Why is this sum of differential operators not elliptic if the highest order of the sum is less than the dimension?

Question

Let $P_{1},...,P_{k}$ be first order differential operators on $\mathbb{R}^{n}$. I want to show that $P_{1}^{2}+\cdots+P^{2}_{k}$ is never elliptic if $k

We have

$$ P_{1}^{2}+\cdots+P^{2}_{k}=\sum_{i=1}^{k}P^{2}_{i}=\sum_{i=1}^{k}\sum_{|\alpha|\le 1}a_{\alpha}\left(\frac{\partial^{2\alpha}}{\partial x_{i}^{2\alpha}}\right). $$ Hence $$ \begin{aligned} \mathcal{F}\left(\sum_{i=1}^{k}P^{2}_{i}u\right)(\xi)&=\sum_{n=1}^{k}\sum_{|\alpha|\le 1}a_{\alpha}\xi_{i}^{2\alpha}\hat{u}(\xi). \end{aligned} $$ Thus $\sum_{i=1}^{k}P_{i}$ is elliptic if $$ p_{1}(\xi)=\sum_{i=1}^{k}\sum_{|\alpha|=1}a_{\alpha}\xi_{i}^{2\alpha} $$ has no real zeroes except $\xi=0$. I don't really see it though.

Answer 1

The symbol of a first order differential operator is a linear form; its zero set is a hyperplane of codimension $1$. Squaring does not change the zero set. For $k0$.

(If it helps, one can translate all statements about codimension into statements about the dimension of orthogonal complement.)