Let $g$ be an idempotent endomorphism. Show that there exists a basis $B$ for $V$ such that for some $v_i\in B$: $g(v_i) = v_i$ for $1 \leq i \leq r$ and $g(v_i) = 0$ for $r+1\leq i \leq n$, where $r = \text{rank}(g)$ and $n = \text{dim}(V)$.
I know $g: V \to V$ as an idempotent endomorphism means $g(g(v)) = g(v)$, but not sure how to proceed with this problem.
The fact that $\text{rank}(g) = r$ means that the image of $g$ will be an $r$-dimensional vector subspace of $V$. Therefore, it can be spanned by $r$ linearly independent vectors of $V$. Let $v_1,\dots,v_r$ be such vectors. Then, as they lay in the image of $g$, there are $w_1,\dots,w_r$ in $V$ such that $v_i =g(w_i)$ for $i=1,\dots,r$. Thus
$g(v_i) = g(g(w_i)) = g(w_i) = v_i$ for $i=1,\dots,r$.
The dimension theorem also shows that the kernel of $g$ has to be $n-r$ dimensional, and therefore can be spanned by $n-r$ linearly independent vectors of $V$, say $v_{r+1},\dots,v_n$. Then by definition,
$g(v_i) = 0$ for $i=r+1,\dots,n$.
Since $\text{im} (g) \cap \text{ker}(g) = {0}$, the set $v_i$ for $i=1,\dots, n$ then also is linearly independent, and therefore form a basis of $V$ with the claimed properties.