Proof: TS = 0 when Range(S) is a Subset of Ker(T)

To prove that TS = 0, where S, T ∈ L(V) and Range S ⊂ Ker T, we will show that for any vector v in V, (TS)(v) = 0.

Let v be any vector in V. Since Range S ⊂ Ker T, we have S(v) ∈ Ker T. This implies that T(S(v)) = 0, as S(v) is in the kernel of T.

Now, consider (TS)(v). By the definition of function composition, we have (TS)(v) = T(S(v)).

Since we know that T(S(v)) = 0, we conclude that (TS)(v) = 0 for any vector v in V.

Thus, we have shown that TS = 0, as (TS)(v) = 0 for all v in V.