In order to prove the uniqueness part, it suffices to show that ker f (A) ∩ ker g(A) = {0}. Assume that x lies in the intersection of ker f (A) and ker g(A), so that f (A) x = 0 and g(A) x = 0. This implies p(A) f (A) x = 0
and q(A) g(A) x = 0. Adding both equations, we obtain x = (p(A) f (A) + q(A) g(A)) x = 0. This shows that show that ker f (A) ∩ ker g(A) = {0}, as claimed.