A second order PDE can be expressed in the form $$ F(D^{2}u , Du ,u , x) = 0.$$ if $F$ is linear in $D^{2}u$ then we can express $$ F(D^{2}u , Du , u ,x) = L [u] + G(Du, u ,x),$$ with $$ L[u] := -\text{tr}(A(x)D^{2}u),$$ for which $A(x)$ is a symmetric $n\times n $ matrix.
Problem statement
Let $u:\mathbb{R}^{n} \to \mathbb{R}$. Show that $$ -\Delta u =0$$ is an elliptic PDE, using the fact that a PDE is elliptic if for every $x, A(x) = A_{ij}(x)$ has nonzero eigenvalues all of the same sign.
Attempt at a solution
From what I gather online, we say a PDE is elliptic if $B^{2} - 4AC < 0$ for $$A(x,y) \frac{\partial^{2}u}{\partial x^{2}} +B(x,y)\frac{\partial^{2}u}{\partial x \partial y} + C(x,y)\frac{\partial^{2}u}{\partial y^{2}} + f(x,y,u,u_{x} , u_{y}) = F(x,y).$$
In the given problem statement $u=u(\vec{x})$ with $\vec{x} = (x_{1} , \dots , x_{n})$, therefore, $$ -\Delta u = -\left(\frac{\partial^{2}u}{\partial x_{1}^{2}}+ \dots + \frac{\partial^{2}u}{\partial x_{n}^{2}} \right).$$ I note that this is precisely the $\text{tr}(D^{2}u)$.But from here I'm lost. I don't know where the $A(x)$ comes from nor what it exactly represents, more precisely what $A(x) = A_{ij}(x)$ means.
Any indications or directions would be much appreciated.
By $A(x) = A_{ij}(x)$, we mean that $A(x)$ is an $n\times n$ matrix, where the $ij$th element is the scalar-valued function $A_{ij}(x)$.
You've noted that $-\Delta u$ is precisely $\text{tr}(D^2u)$. Then in fact, we have that $-\Delta u = \text{tr}(ID^2u)$, where $I$ is the $n\times n$ identity matrix. Hence, we have that the matrix $A(x)$ is just the constant identity matrix. Indeed, $I$ is a symmetric matrix, with all eigenvalues equal to $+1$, and hence $-\Delta$ is elliptic.