Interesting Result (Question from One of My Finals)

Claim: There is no Kähler-Einstein metric on $\mathbb P^2$  blown up at two distinct points.
Proof: Without loss of generality, let $\widetilde{\widetilde{\mathbb{P}^2}}$ be the blow up of $\mathbb{P}^2$ at the points $\left[1,0,0\right]$ and $\left[0,1,0\right]$. Then$$\text{Aut}_{\mathcal{O}}\left(\widetilde{\widetilde{\mathbb{P}^2}}\right)=\left\{\begin{pmatrix}1&0&*\\0&*&*\\0&0&*\end{pmatrix}\in\text{GL}_3\left(\mathbb{C}\right)\right\}\ni\begin{pmatrix}1&0&*\\0&a&*\\0&0&b\end{pmatrix}.$$Observe that$$\begin{pmatrix}1&0&*\\0&a&*\\0&0&b\end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}\qquad\text{and}\qquad\begin{pmatrix}1&0&*\\0&a&*\\0&0&b\end{pmatrix}\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\a\\0\end{pmatrix}=\begin{pmatrix}0\\1\\0\end{pmatrix}.$$Let$$H=\left\{\begin{pmatrix}1&0&x\\0&1&0\\0&0&1\end{pmatrix}\in\text{GL}_3\left(\mathbb{C}\right)\right\}.$$Then$$\begin{pmatrix}1&0&a\\0&b&c\\0&0&d\end{pmatrix}\begin{pmatrix}1&0&*\\0&1&0\\0&0&1\end{pmatrix}=\begin{pmatrix}1&0&*+a\\0&b&c\\0&0&d\end{pmatrix}=\begin{pmatrix}1&0&a+*\\0&b&c\\0&0&d\end{pmatrix}=\begin{pmatrix}1&0&*\\0&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&a\\0&b&c\\0&0&d\end{pmatrix},$$which implies $H$ is an Abelian and thus normal subgroup of $\text{Aut}_{\mathcal{O}}\left(\widetilde{\widetilde{\mathbb{P}^2}}\right)$. Let $h\in H$. Then$$\left(h-I_3\right)^2=\begin{pmatrix}0&0&x\\0&0&0\\0&0&0\end{pmatrix}^2=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix},$$which implies $\left\{I_3\right\}$ is not the only unipotent radical. Therefore, $\text{Aut}_{\mathcal{O}}\left(\widetilde{\widetilde{\mathbb{P}^2}}\right)$ is not reductive, which, by Matsushima, implies that there is no Kähler-Einstein metric on $\widetilde{\widetilde{\mathbb{P}^2}}$. $\blacksquare$