Show that any two distinct lines in $\Bbb P^2$ intersect in one point.
Proof(My attempt).
Let $L_1, L_2$ be any two distinct lines in $P^2$.
Write $L_i = V (a_iX + b_iY + c_iZ), i = 1,2$.
It suffices to show that $L_1 ∩L_2 = V(a_1X +b_1Y +c_1Z,a_2X +b_2Y +c_2Z)=V$ is a point.
Now I know that there is a projective change of coordinates $T$ such that $V^T =V(Z)$ or $=V(Y,Z)$
If $V^T =V(Y,Z)$ is an unique point.
Otherwise $V^T =V(Z)$,then can I say $L_1 =L_2$?? Why?
This is my problem. If this is true then I am done
Best Answer
Finding a point $[X_0:Y_0:Z_0]\in L_1\cap L_2$ amounts to finding (up to multiplication by a non-zero factor) a non-zero triplet $(X_0,Y_0,Z_0)$ solving the system $$\begin{cases}a_1X+b_1Y+c_1Z=0\\a_2X+b_2Y+c_2Z=0\end{cases}$$ or, which is the same, a non-zero vector $\begin{pmatrix}X_0\\Y_0\\Z_0\end{pmatrix}\in\ker\begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\end{pmatrix}$. But by rank-nullity theorem, such a vector exists, because for a $(2\times3)$ matrix $A$ it holds $\dim\ker A=3-\operatorname{rk}A\ge3-2=1$.
In order for each equation to identify a line in $\mathbb P^2$, both rows must be non-zero. In order for the two lines to be dinstinct, the two rows must be linearly independent. So actually $\dim \ker A=1$.
This means that, if $(X_0,Y_0,Z_0)$ and $(X_1,Y_1,Z_1)$ are both non-zero solutions of the system, then $(X_0,Y_0,Z_0)=(\lambda X_1,\lambda Y_1,\lambda Z_1)$, i.e. that $[X_0:Y_0:Z_0]=[X_1:Y_1:Z_1]$. This proves uniqueness.