= Solution
The cone condition implies $u(\lambda x)=\lambda u(x)$, so $Du(\lambda x)=Du(x)$. The <gradient> is homogeneous of degree zero, and $v=|Du|^2$ attains its global maximum on the compact unit sphere. For $n\geq2$ the punctured space is connected. Part (b) gives $\mathcal Lv\geq0$ for a locally <uniformly elliptic operator> with no zeroth-order term. The <strong maximum principle for elliptic operators> makes $v$ constant. Hence
$$
0=\mathcal Lv=2\sum_k a^{ij}u_{ki}u_{kj}.
$$
Positive definiteness gives $D^2u=0$. Thus $u=a\cdot x+b$, and degree-one homogeneity forces $b=0$: $\boxed{u(x)=a\cdot x}$. This proves that a <minimal graphical cone is a plane>. Strictly, the graph restricted to $x\ne0$ is the plane with its origin omitted; its closure is the full plane through the origin.
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