TITLE: How to show the sum of the images of such $m$ projections is direct and is the whole space?
QUESTION [6 upvotes]: There are $m$ projections (whose square are themselves) $\phi_1,\cdots,\phi_m$ acting on a  finite-dimensional vector space $V$ such that
$$\phi_i\phi_j=0\quad i\ne j\tag{1}$$
where $0$ denotes the zero transformation, and that
$$\bigcap_{1}^m \text{Ker}\phi_i=0\tag{2}$$
where $0$ denotes the zero vector space. Prove that
$$\bigoplus_{1}^m \text{Im}\phi_i=V$$
I have successfully shown that the sum is direct, but I failed to show the sum is exactly $V$. Since I only used  argument (1) to show the directness, it appears that I must use the (2) to show the sum is indeed the whole space. But for me I can do little about that. 
One possible approach is try to show that $\phi_1+\cdots+\phi_m=\text{id}$ or $\text{id}-\phi_1=\phi_2+\cdots+\phi_m$, since  $\text{Im}\phi_i\oplus\text{Ker}\phi_i(=\text{Im}(\text{id}-\phi_i))=V$ and maybe we can get $\text{Ker}$ involved somehow  in this way. But I've made no progress either. 
Any help or hint will be appreciated, thanks.

REPLY [3 votes]: Take a hypothetical vector that is not in the direct sum of the images. Applying a projection yields a vector in the corresponding image.  Subtracting this from the original vector yields a nonzero element in that projection's kernel. We can do this for all of the projections at once to obtain a nonzero vector in the intersection of the kernels,  which is a contradiction. 
More succinctly, 
$$v-\phi_1(v)-\cdots-\phi_n (v) $$
is always in the intersection of all of the kernels,  so it is 0. Thus
$$v=\phi_1(v)+\cdots+\phi_n (v) $$
So the vector is visibly in the span of the images.  This is essentially an idea you've already had.