= Solution
Let $\lambda$ be the single <eigenvalue>. By <diagonalizability>, there is a <basis> $v_1,\ldots,v_n$ of <eigenvectors>, and every corresponding <eigenvalue> is $\lambda$. For any $v=\sum_i a_iv_i$, <linearity> gives
$$
T(v)=\sum_i a_iT(v_i)=\sum_i a_i\lambda v_i=\lambda v.
$$
Thus $\boxed{T=\lambda I}$, a <scalar multiple> of the <identity map>.
A counterexample to the conclusion without <diagonalizability> is
$$
\boxed{J=\begin{pmatrix}2&1\\0&2\end{pmatrix}.}
$$
Its <characteristic polynomial> is $(t-2)^2$, but $(J-2I)(a,b)^\mathsf T=(b,0)^\mathsf T$. The <eigenspace> for $2$ is therefore $\operatorname{span}\{(1,0)^\mathsf T\}$, of <dimension> one. There is no <basis> of <eigenvectors> in <dimension> two, so $J$ is not <diagonalizable>.
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