Set W1=imϕ, choose a direct-sum complementU such that kerϕ=U⊕W1, and choose a complement W2 such that V=kerϕ⊕W2.
The restriction ϕ∣W2:W2→W1 is injective because W2∩kerϕ=0, and it is surjective because every image ϕ(v) equals ϕ(w2) after decomposing v=k+w2. It is therefore a isomorphism, so dimW2=dimW1 and ϕ(W2)=W1. Also ϕ(U)=ϕ(W1)=0. Hence