solution
we have f(λ)=mA(λ)g(λ)+r(λ)(1)
deg(r(λ))<deg(mA(λ))(2)
+Supposse that r(λ)=0 From (1). we have
⟹f(A)=mA(A)q(A)+r(A)
Since. f(A)=0 and mA(A)=0
⟹r(A)=0, with (2), contradiction to the definition of minimal polynomial. Then, r(λ)=0
Answer
Therefore f(λ) is divisible by mA(λ).