Ellipticity of the Laplacian and Positive-Definiteness In page 14, example 2.1 of these notes https://www.math.upenn.edu/~kazdan/japan/japan.pdf on geometric analysis we note that the symbol of a second order differential operator $$Pu = \sum_{i,j} a^{ij}(x) \frac{\partial ^2 u}{\partial x_i \partial x_j} + \sum_j b^j(x) \frac{\partial u}{\partial x_j} + c(x)u$$ 
is given by the $1 \times 1$ matrix $$\sigma_\xi(P;x) = -\sum_{i,j} a^{ij}(x) \xi_i \xi_j.$$ It then says that $P$ is elliptic at $x$ if and only if the matrix $(a^{ij}(x))$ is positive (or negative) definite. 
I have thought about this for a few hours and do not see where it comes from. Ellipticity means that the symbol is invertible. But invertibility is quite different from positive-definiteness. 
Honestly, I do not see why $\sigma_\xi$ is a $1 \times 1$ matrix either given that there  are covectors of different induces $\xi_i$ and $\xi_j$ that show up in the formula.
 A: Kazdan is defining differential operators acting on sections of vector bundles: note that he writes $$P : C^\infty(E) \to C^\infty(F),\;P(u)(x)=\sum_\alpha a^\alpha(x)\partial_\alpha u(x)$$ where let's say $E$ is a bundle of dimension $k$ and $F$ is a bundle of dimension $l.$ The coefficient $a^{\alpha}(x)$ (after fixing $\alpha$!) is in fact a $l \times k$ matrix, which is just the coordinate representation of a linear map $E_x \to F_x.$ Thus in the case of second-order operators $|\alpha|=2$, $a$ is really a matrix of matrices; i.e. something with components $a^{ij}{}^c_d(x)$ where $1\le i,j \le n$, $1\le c \le l$, $1 \le d \le k.$
Since you're working with an operator acting on scalar functions, you're in the special case where $E$ and $F$ are both the 1-dimensional trivial bundle $M \times \mathbb R$ over $M;$ so $a^{ij}(x)$ is a $1 \times 1$ matrix, or equivalently just a scalar.
As to the ellipticity, remember the definition: $P$ is elliptic if for every x and every non-zero $\xi$, the symbol $\sigma_\xi(P;x)\in \mathbb R^{l\times k}$ is an invertible matrix. In our case we have a $1\times 1$ matrix, which is invertible if and only if it is non-zero. Thus we need $\sigma_\xi(P;x)=-\sum a^{ij}(x)\xi_i \xi_j\ne0$ for every $\xi \ne 0$, which is equivalent to the $n \times n$ matrix $a^{ij}(x)$ being definite.
A: The symbol is a real number, which they think of as a 1x1 matrix. As a mapping it is invertible if the symbol is not zero for all $x$ and $\xi$. If $a^{ij}$ is not positive or negative definite, then it is possible to find nonzero $\xi$ for which the symbol is zero (i.e., not invertible). 
For example, if $a^{ij}$ is not positive or negative definite,  take $\xi_1$ and $\xi_2$ for which the symbol is positive and negative, respectively, and consider the line between $\xi_1$ and $\xi_2$. By the mean value property, somewhere on this line you will find a $\xi$ that makes the symbol zero. You can move $\xi_1$ and $\xi_2$ slightly, if necessary, to ensure $\xi\neq 0$ (by continuity). 
