Final answer:
The proof shows that any vector in V can be expressed uniquely as the sum of a vector in Image(T) and a vector in ker(T), satisfying the requirements for a direct sum. This is done by demonstrating that every vector in V decomposes uniquely into parts from Image(T) and ker(T), and that the intersection of these subspaces is the zero vector.
Step-by-step explanation:
Proof of Direct Sum of Image and Kernel
For a linear map T:V\u2192V such that T^2 = T, we must show that V = Image(T) \u2295 ker(T). First, let's prove that every vector v in V can be written as the sum of two vectors, one in Image(T) and one in ker(T). Take any vector v in V. When we apply T to v, we get T(v), which is in Image(T). Now, let us consider the vector v - T(v), which will be in ker(T) because T(v - T(v)) = T(v) - T^2(v) = T(v) - T(v) = 0. Therefore, we can express v as T(v) + (v - T(v)), where T(v) is in Image(T) and (v - T(v)) is in ker(T).
Next, we must show that the intersection of Image(T) and ker(T) is {0}, which means they have no non-zero vectors in common. Suppose there is a non-zero vector u which is in both Image(T) and ker(T). Then, by definition of the kernel, T(u) = 0. However, because u is also in the Image(T), there exists some vector w such that T(w) = u. Applying T again, we have T(T(w)) = T^2(w) = T(w) = u, which implies u=0, a contradiction. Thus, Image(T) and ker(T) must only intersect at the zero vector.
Having established these two properties, we can conclude that V is the direct sum of Image(T) and ker(T), as required. This completes the proof.