Invariant point (flex) invariant under projective transformations. Let $C$ be a projective curve in $\mathbb{P}_2$ defined by a homogeneous polynomial $P(x, y, z)$ and let $\alpha$ be a linear transformation of $\mathbb{C}^3$. Let $Q$ be the homogeneous polynomial $Q = P \circ \alpha^{-1}$ which defines the image of $C$ under the projective transformation given by $\alpha$. Show that the matrix of second derivatives of $Q$ at a point of $\mathbb{P}_2$ represented by $v \in \mathbb{C}^3 - \{0\}$ is given by pre- and post-multiplying the matrix of second derivatives of $P$ at the point represented by $\alpha^{-1}(v)$ by the matrix of the linear transformation $\alpha^{-1}$ and its transpose, and hence that$$\mathcal{H}_P \circ \alpha^{-1} = (\text{det}\,\alpha)^2 \mathcal{H}_Q.$$Deduce that the definition of an inflection point is invariant under projective transformations.

The Hessian $\mathcal{H}_P$ of $P$ is the polynomial defined by$$\mathcal{H}_P(x, y, z) = \det\begin{pmatrix} P_{xx} & P_{xy} & P_{xz} \\ P_{yx} & P_{yy} & P_{yz} \\ P_{zx} & P_{zy} & P_{zz} \end{pmatrix}.$$

Write $P(\alpha(X)) = Q(X)$, with $X=(x_1,x_2,x_3)$, and use Taylor formula :
$$(1) \quad Q(X+H) = Q(X) + DQ(X).H + \underbrace{ {}^t H.D^2Q(X).H } + O(H^3).$$
where : $H=(h_1,h_2,h_3)$, $DQ(X).H = \sum_{i=1}^3 \tfrac{\partial Q}{\partial x_i}(X).h_i$, $D^2Q(X) = \mathcal{H}_Q(X)$ and $O(H^3)$ is some polynomial in $X$ and $H$ of total degree $\geq 3$ in $(h_1,h_2,h_3)$.
Putting $Y=\alpha(X)$ and $K=\alpha(H)$, we have $P(Y+K) = Q(X+H).$
But
$$(2) \quad P(Y+K) = P(\alpha(X)) + DP(\alpha(x)).\alpha(H) + \underbrace{ {}^t (\alpha(H)).D^2P(\alpha(X)).(\alpha(H)) }_{\text{order } 2} + O(H^3).$$
Identifying the homogenuous part of order $2$ in $H$ in eq. $(1)$ and $(2)$ we get
$$\mathcal{H}_Q(X) ={}^t\alpha.\mathcal{H}_P(\alpha(X)).\alpha.$$
Then take the determinant.
