeated-eigenvalue classification in dimension two proves the required result without assuming the general Jordan normal form theorem.