Final answer:
To prove that the range of T1 is equivalent to the range of T2 if and only if there is an invertible operator S such that T1 = T2S, we involve concepts of linear transformations and invertible linear maps and demonstrate a one-to-one correspondence between vectors mapped by T1 and T2 for their shared range.
Step-by-step explanation:
The student is asking a question related to linear algebra, particularly about linear operators and their ranges in finite-dimensional vector spaces. To address the proof that the range of T1 is equal to the range of T2 if and only if there is an invertible operator S such that T1 = T2S, one must engage with concepts such as linear transformation, range of an operator, and invertible (bijective) linear maps.
If the range of T1 is equal to the range of T2, then for each vector w in the range of T1, there must be a vector v in V such that T1(v) = w. Since T2 is also mapping to the same range, there must also be a vector u in V such that T2(u) = w. The invertible operator S comes into play by ensuring that for each v, S(v) = u, i.e., S maps vectors in V to the vectors in V that T2 needs to reach every w in the range. S being invertible means that it creates a one-to-one correspondence between the vectors used by T1 and T2 to achieve their shared range.
For the other direction, if there exists an invertible operator S such that T1 = T2S, then every vector in the range of T1 can be written as T1(v) for some v in V, and because T1 = T2S, this is also T2(S(v)). Since S is invertible, every vector in V is the image of some vector under S, and thus the range of T2 covers the range of T1. Since T1 can be expressed as a composition of T2 and an invertible operator, it does not extend the range beyond that of T2, and thus the ranges of T1 and T2 are equal.