Assume $P$ solves the Riccati equation
$${A^T}P + PA - PB{B^T}P + {C^T}C = 0$$
show that nullspace($P$) is invariant up to the application of the state matrix and show that $kernel(P) \subset C$
Attempt: I have no idea how to prove first part, but for second part: To show that kernel$(P) \subset $kernel$(C)$ multiply the algebraic Riccati equation by $x\in$kernel$(P)$ \begin{eqnarray} {A^T}Px + PAx - PB{B^T}Px + {C^T}Cx = 0 \end{eqnarray} and use the fact that $Px=0$ and $PAx=0$ which gives \begin{eqnarray} {C^T}Cx = 0 \end{eqnarray} and hence $x\in$kernel$(C)$
Assume that the Riccati equation is satisfied. Then, we have that
$$x^T\left({A^T}P + PA - PB{B^T}P + {C^T}C\right)x = 0$$
for all $x\in\mathbb{R}^n$. Now if we assume that $x\in\mathrm{ker}(P)$, then we get
$$x^TC^TCx = 0$$ for all $x\in\mathrm{ker}(P)$. Since $x^TC^TCx=||Cx||^2$, then $x^TC^TCx=0$ if and only if $Cx=0$. This implies that $x\in\mathrm{ker}(C)$ and therefore that $\mathrm{ker}(P)\subset\mathrm{ker}(C)$. This proves the inclusion.
For the invariance, we want to show that the satisfiability of the Riccati equation implies that for all $x\in\mathrm{ker}(P)$ we have that $PAx=0$, i.e. $Ax\in\mathrm{ker}(P)$. So, we have that
$$\left({A^T}P + PA - PB{B^T}P + {C^T}C\right)x=0$$ for all $x\in\mathrm{ker}(P)$. This yields
$$\left(PA + {C^T}C\right)x=0.$$
But we know that $\mathrm{ker}(P)\subset\mathrm{ker}(C)$, so we get that
$$PAx=0,$$
which shows implies that $\mathrm{ker}(P)\subset A\mathrm{ker}(P)$. This proves the result.