Problem. If a projective transformation$\varphi: \mathbb{RP^2}\to \mathbb{RP}^2$ satisfies$\varphi\circ\varphi = \text{id}$, it is called a projective involution. If a point $P$ satisfies $\varphi(P)=P$, it is called a fixed point. Proof: For any projective involution $\varphi$, there always exists a line $l$ and a point $P$ not on $l$ such that both $P$ and every point on $l$ are fixed points of $\varphi$.
My Approach. I think this problem is more like a linear algebra problem. For the projective involution $\varphi = \bar{\Phi}_{A}$, assuming without loss of generality that $A^2=I_3$. If $A = \pm I_3$, , the conclusion is trivial. Otherwise, from $(A-I_3)(A+I_3)=O$, and Sylvester inequality, we have $$3=3+r(A^2-I_3)\geq r(A-I_3)+r(A+I_3)\geq 2,$$Therefore, without loss of generality, we can assume that $r(A-I_3)=1$. Therefore, the dimension of the solution space of $(A-I_3)\alpha =O$ is $2$. Suppose that $\{(x,y,z)':ax+by+cz=0\}$ is the solution set. $a^2+b^2+c^2>0$. Then $l:\{(x:y:z):ax+by+cz=0\}$ satisfies the given condition.
My Problem. I cannot find out the point $P$ which is not on $l$. By considering the special case of the line at infinity $l_{\infty}$ and $P(0:0:1)$, I conjecture that another point could possibly be $(a:b:c)$. However, I encountered some issues while trying to prove it. Fix $$A=\left(\begin{matrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ \end{matrix}\right).$$ For any triple $(x,y,z)$ that satisfies $ax+by+cz=0$, we have $$x=a_{11}x+a_{12}y+a_{13}z,$$ $$y=a_{21}x+a_{22}y+a_{23}z,$$ $$z=a_{31}x+a_{32}y+a_{33}z.$$ Hence $b_{11}-1:b_{12}:b_{13}=b_{21}:b_{22}-1:b_{23}=b_{31}:b_{32}:b_{33}-1=a:b:c$. So let$$B-I_3=\left(\begin{matrix} \lambda_1\\ \lambda_2\\ \lambda_3 \end{matrix}\right)\left(\begin{matrix}a & b & c\end{matrix}\right):=C.$$ Then$$B\left(\begin{matrix} a\\ b\\ c \end{matrix}\right)=\left(\begin{matrix} a+\lambda_1(a^2+b^2+c^2)\\ b+\lambda_2(a^2+b^2+c^2)\\ c+\lambda_3(a^2+b^2+c^2) \end{matrix}\right).$$ $(a:b:c)$ is a fixed point $\iff a:b:c=\lambda_1:\lambda_2:\lambda_3.$ $B^2=(C+I_3)^2=I_3$, so $C^2+2C=O$. Thus, $$\left(\begin{matrix} \lambda_1\\ \lambda_2\\ \lambda_3 \end{matrix}\right)(a\lambda_1+b\lambda_2+c\lambda_3)\left(\begin{matrix}a & b & c\end{matrix}\right)+2\left(\begin{matrix} \lambda_1\\ \lambda_2\\ \lambda_3 \end{matrix}\right)\left(\begin{matrix}a & b & c\end{matrix}\right)=(a\lambda_1+b\lambda_2+c\lambda_3+2)(A-I_3)=O.$$Then $a\lambda_1+b\lambda_2+c\lambda_3+2=0$. Then I haven't made any progress. Maybe the claim that $(a:b:c)$ is a fixed point is wrong. Can anyone help me?