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First, we will show that the Image of a linear transformation F is a subspace of U. In order to prove this, we must show that the Image of F satisfies the subspace test. That is: 1. The zero vector of U is in the Image of F. 2. The Image of F is closed under vector addition. 3. The Image of F is closed under scalar multiplication. Let's check each condition: 1. The zero vector of U is in the Image of F: Since F is a linear transformation, it must satisfy F(0_V) = 0_U, where 0_V is the zero vector in V and 0_U is the zero vector in U. Thus, the zero vector in U is in the Image of F. 2. The Image of F is closed under vector addition: Let u1 and u2 be two arbitrary vectors in the Image of F. Then, there exist v1 and v2 in V such that F(v1) = u1 and F(v2) = u2. Now, we need to show that the sum u1 + u2 is still in the Image of F. Since F is a linear transformation, we have F(v1 + v2) = F(v1) + F(v2) = u1 + u2. Thus, the sum u1 + u2 is also in the Image of F. 3. The Image of F is closed under scalar multiplication: Let u be an arbitrary vector in the Image of F, and let c be an arbitrary scalar. Then, there exists a vector v in V such that F(v) = u. We must show that cu is also in the Image of F. Since F is a linear transformation, we have F(cv) = cF(v) = cu. Thus, the product cu is also in the Image of F. Since all three conditions of the subspace test are satisfied, the Image of F is a subspace of U.

Now, we will show that the Kernel of a linear transformation F is a subspace of V. In order to prove this, we must show that the Kernel of F satisfies the subspace test. That is: 1. The zero vector of V is in the Kernel of F. 2. The Kernel of F is closed under vector addition. 3. The Kernel of F is closed under scalar multiplication. Let's check each condition: 1. The zero vector of V is in the Kernel of F: Since F is a linear transformation, it must satisfy F(0_V) = 0_U. By definition of the Kernel, the zero vector 0_V is in the Kernel of F. 2. The Kernel of F is closed under vector addition: Let v1 and v2 be two arbitrary vectors in the Kernel of F. Then, F(v1) = 0_U and F(v2) = 0_U. We must show that the sum v1 + v2 is still in the Kernel of F. Since F is a linear transformation, we have F(v1 + v2) = F(v1) + F(v2) = 0_U + 0_U = 0_U. Thus, the sum v1 + v2 is also in the Kernel of F. 3. The Kernel of F is closed under scalar multiplication: Let v be an arbitrary vector in the Kernel of F, and let c be an arbitrary scalar. Then, F(v) = 0_U. We must show that cv is also in the Kernel of F. Since F is a linear transformation, we have F(cv) = cF(v) = c0_U = 0_U. Thus, the product cv is also in the Kernel of F. Since all three conditions of the subspace test are satisfied, the Kernel of F is a subspace of V. In conclusion, we have proved both parts (a) and (b) of Theorem 5.3. The Image of a linear transformation is a subspace of the codomain, and the Kernel of a linear transformation is a subspace of the domain.