If Ricci tensor has a null eigenvector in dimension $3$, then it has at most two nonzero eigenvalues and $|\text{Ric}|^2 \geq \dfrac{R^2}{2}$

Question

Here's the necessary context:

I assume a null eigenvector would be a vector $v$ such that $\operatorname{Ric}(v, \cdot) = 0$ (i.e $\text{Ric}_{ij}v^{j} = 0$). But I don't understand why the claim "where the Ricci tensor has a null eigenvector, it has at most two nonzero eigenvalues" is so obvious - and why that implies the inequality in the title. I would really appreciate it if someone gave a detailed explanation of why this is true. Thanks in advance.

Also, it is relevant to point out that $R$ stands for the scalar curvature.

Answer 1

The Ricci tensor is a $3 \times 3$ matrix, so it has at most 3 eigenvalues. If it has a null eigenvector, then $0$ is an eigenvalue; so it has at most 2 non-zero eigenvalues. If we call these eigenvalues $a,b,$ then we have $$|\mathrm{Rc}|^2 = a^2 + b^2, \; R = \mathrm{tr}\,\mathrm{Rc}=a+b;$$ so applying the inequality $2ab \le a^2 + b^2$ we see $$R^2 = a^2 + 2ab + b^2 \le 2(a^2 + b^2) = 2|\mathrm{Rc}|^2.$$