Let $U_1$, $U_2$ be vector subspaces of a vector space $V$. Let $J \subset \mathbb{P}(V)$ be the union of all lines in $\mathbb{P}(V)$ joining each $x \in \mathbb{P}(U_1)$ and $y \in \mathbb{P}(U_2)$. Show that $J$ is a linear subspace of $\mathbb{P}(V)$, and describe it in terms of $U_1$, $U_2$.
I am very new to Projective Geometry, so please enlighten me if you find any mistakes.
Let $v_x \in U_1$, $v_y \in U_2$ be arbitrary vectors corresponding to $x,y \in U_1, U_2$.
A line joining $x,y$ is a set of lines joining $\mu v_x, \lambda v_y$ for any nonzero scalar $\mu, \lambda$.
So, a point in $J$ has a corresponding line, and the line belongs to a plane spanned by $v_x,v_y$.
Let $a \in J$. Let $\text{span}(U_1,U_2) = \{ \text{union of }\text{span}(v_1,v_2) | v_1 \in U_1, v_2 \in U_2 \}.$
As mentioned, $a$ corresponds to a line and the line belongs to a plane spanned by some vectors in $U_1,U_2$.
So, the line is in $\text{span}(U_1,U_2)$, and thus $a \in \mathbb{P}(\text{span}(U_1,U_2))$. Therefore, $J \subset \mathbb{P}(\text{span}(U_1,U_2))$.
Let $a \in \mathbb{P}(\text{span}(U_1,U_2))$, then there is a vector $v_a$ corresponds to $a$. The line $\text{span}(v_a)$ is in $\text{span}(U_1,U_2)$ containing all lines joining all vectors in $U_1, U_2$. And, by the definition of $J$, $a \in J$. Thus, $J = \mathbb{P}(\text{span}(U_1,U_2))$.
By the fact that $U_1,U_2$ are subspaces of $V$, $J=\mathbb{P}(\text{span}(U_1,U_2))$ is a linear subspace of $\mathbb{P}(V)$.