Final answer:
Col(A) is a subspace of R⁵ as it satisfies all subspace properties. Nul(A) is also a subspace but is limited to the zero subspace of R³ since A is of rank 3. The Rank-Nullity Theorem provides a relationship between the dimensions of Col(A) and Nul(A), with Col(A) having a dimension of 3 while Nul(A) has a dimension of 0.
Step-by-step explanation:
To address the student's question, let's consider each part individually:
(a) Showing Col(A) is a subspace of R⁵Column space, Col(A), consists of all linear combinations of the columns of A. To prove Col(A) is a subspace of R⁵, we must show it is closed under addition and scalar multiplication and includes the zero vector. Since A is a 5×3 matrix, any linear combination of its columns will result in a vector in R⁵, verifying the subspace properties.
(b) Demonstrating Nul(A) is a subspace of R³
The null space, Nul(A), is the set of all vectors X in R³ such that AX=0. A is of rank 3, meaning all of its columns are linearly independent and span R³. However, since A is a 5×3 matrix, there can be no non-zero vector in R³ that, when multiplied by A, gives the zero vector in R⁵. Thus, we can only have the trivial solution (the zero vector), and Nul(A) must be the zero subspace of R³, which is indeed a subspace.
(c) Relationship between Col(A) and Nul(A)
The Rank-Nullity Theorem tells us the dimension of Nul(A) plus the rank of A equals the number of columns in A. Here, the rank of A is 3, and since Nul(A) must be the zero subspace (because there are no non-trivial solutions to AX=0), its dimension is 0. Therefore, the dimension of Col(A) can be understood as the dimension of the image of A, which is also the rank of A, indicating a dimension of 3 in R⁵.