Please note that this is not a duplicate and is related to the finite-dimensional case (right now I've just started learning the subject from Serge Lang's Linear Algebra). I just have a question specific to my approach, which is as follows:
Let $v_1 \in V$ be a non-zero vector such that $P(v_1) = v_2 \neq 0$. Assume that $P(v_2) = 0$. Then $P(v_1) = v_2 \implies P^2(v_1) = P(v_2) = 0$. But since $P^2 = P$, $P^2(v_1) = P(v_1) = v_2 \neq 0$, contradicting the assumption that a vector in the image of $P$ can also lie in $\ker P$.
So the kernel and image of $P$ are both disjoint subspaces of $V$. Suppose their bases are respectively $\{u_1, \ldots, u_m\}$ and $\{w_1, \ldots, w_n\}$. Then $\dim V = \dim \ker P + \dim \text{im } P = m+n$. All elements in the two bases are mutually independent, for otherwise one could express one of the two bases' elements in terms of the other basis, which would contradict the fact that the two subspaces are disjoint.
$\{u_1, \ldots, u_m, w_1, \ldots, w_n\}$ forms maximal independent subset, and hence a basis, of $V$. If $v \in V$, it can be written as
$v = x_1u_1 + \ldots + x_mu_m+ y_1w_1 + \ldots + y_nw_n = u + w,$
for some $u \in \ker P$ and $w \in \text{im } P$.
Is there a problem with the above reasoning? I'm asking this because the representation of $v$ in this way seems like a unique one (as a sum of vectors in kernel and image). So $V$ should be a direct sum of $\ker P$ and $\text{im } P$, which is a stronger result than the one required to be proved (which of course makes me skeptical of my approach or concept).
Best Answer
For each $v \in V$ we have
$$ \quad v=Pv+(I-P)v \in \text{im} P+ \ker P,$$
since $(I-P)v \in \ker P$.