91Ó°ÊÓ

From the given information, we know that the matrix \(A\) has a rank of \(2\). This means the column space of \(A\), denoted by \(R(A)\), has a dimension of 2. A 2-dimensional subspace in \(\mathbb{R}^3\) will be a plane, since it is described by two linearly independent vectors in 3 dimensions. So, geometrically, \(R(A)\) represents a plane in \(\mathbb{R}^3\).

Now, we will find the null space of the transpose of \(A\), \(N\left(A^{T}\right)\). To understand the dimensions of the kernel, we can use the rank-nullity theorem. The rank-nullity theorem states that for any linear transformation \(T: V \rightarrow W\), the dimensions of the domain of definition (\(V\)) and the codomain (\(W\)) are related by the following equation: $$\text{dim}(V) = \text{rank}(T) + \text{nullity}(T)$$ Apply this theorem to the transpose of A, \(A^T\), which maps from \(\mathbb{R}^3\) to \(\mathbb{R}^2\). Also, notice that the row space of \(A^T\) is equivalent to the column space of A, which has rank = 2 already. So, \(\text{dim}(\mathbb{R}^3) = \text{rank}(A^T) + \text{nullity}(A^T)\) Then, \(3 = 2 + \text{nullity}(A^T)\) So, \(\text{nullity}(A^T) = 1\) Since the nullity of the transpose of \(A\) is 1, it means that the kernel of the transpose of \(A\), denoted by \(N\left(A^{T}\right)\), is a 1-dimensional subspace (line) in \(\mathbb{R}^3\).

As mentioned earlier, \(R(A)\) is a plane in \(\mathbb{R}^3\) and \(N\left(A^{T}\right)\) is a line in \(\mathbb{R}^3\). The kernel of the transpose of \(A\) represents the set of vectors that are orthogonal to the rows of \(A^T\). Since the rows of \(A^T\) are the columns of \(A\), the kernels of \(A^T\) represent the vectors that are orthogonal to the column space of \(A\). Thus, geometrically, the relationship between \(R(A)\) and \(N\left(A^{T}\right)\) is that \(N\left(A^{T}\right)\) represents a line in \(\mathbb{R}^3\) that is orthogonal (perpendicular) to the plane formed by \(R(A)\).