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.
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.