= Solution
Let the repeated <eigenvalue> be $a$, and choose an <eigenvector> $v_1\ne0$. 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 $T=\begin{pmatrix}a&c\\0&d\end{pmatrix}$. 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 $c\ne0$, rescale the first basis vector by $k=c$ and apply part (a)(iii); the upper-right entry becomes one. Thus
$$
\boxed{M\sim aI\quad\text{or}\quad M\sim\begin{pmatrix}a&1\\0&a\end{pmatrix}.}
$$
The 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.
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