Value of the matrix of 2nd order partial derivatives of the quadratic form?

Question

Let $$A=\begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & -2 \\ 0 & 0 & 1 \end{pmatrix}$$ Define $$Q(x,y,z)=\begin{pmatrix} x & y & z \\ \end{pmatrix}A\begin{pmatrix} x \\ y \\ z \end{pmatrix}$$ for $x,y,z\in \mathbb R$. Then what is the value of the matrix of 2nd order partial derivatives of the quadratic form?

Answer 1

Consider \begin{align*} Q(x, y, z) &= \begin{pmatrix} x & y & z \end{pmatrix}\begin{pmatrix}1 & 2 & 0 \\ 0 & 0 & -2 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} \\ &= \begin{pmatrix} x & y & z \end{pmatrix}\begin{pmatrix} x + 2y \\ -2z \\ z \end{pmatrix} \\ &= x(x + 2y) + y(-2z) + z^2 \\ &= x^2 + z^2 + 2xy - 2yz. \end{align*} We can now differentiate these partially with respect to $x, y, z$: \begin{align*} \frac{\partial Q}{\partial x} &= 2x + 2y \\ \frac{\partial Q}{\partial y} &= 2x - 2z \\ \frac{\partial Q}{\partial z} &= 2z - 2y. \end{align*} Differentiating again, \begin{align*} \frac{\partial^2 Q}{\partial x^2} &= 2 \\ \frac{\partial^2 Q}{\partial y^2} &= 0 \\ \frac{\partial^2 Q}{\partial z^2} &= 2 \\ \frac{\partial^2 Q}{\partial x \partial y} = \frac{\partial^2 Q}{\partial y \partial x} &= 2 \\ \frac{\partial^2 Q}{\partial x \partial z} = \frac{\partial^2 Q}{\partial z \partial x} &= 0 \\ \frac{\partial^2 Q}{\partial y \partial z} = \frac{\partial^2 Q}{\partial z \partial y} &= -2. \end{align*} As a matrix, we get $$\begin{pmatrix} \frac{\partial^2 Q}{\partial x^2} & \frac{\partial^2 Q}{\partial x \partial y} & \frac{\partial^2 Q}{\partial x \partial z} \\ \frac{\partial^2 Q}{\partial y \partial x} & \frac{\partial^2 Q}{\partial y^2} & \frac{\partial^2 Q}{\partial y \partial z} \\ \frac{\partial^2 Q}{\partial z \partial x} & \frac{\partial^2 Q}{\partial z \partial y} & \frac{\partial^2 Q}{\partial z^2}\end{pmatrix} = \begin{pmatrix} 2 & 2 & 0 \\ 2 & 0 & -2 \\ 0 & -2 & 2 \end{pmatrix}.$$