Final answer:
To prove that the ranges of two linear transformations T1 and T2 are equal if and only if there exists an invertible linear operator S such that T1 = T2S, one can show that S can be constructed by mapping basis vectors of V under T1 to T2 and then extending S linearly, ensuring that S is injective and thus invertible.
Step-by-step explanation:
The question asks to prove that the range of two linear transformations T1 and T2 from a finite-dimensional vector space V to another vector space W is equal if and only if there exists an invertible linear operator S such that T1 = T2S. First, let's assume that T1 and T2 have the same range. Since V is finite-dimensional, the range of T1 and T2 also has finite dimension. Consider a basis of the range of T1 (which is also the basis for range of T2). Each vector in this basis is an image of some vector in V under T1 and under T2. Define a linear operator S on the basis vectors in V such that T1(v) = T2S(v) for each basis vector v. Extend S linearly to the whole space V. S is invertible if it is injective, which is guaranteed if the dimension of the range is the same as the dimension of V; if that's not the case, then we use the property of linear independence of the image under T1 and T2 to show that S must be an injection and thus invertible. Conversely, if there exists an invertible operator S such that T1 = T2S, the equality T1(v) = T2S(v) ensures that the ranges of T1 and T2 coincide.