Let V and W be vector spaces over a field and let T:V->W be a linear transformation then V/kerT =R(T)

(A) True
(B) False
(C) Indifference
(D) None of the above

(A) (B) (C) (D)

The correct answer is (A) True.

To understand why this is true, let's break down the statement:

V/kerT denotes the quotient space of V by the kernel of T. This means that we divide V into equivalence classes, where two vectors v1 and v2 are considered equivalent if and only if v1 - v2 belongs to the kernel of T.

R(T) denotes the range (or image) of the linear transformation T, which is the set of all vectors in W that can be obtained by applying T to vectors in V.

The statement "V/kerT = R(T)" means that the quotient space V/kerT is equal to the range of T.

To prove this, we need to show two things:
1. Every vector in R(T) is represented by at least one equivalence class in V/kerT.
2. Two equivalence classes in V/kerT represent the same vector if and only if they belong to the same equivalence class.

Let's start with the first point. To show that every vector in R(T) is represented by at least one equivalence class in V/kerT, we can take any vector w in R(T) and find a vector v in V such that T(v) = w. Since T is a linear transformation, we know that v and v + k, where k belongs to the kernel of T, will produce the same result when applied T. Therefore, both v and v + k belong to the same equivalence class in V/kerT, representing the vector w in R(T).

Now, let's prove the second point. Suppose we have two vectors v1 and v2 in V that belong to different equivalence classes in V/kerT. This means that v1 - v2 does not belong to the kernel of T. If we apply T to both v1 and v2, we obtain T(v1) and T(v2) in R(T). Since v1 - v2 is not in the kernel of T, T(v1 - v2) is not the zero vector.

Now, let's consider the vector T(v1) - T(v2). Since T is linear, we have T(v1 - v2) = T(v1) - T(v2). Since T(v1 - v2) is not the zero vector, T(v1) - T(v2) is not the zero vector as well. This implies that T(v1) and T(v2) are different vectors in R(T), and therefore, v1 and v2 represent different equivalence classes in V/kerT.

This proves that two equivalence classes in V/kerT represent the same vector if and only if they belong to the same equivalence class.

Therefore, we have shown that every vector in R(T) is represented by at least one equivalence class in V/kerT, and two equivalence classes represent the same vector if and only if they belong to the same equivalence class. Hence, V/kerT = R(T).

Thus, the answer is (A) True.