Final answer:
The column space of a 4 x 7 matrix A with three pivot columns is R³ because the number of pivot columns determines the dimensions of the column space. Using the Rank-Nullity Theorem, the dimension of the null space (Nul A) is calculated to be 4, since there are 4 non-pivot columns (7 total columns minus 3 pivot columns).
Step-by-step explanation:
The question asks whether the column space of a 4 x 7 matrix A, with three pivot columns, is equal to R³, and what the dimension of the null space (Nul A) is. To answer this, we consider the pivot columns, which represent the independently varying directions in the column space of A. Given that A has three pivot columns, the column space of A, denoted as Col A, is a subspace of R⁴ because A has 4 rows, but the dimension of Col A is equal to the number of pivot columns, which is 3. Hence, we can infer that Col A is indeed R³.
For the dimension of the null space, we apply the Rank-Nullity Theorem, which states that the dimension of the column space (rank) plus the dimension of the null space equals the number of columns of the matrix. Symbolically, Rank(A) + Dim(Nul A) = number of columns of A. Since the number of pivot columns (rank) is 3 and there are 7 columns total, the dimension of the null space is 7 - 3 = 4. Therefore, the dimension of Nul A is 4.