You need to show that the 3 properties of a subspace hold. I'll use u and v as arbitrary elements in S.
1. 0 is in {T(s), s in S}
0 is in S because S is a subspace, and T(0) = T(0 u) = 0T(u) [because T is a linear transformation] = 0
2. for any two elements in {T(s), s in S}, any linear combination of the two is in {T(s), s in S}.
Without loss of generality, you may let the two elements be T(u) and T(v). Then any linear combination of the two is of the form aT(u) + bT(v) for some a and b. By properties of a linear transformation, aT(u) + bT(v) = T(a u) + T(b v) = T(a u + b v). a u + bv is a linear combination of u and v, so it's in S. Thus, T(a u + b v) is in {T(s), s in S}.
3. for any element in {T(s) [call in T(u) for u in S], s in S} and c is a scalar, cT(u) is in {T(s), s in S}.
cT(u) = T(c u). c u is in S because u is in S, c is a scalar, and S is a vector space. Thus, cT(u) is in {T(s), s in S}.
no offense but it's mostly pretty straightforward stuff that follows from definitions.
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